1. The chemical formula for sodium bromide is NaB.*
True
False

Answers

Answer 1

Answer:

NaBr

Explanation:

Bromine's symbol is Br, not B, for B is for Boron


Related Questions

Atoms of the same element are alike because they must have the same number of

Answers

Answer:

They must have the same number of protons.

Explanation:

Protons determine the identity of an element. However, the neutrons can vary, resulting in different masses.

Answer:

Atoms of the same element can have different numbers of neutrons, however. Atoms of the same element (i.e., atoms with the same number of protons) with different numbers of neutrons are called isotopes. Most naturally occurring elements exist as isotopes.

Explanation:

Amylose is a "polysaccharide" that plants use to store energy. It is made of repeating subunits of C6H10O5. If a particular amylose molecule has 2537 of these subunits, what is its molecular formula? What is its molar mass? What is the empirical formula of amylose?

Answers

Answer:

See exaplanation

Explanation:

For this question, we several sub-questions, lets start with the first one:

What is its molecular formula?

For this question we have to multiply the subunit by the number of subunits in the polysaccharide, so:

[tex](C_6H_1_0O_5)*2537=C_1_5_2_2_2H_2_5_3_7_0O_1_2_6_8_5[/tex]

What is its molar mass?

For this question, we have to find the molar mass of 1 subunit and then multiply by the number of subunits:

Atomic masses: C: 12 g/mol H: 1 g/mol O: 16 g/mol

Now we can multiply, the atomic masses by the number of atoms, so:

[tex](12*6)+(10*1)+(16*5)=~162~g/mol[/tex]

If we take into account the number of subunits:

[tex](162~g/mol)*2537=410994~g/mol[/tex]

What is the empirical formula of amylose?

In this case, we have to remember that the empirical formula is the smallest number of atoms. In other words, we have to simplify the formula. Therefore, the smallest formula is the subunit formula: [tex]C_6H_1_0O_5[/tex].

I hope it helps

Tenormin, a member of the group of drugs known as beta-blockers, is used to treat high blood pressure and improve survival after a heart attack. It works by slowing down the heart to reduce its workload. Which atom in Tenormin is the most basic?

Answers

Answer:

Oxygen

Explanation:

In the compound tenormin, there are two highly electronegative atoms capable of accepting electrons; oxygen and nitrogen. Oxygen is more electronegative than nitrogen.

However, the oxygen atom in tenormin is bonded to carbon in a carbonyl bond. Recall that the carbonyl bond is polar and the direction of the dipole is towards the oxygen atom. Looking at the structure of tenormin, it is clear that the electron density of the bond tends towards the oxygen atom of the carbonyl group. Electron density is withdrawn from the adjacent nitrogen atom of the amine group via mesomeric and inductive mechanism towards the more electronegative oxygen atom.

On the other side of the structure, there are two oxygen atoms. These oxygen atoms are more electronegative than nitrogen thus they are more basic.

Hence the oxygen atom is the most basic atom in the compound tenormin.

2 MnO4−(aq) + 10 Cl−(aq) + 16 H+(aq) → 5 Cl2(g) + 2 Mn2+(aq) + 8 H2O(l) Oxidizing agent: Reducing agent:

Answers

Answer:

MnO4 is the oxidizing agent.

Cl is the reducing agent.

Explanation:

To know which is oxidizing agent and which is reducing agent, let us calculate the change in oxidation number of each element in the equation below:

2MnO4−(aq) + 10Cl−(aq) + 16H+(aq) → 5Cl2(g) + 2Mn2+(aq) + 8H2O(l)

Please note:

1. The oxidation state of oxygen is always –2 except in peroxide where it is –1

2. The oxidation state of Hydrogen is always +1 except in hydrides where it is –1

3. The oxidation state of chlorine is always –1

For Mn:

At left hand side

MnO4 = –1

Mn + 4O = –1

O = –2

Mn + (4 x –2) = –1

Mn – 8 = –1

Collect like terms

Mn = –1 + 8

Mn = +7

At the right hand side, Mn is +2.

The oxidation state of Mn changes from +7 to +2.

For Cl:

At the left hand side

Cl = –1

At the right hand side

Cl = 0

The oxidation state of Cl changes from – 1 to 0.

The oxidation state of O and H are unchanged.

Since the oxidation state of Mn changes from +7 to +2 i.e reduced, MnO4 is the oxidizing agent.

Since the oxidation state of Cl changes from – 1 to 0 i.e increase, Cl is the reducing agent.

How many molecules are in 0.3672 moles of LiBr?

Answers

Answer:

2.21×10²³molecules

Explanation:

Hello,

The question requires us to calculate the number of molecules present in a given number of mole.

Numbe of mole = 0.3672 moles

According to mole ratio concept, one mole of any substance is equal to Avogadro's number of particles, atoms or molecules. I.e,

1 mole = Avogadro's number

Avogadro's number = 6.022×10²³ atoms/molecules/particles or ions

1 mole = 6.022×10²³molecules

0.3672 mole = x molecules

x = 0.3672 × 6.022×10²³

X = 2.21×10²³molecules

Therefore, 0.3672moles of LiBr contains 2.21×10²³molecules

A 1.8 g mass of fructose is added to 0.100 kg of water and it is
found that the freezing point has decreased by 0.186 °C. Given
that the Kf value of water is 1.86 °C kg/mol, what is the molar
mass of fructose (van't Hoff factor, i = 1)?
g/mol
Round your answer to the nearest whole number. Do not include units in your answer​

Answers

The molar mass of fructose will be "180 g/mol". To understand the calculation check below.

Molar mass and Molality

According to the question,

Fructose mass = 1.8 g

Water's mass = 0.100 kg

Molal freezing point depression constant, [tex]K_f[/tex] = 1.86°C/m

Freezing point change, Δ[tex]T_f[/tex] = 0.186°C

Freezing point constant, [tex]K_f[/tex] = 1.86°C/m

We know the relation,

→ Δ[tex]T_f[/tex] = i × [tex]K_f[/tex] × m

or,

         = i × [tex]K_f[/tex] × [tex]\frac{Fructose \ mass}{Fructose \ molar \ mass\times Solvent's \ mass}[/tex]

By substituting the values, we get

0.186 = 1 × 1.86 × [tex]\frac{1.8}{Fructose \ molar \ mass\times 0.100}[/tex]

By applying cross-multiplication,

Molar mass = 180 g/mol

Thus the above approach is right.          

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A 1.8 g mass of fructose is added to 0.100 kg of water and it is found that the freezing point has decreased by 0.186 °C. The molar mass of fructose is 180 g/mol.

Given:

ΔTf = 0.186 °C

Kf = 1.86 °C kg/mol

m = mass of fructose/mass of water

mass of fructose = 1.8 g

mass of water = 0.100 kg = 100 g

Use the equation:

ΔTf = Kf mi

Where:

ΔTf is the change in freezing point (in °C)

Kf is the cryoscopic constant of water (in °C kg/mol)

m is the molality of the solute (in mol/kg)

i is the van't Hoff factor

m = (mass of fructose) / (mass of water)

m = 1.8 g / 100 g

m = 0.018 mol/kg

Rearrange the equation to solve for the molar mass (M):

ΔTf = Kfmi

Substitute the values in the above equation:

0.186 = 1.86 × 0.018 × 1 / M

By applying cross multiplication,

Molar mass (M) = 180 g/mol

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Nitric oxide, NO, is made from the oxidation of NH3, and the reaction is represented by the equation: 4NH3 + 5O2 → 4NO + 6H2O What mass of O2 would be required to react completely with 6.87 g of NH3?

Answers

Answer:

16.16g of O2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

4NH3 + 5O2 → 4NO + 6H2O

Next, we shall determine the mass of NH3 and O2 that reacted from the balanced equation. This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 5 x 32 = 160g.

From the balanced equation above,

68g of NH3 reacted with 160g of O2.

Now, we can calculate the mass of O2 that will be required to react completely with 6.87 g of NH3. This is illustrated below:

From the balanced equation above,

68g of NH3 reacted with 160g of O2

Therefore, 6.87g of NH3 will react with = (6.87 x 160)/68 = 16.16g of O2.

Therefore, 16.16g of O2 is needed for the reaction.

Mass of oxygen required for the reaction is 16.16 g.

The equation of the reaction is;

4NH3 + 5O2 → 4NO + 6H2O

Since Number of moles = Mass/molar mass

Molar mass of ammonia = 17 g mol

Number of moles of NH3 reacted =  6.87 g/17 g mol = 0.404 moles

From the reaction equation;

4 moles of ammonia reacts with 5 moles of oxygen

0.404 moles reacts with 0.404 moles × 5 moles/4 moles  = 0.505 moles

Mass of oxygen = 0.505 moles × 32 g/mol = 16.16 g

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You decide to focus on pickling cucumbers. The cucumbers that you have are all approximately 12 cm in length and 3 cm in diameter. If the diffusion coefficient for H+ ions in cucumbers is the same as in water, 7 x 10-5 cm2/sec, how long will it take for the protons to reach the center of the cucumbers? Enter your answer in minutes without including the units.

Answers

Answer:

0.0036

Explanation:

SEE THE ATTACHMENT BELOW FOR DETAILED EXPLANATION

Consider four beakers labeled A, B, C, and D, each containing an aqueous solution and a solid piece of metal. Identity the beakers in which a chemical reaction will occur and those in which no reaction will occur. Drag each item to the correct bin.
1. Mn(s) Ca(NO3)2(aq)
2. KOH(aq) Fe(s)
3. Pt(NO3)2(aq) Cu(s)
4. Cr(s) H2SO4(aq)
A. Reaction
B. Non-reaction
Consider the following data for five hypothetical element: Q, W, X, Y, and Z. Rank the elements from most reactive to least reactive.
Combination Observation of reaction
Q + W+ Reaction occurs
X + Z+ No reaction
W + Z+ Reaction occurs
Q+ + Y Reaction occurs
Place the element symbol from most to least reactive.
1. Q
2. W
3. X
4. Y
5. Z
A. Most reactive
B. Least reactive
Use the following reactions to arrange the elements A, B, C, and D in order of their rodox reactivity from most reactive to least reactive. Rank from most reactive to least reactive.
1. A + D+ righ tarrow A+ + D
2. C+ + D righ tarrow C + D+
3. B+ + D righ tarrow B + D+
4. B + C+ righ tarrow B+ + C
A. Most reactive
B. Least reactive

Answers

Answer:

A Reaction

3. Pt(NO₃)₂(aq) + Cu(s)

4. Cr(s) + H₂SO₄(aq)

B Non Reaction

1. Mn(s) + Ca(NO₃)₂(aq)

2. KOH(aq) +  Fe(s)

Y > Q > W > Z > X

Explanation:

The first question is whether a reaction will occur base on the chemical equation below.

1. Mn(s) + Ca(NO₃)₂(aq)

2. KOH(aq) +  Fe(s)

3. Pt(NO₃)₂(aq) + Cu(s)

4. Cr(s) + H₂SO₄(aq)

Firstly, some element are more reactive than others , base on this criteria element can be arranged  base on it reactivity .

1. Mn(s) + Ca(NO₃)₂(aq)

This reaction will not occur because Mn cannot displace Ca in it compound. Usually, more reactive element displaces less reactive element.

2. KOH(aq) +  Fe(s)

The reaction will not occur since Iron is less reactive and lower in the reactivity series than potassium . So iron won't be able to displace potassium.

3. Pt(NO₃)₂(aq) + Cu(s)

Copper is more reactive than platinum so it will displace platinum easily . The reaction will definitely occur.

4. Cr(s) + H₂SO₄(aq)

Chromium is higher up in the reactivity series than hydrogen so, it will definitely displace hydrogen in it compound . The reaction will occur in this case.

Base on the reaction

Q + W+ Reaction occurs

Since the reaction occurred element Q is more reactive as it displace element w from it compound.  

X + Z+ No reaction

No reaction occurred because element x is less reactive than z therefore, it cannot displace z from it compound.

W + Z+ Reaction occurs

Element w is more reactive than z as it displaces z form it compound.

Q+ + Y Reaction occurs

Element Y is more reactive than element  Q as it displaces Q from it compound.

Therefore, the  order of reactivity from the most reactive to the least reactive will be Y > Q > W > Z > X

A. The beakers in which a chemical reaction will occur:

3. Pt(NO₃)₂(aq) + Cu(s)

4. Cr(s) + H₂SO₄(aq)

B. The beakers in which there is no reaction:

1. Mn(s) + Ca(NO₃)₂(aq)

2. KOH(aq) +  Fe(s)

C. The elements from most reactive to least reactive is:

Y > Q > W > Z > X

Solving for each part:

A.

1. Mn(s) + Ca(NO₃)₂(aq)

2. KOH(aq) +  Fe(s)

3. Pt(NO₃)₂(aq) + Cu(s)

4. Cr(s) + H₂SO₄(aq)

Elements can be arranged on the basis of reactivity:

1. Mn(s) + Ca(NO₃)₂(aq)

This reaction will not occur because Mn cannot displace Ca in it compound. Usually, more reactive element displaces less reactive element.

2. KOH(aq) +  Fe(s)

The reaction will not occur since Iron is less reactive and lower in the reactivity series than potassium . So iron won't be able to displace potassium.

3. Pt(NO₃)₂(aq) + Cu(s)

Copper is more reactive than platinum so it will displace platinum easily . The reaction will definitely occur.

4. Cr(s) + H₂SO₄(aq)

Chromium is higher up in the reactivity series than hydrogen so, it will definitely displace hydrogen in it compound . The reaction will occur in this case.

According to reactions:

Q + W→ Reaction occurs

Since the reaction occurred element Q is more reactive as it displace element w from it compound.  

X + Z →No reaction

No reaction occurred because element X is less reactive than Z therefore, it cannot displace z from it compound.

W + Z→ Reaction occurs

Element W is more reactive than Z as it displaces Z form it compound.

Q + Y →Reaction occurs

Element Y is more reactive than element Q as it displaces Q from it compound.

Thus, the order of reactivity from the most reactive to the least reactive will be: Y > Q > W > Z > X

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A solution consists of 0.27 M MgCl2 and 0.7 M CuCl2. Calculate the concentration of hydroxide ion needed to separate the metal ions. The Ksp of Mg(OH)2 is 6.3x10-10, and the Ksp of Cu(OH)2 is 2.2x10-20. Multiply your answer by 106 and enter that number to 2 decimal places.

Answers

Answer:

The concentration of hydroxide ion needed to precipitate Cu²⁺ whereas Mg²⁺ still in solution is 1.77x10⁻¹⁰

Explanation:

Based on the different Ksp of the hydroxides of the metal ions it is possible to precipitate 1 metal as hydroxide whereas the other still in solution.

Ksp of both hydroxides is:

Mg(OH)₂ ⇄ Mg²⁺ + 2OH⁻

Ksp = [Mg²⁺] [OH⁻]² = 6.3x10⁻¹⁰

Cu(OH)₂ ⇄ Cu²⁺ + 2OH⁻

Ksp = [Cu²⁺] [OH⁻]² = 2.2x10⁻²⁰

Replacing with the concentration of the metals:

[0.27] [OH⁻]² = 6.3x10⁻¹⁰

[OH⁻] = 4.83x10⁻⁵M

That means the precipitation of Mg²⁺ begins when [OH⁻] is 4.83x10⁻⁵M

[0.7] [OH⁻]² = 2.2x10⁻²⁰

[OH⁻] = 1.77x10⁻¹⁰

As the Cu²⁺ needs a low concentration of OH⁻ to begin precipitation,

the concentration of hydroxide ion needed to precipitate Cu²⁺ whereas Mg²⁺ still in solution is 1.77x10⁻¹⁰

A metal M react with sulphur ro form MS. If 3.6g of M reacts with 0.09mol of sulphur to form MS . what is the name of M.

Answers

Answer:

Calcium

Explanation:

We must first put down the reaction equation as this will serve as a guide in solving the question.

M(s) + S(s) -----> MS(s)

According to the reaction equation; 1 mole of metal reacts with 1 mole of sulphur. Hence 0.09 moles of sulphur reacts with 0.09 moles of metal.

Now recall that;

Number of moles (n) = mass(m)/ molar mass(M)

Since

mass of metal reacted= 3.6g

Number of moles of metal= 0.09 moles

Then;

Making molar mass of metal the subject of the formula;

Molar mass of metal = mass of metal / number of moles of metal

Molar mass of metal = 3.6g /0.09 moles

Molar mass of metal= 40 gmol-1

The metal having a molar mass of 40gmol-1 is calcium, therefore the metal is calcium.

1) How many kJ are absorbed when 45.2 g of water at 31.3 oC is heated to 76.9 oC? 2) Calculate the total heat released in kcal when 72.1 g water at 25.2 oC is cooled to 0 oC and freezes. 3) How many kilojoules are required to heat 55,500 mg of gold with specific heat = 0.129 J/g oC is heated from 24.6 oC to 123.4 oC? 4) Calculate the heat needed in kcal to change 45.6 g of water at 100 oC to change into steam.

Answers

Answer:

1. Q = 8.66 KJ

2. Q = 7.58 Kcal

3. Q = 0.71 KJ

4. Q = 24.31 Kcal

Explanation:

1. The quantity of heat absorbed can be determined by:

       Q = mcΔθ

where: Q is the quantity of heat absorbed or released, m is the mass of the substance, c is the specific heat capacity of water = 4.2 j/g[tex]^{0}C[/tex] and Δθ is the change in temperature.

           = 45.2 × 4.2  × (76.9 - 31.3)

           = 8656.704

∴       Q = 8.66 KJ

The quantity of heat absorbed is 8.66 KJ.

2. Q = mcΔθ + mL

Where L is the latent heat of fusion of ice = 334 J.

       = m(cΔθ + L)

       = 72.1(4.2 × 25.2 + 334)

   Q = 31712.464 J

       = 7579.466 calories

The total heat released is 7.58 Kcal.

3. Q = mcΔθ

      = 55.5 × 0.129 × (123.4 - 24.6)

     = 707.3586

The quantity of heat required to increase the temperature of gold is 0.71 KJ.

4. Q = mL

Where: L is the specific latent heat of vaporization = 533 calories.

     Q  = 45.6 × 533

      = 24304.8

The quantity of heat required to change water to steam is 24.31 Kcal.

Can the ph scale be utilized for all acids (Arrhenius, Bronsted-Lowery, and Lewis)? Explain reasoning.

Answers

Answer:

No, the pH scale can not be utilized for all types of acids known

Explanation:

The pH scale is a scale that shows the degree of acidity or alkalinity of a substance. The pH scale is graduated from 0-14. The pH scale is mostly applied to Arrhenius acids. Recall that in the Arrhenius definition, an acid produces hydrogen ion in solution while a base produces hydroxide ion in solution. Hence we could easily measure or calculate the pH and pOH of Arrhenius acids. Arrhenius acids apply mostly to acid-base discussion in Aqueous media.

Beyond aqueous media, we can still apply the Brownstead-Lowry and Lewis definitions for acid and base. Some of these substances to which we apply these other definitions of acid and base do not necessarily contain hydrogen or hydroxide ions hence the pH scale can not be utilized in discussing their acid-base properties, hence the answer.

A 38.8 gram piece of metal absorbs 181J as it temperature increases from 25.0 degree celsius to 36.0 degree celsius. What is the specific heat of the metal?


__________________________J/g degree C

Answers

Answer: 0.424 J/g°C

Explanation:

For this problem, we would have to manipulate the equaiton for heat, q=mCT. Specific heat is the C in the equation. Since we are looking for specific heat, we manipulate the equation so that it says C=.

[tex]C=\frac{q}{m(deltaT)}[/tex]

*I didn't know how to type in delta so I just wrote the word delta, but pretend you see a Δ.

Now that we have our equation, we can plug in our values and solve.

[tex]C=\frac{181J}{(38.8g)(36-25°C)}[/tex]

*Please ignore the capital A in the equation. It pops up every time I type in the ° sign.

[tex]C=0.424J/g°C[/tex]

The standard free energy change in physiological conditions (G') for the reaction catalyzed by malate dehydrogenase in the citric acid cycle malate + NAD+ oxaloacetate + NADH + H+ is ~+29 kJmol-1 Calculate the actual G' at 37C if Keq' is 1.02 × 10-5

Answers

Answer: The actual value of [tex]\Delta G[/tex] is -618 J/mol

Explanation:

Relation of ree energy change and equilibrium constant

[tex]\Delta G=\Delta G^0+2.303\times RT\times \log K_{eq}[/tex]

where,

[tex]\Delta G[/tex] = Free energy change

[tex]\Delta G^o[/tex] = standard free energy change = +29 kJ/mol =

R = universal gas constant

T = temperature = [tex]37^0C=(37+273)K=310K[/tex]

[tex]K_{eq}[/tex] = equilibrium constant = [tex]1.02\times 10^{-5}[/tex]

[tex]\Delta G=+29000J/mol+2.303\times 8.314J/Kmol\times 310K\times \log (1.02\times 10^{-5})[/tex]

[tex]\Delta G=29000J/mol-29619J/mol=-618J/mol[/tex]

The actual value of [tex]\Delta G[/tex] is -618 J/mol

If 200 mL of 3M CaCO3 is diluted to 250 mL, what is the new molarity?

Answers

Answer: The new molarity is 2.4 M

Explanation:

According to the dilution law,

[tex]C_1V_1=C_2V_2[/tex]

where,

[tex]C_1[/tex] = concentration of pure solution = 3 M

[tex]V_1[/tex] = volume of pure solution = 200 ml

[tex]C_2[/tex] = concentration of diluted solution= ?

[tex]V_2[/tex] = volume of diluted solution= 250 ml

Putting in the values:

[tex]3M\times 200ml=M_2\times 250ml[/tex]

[tex]M_2=2.4M[/tex]

Thus the new molarity is 2.4 M

A gas at 350 K and 12 atm has a molar volume 12 per cent larger than that calculated from the perfect gas law. Calculate (i) the compression factor under these conditions and (ii) the molar volume of gas. Which are dominating in the sample, the attractive or repulsive forces

Answers

Answer:

(i)The compression factor is 1.12

(ii) The molar volume of the gas is 2.68 L/mol

Since the compression factor is greater than 1, the attractive forces are dominating.

Explanation: Please see the attachments below

Kate’s blood volume is 3.8L. Before treatment,if her blood glucose was 170 mg/dL, how many grams of glucose were in her blood?

Answers

Answer:

6.460 gm  is the correct answer .

Explanation:

Given

Kate’s blood volume=[tex]3.8L[/tex]

blood glucose=[tex]170\ mg/dL[/tex]

As we know that

[tex]1\ dL= 10^{-1}\ litre[/tex]

The [tex]10^{-1}[/tex]   includes the blood glucose of the [tex]170\ mg/dL,[/tex]

So [tex]3.8L.[/tex] of the Kate’s blood volume  can be determined by

[tex]=\ \frac{170 * 3.8 }{10^{-1} }[/tex]

[tex]=170 * 38 \ mg[/tex]

[tex]=6460\ mg[/tex]

To convert into the gm we have divided by the 1000

[tex]6.460 \ gm[/tex]

Therefore the glucose in her blood=

[tex]6.460 \ gm[/tex]

Answer:

6.460 g

Explanation:

Glucose is a simple carbohydrate molecule which circulates in the blood along with other substances. The amount of glucose is measured to treat the medical conditions of patients which is measured as mg/dL.

1 liter of liquid contains 10 deciliters therefore 1 dL will have 0.1 liter or

1 dL= 10^-1 liter (0.1)

In the given question,  

the total volume of blood = 3.8 L

1 dL of blood contains = 170 mg

3.8 L will have = (3.8 x 170)/0.1

=6460 mg

= 6.460 g

Thus, 6.460 g is the correct answer.

What is the chemical formula for a compound between Al and F? AlF AlF 3 AlF 2 Al 3F

Answers

Answer:

AlF3

Explanation:

Aluminum fluoride (AlF3) is an inorganic compound in which aluminum and fluoride has ionic bond between each other.

Aluminum ( atomic number - 13) has 3 valence electrons and can lose 3 electrons to get a stable configuration while fluorine (atomic number - 9) can gain one electron to make a stable configuration. So in order to attain a stable configuration one aluminum (Al) atom form ionic bond with three fluorine atoms and form AlF3.

Hence, the chemical formula between  Al and F is AlF3.

Answer:

Explanation:

mmmmmmmmmmmmmh,mm

When I was a boy, Uncle Wilbur let me watch as he analyzed the iron content of runoff from his banana ranch. A 25.0-mL sample was acidified with nitric acid and treated with excess KSCN to form a red complex. (KSCN itself is colorless.) The solution was then diluted to 100.0 mL and put in a variable-path length cell. For comparison, a 10.0-mL reference sample of 6.74 times 10^-4 M Fe^3+ was treated with HNO_3 and KSCN and diluted to 50.0 mL, The reference was placed in a cell with a 1.00-cm light path. The runoff sample exhibited the same absorbance as the reference when the path length of the runoff cell was 2.41 cm. What was the concentration of iron in Uncle Wilbur's runoff?

Answers

Answer:

C = 2.24x10⁻⁴ M

Explanation:

The concentration of iron in Uncle Wilbur's runoff can be calculated using Beer-Lambert law:

[tex] A = \epsilon*C*l [/tex]   (1)

Where:

A: is the absorbance of the compound

ε: is the molar absorptivity of the compound

C: is the concentration of the compound

l: is the optical path length

Since the runoff sample exhibited the same absorbance as the reference sample, we can find the concentration using equation (1):      

[tex] \epsilon*C_{1}*l_{1} = \epsilon*C_{2}*l_{2} [/tex]    (2)

Where:

Subscripst 1 and 2 refer to Uncle Wilbur's runoff and to reference sample, respectively.

l₁ = 2.41 cm

l₂ = 1.00 cm

We can find C₂ as follows:

[tex] C_{2} = \frac{C_{2i}*V_{i}}{V_{f}} [/tex]    (3)

Where:

[tex]C_{2i}[/tex]: is the initial concentration of the reference sample = 6.74x10⁻⁴ M

[tex]V_{i}[/tex]: is the initial volume = 10.0 mL

[tex]V_{f}[/tex]: is the final volume = 50.0 mL

[tex] C_{2} = \frac{6.74 \cdot 10^{-4} M*10.0 mL}{50.0 mL} = 1.35 \cdot 10^{-4} M [/tex]

Now, we can find C₁ using equation (2):

[tex] C_{1} = \frac{C_{2}*l_{2}}{l_{1}} = \frac{1.35 \cdot 10^{-4} M*1.00 cm}{2.41 cm} = 5.60 \cdot 10^{-5} M [/tex]

Finally, since the runoff solution was diluted to 100.0 mL, the initial concentration can be calculated using equation (3) for [tex]C_{1i}[/tex]:

[tex]C_{1i} = \frac{C_{1}*V_{f}}{V_{i}} = \frac{5.60 \cdot 10^{-5} M*100.0 mL}{25.0 mL} = 2.24 \cdot 10^{-4} M[/tex]

Therefore, the concentration of iron in Uncle Wilbur's runoff is 2.24x10⁻⁴ M.

I hope it helps you!

Classify each amino acid according to whether its side chain is predominantly protonated or deprotonated at a pH of 7.40. The pKa values of the Asp, His, and Lys side chains are 3.65, 6.00, and 10.53, respectively.

Answers

Answer and Explanation:

7.40 becomes lower or shorter than the pKa of functional groups or chains of all the 3 amino acids respectively Lys, His, as well as Asp, although since pH provided throughout the statement. However, all of the covalent chains or bonds of the three compounds or acids (amino) will serve a purpose in the pro-toned pattern.

The pH during which everyone's binding sites will mostly keep track of customer for someone's comparison is:

Lys:

[tex]pH > 10.53[/tex]

His:

[tex]pH > 6.00[/tex]

Asp:

[tex]pH > 3.65[/tex]

Which describes the correct procedure when converting a number from scientific notation to standard notation? If the power of 10 is positive, move the decimal point to the left. If the power of 10 is positive, move the decimal point to the right. If the number being converted is greater than 10, move the decimal point to the left. If the number being converted is greater than 10, move the decimal point to the right.

Answers

Answer:

option 2

Explanation:

If the power of ten is positive, you move the decimal place to the right when converting  a number from scientific to standard notation

For example:

5.6894 * 10^9

10^9 is 1 with 9 zeros

5.6894 * 1000000000

5689400000.

it moved right

If the power of 10 is positive, move the decimal point to the left.

If the power of 10 is positive, move the decimal point to the right.

If the number being converted is greater than 10, move the decimal point to the left.

If the number being converted is greater than 10, move the decimal point to the right.

This is option 2

The correct answer is if the power of 10 is positive, move the decimal point to the right.

What is scientific notation?

A frequently-used floating-point system in which integers are expressed as products of a number between 1 and 10 multiplied by a power of 10.

Learn more about scientific notation here:-

https://brainly.com/question/2890945

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Fe3O4(s) + CO(g) → FeO(s) + CO2(g) What coefficients will balance the equation? 3, 1, 1, 1 1, 1, 3, 1 3, 3, 1, 1 2, 2, 6, 4

Answers

Answer:

1, 1, 3, 1.

Explanation:

When balancing chemical equations, the amount of moles of each element on both sides of the equation should be equal.

Looking at the original equation, we can see that there are 3 mols of iron on the reactants side and 1 on the products. We can simply add a coefficient of '3' to 'FeO' to balance the iron.

For Oxygen, we can see 5 on the reactants side. However, since we already added a coefficient of '3' to 'FeO', this already balanced out the oxygen for us. We have 5 mols in the reactants, and 5 in the products.

Carbon is already balanced on both sides.

Therefore, the final formula is [tex]Fe_{3}O_{4} + CO -> 3FeO + CO_{2}[/tex], with coefficients of 1, 1, 3, and 1.

Answer:

B

Explanation:

Fe3O4(s) + CO(g) →FeO(s) + CO2(g)

Left side

Fe =3

O = 5

C = 1

Right side

Fe =1

O = 3

C = 1

Balance by finding common denominator

We'll make Fe 3 on the left side

Fe3O4(s) + CO(g) →3FeO(s) + CO2(g)

Left side

Fe =3

O = 5

C = 1

Right side

Fe =3

O = 5

C = 1

Draw the reaction mechanism the reaction between 2-methylpent-2-ene and chlorine. State the type of reaction. Label the electrophile, intermediate and final product.​

Answers

Answer:

Explanation:

CH₃C(CH₃)=CH-CH₂-CH₃ ( 2 - methylpent-2-ene)

Cl₂ ⇒ Cl⁺ + Cl⁻

Cl⁺( electrophile)

CH₃C(CH₃)=CH-CH₂-CH₃  + Cl⁺ ( electrophile)  ⇒ CH₃C⁺(CH₃)-C(Cl)H-CH₂-CH₃ (  intermediate )

CH₃C⁺(CH₃)-C(Cl)H-CH₂-CH₃ + Cl⁻  ⇒ CH₃CCl(CH₃)-C(Cl)H-CH₂-CH₃ ( final product )

It is electrophilic addition reaction.

A 1.34 mole sample of LiCl dissolves in water, The volume of the final
solution is 0.86L. Find the Molarity of the Solution.​

Answers

Answer:

[tex]1.56 \,\,mol/L[/tex]

Explanation:

Molarity of the solution measures number of moles of a solute per litre of solution.

Molarity = volume of solution in litres/number of moles of solute dissolved in solution

Volume of solution in litres = 0.86 L

Also, 1.34 mole sample of LiCl dissolves in water

So,

Molarity of the Solution = [tex]\frac{1.34}{0.86}=1.56 \,\,mol/L[/tex]

A liquid solvent is added to a flask containing an insoluble solid. The total volume of the solid and liquid together is 93.0 mL. The liquid solvent has a mass of 33.7 g and a density of 0.865 g/mL. Determine the mass of the solid given its density is 3.50 g/mL

Answers

Answer:

Mass of solid = 189.141 gram

Explanation:

Given:

Total volume = 93 ml

Mass of liquid = 33.7 gram

Density of liquid = 0.865 g/ml

Density of solid = 3.50 g/ml

Find:

Mass of solid = ?

Computation:

Volume of liquid = Mass of liquid / Density of liquid

Volume of liquid = 33.7 / 0.865

Volume of liquid = 38.9595 ml

Volume of solid = Total volume - Volume of liquid

Volume of solid = 93 - 38.9595

Volume of solid = 54.0405 ml

Mass of solid = Volume of solid × Density of solid

Mass of solid = 54.0405 ml × 3.50 g/ml

Mass of solid = 189.141 gram

PLEAS HELP I HAVE LIMITED TIME!!
Which compound is the limiting reagent?
Select one:
O2
O H20
H2
Cannot be determined.

Answers

Answer:

H2 is the limiting reactant.

Explanation:

From the diagram above:

H2 => White ball

O2 => Red ball

Before the reaction

H2 => White ball => 10

O2 => Red ball => 7

After the reaction

H2O => White and red ball => 10

O2 => 2

From the simple illustration above, we can see that all the H2 were used up in the reaction but there are left over of O2.

This simply means that H2 is the limiting reactant as all of it is used up in the reaction while O2 is the excess reactant as there are leftover.

Convert the average temperatures for each collected data point given below from °C to K. Plot the average cell potentials E (y-axis) vs T (x-axis). The plot should be approximately linear. Add a trendline to find the best linear fit and write down the y-intercept and slope (b and m from the linear equation) for the trendline below.

Average Temperature in °C - Average Cell Potential (V)

15 - 0.465
18 - 0.467
21 - 0.468
24 - 4.69
27 - 0.471
30 - 0.472
33 - 0.474

Answers

Answer:

Explanation:

The equation of above line , y = 0.0005x+ 0.458

This can be compared with y = mx+c

Hence slope, m = 0.0005 and Y-intercept, c = 0.458

Or it can be plotted manually where straight line has to be drawn touching maximum number of data points. After drawing a straight linear line, we need to take any two points from the straight line and slope is calculated

Slope,

[tex]m = \frac{y_2-y_1}{x_2-x_1}[/tex]

and y -intercept is calculated using extraplotting backwards such that it touches the Y-axis. the point where straight line touches Y-axis is Y-intercept (c).

Plot the average cell potentials E (y-axis) vs T (x-axis). image attached

Classify each titration curve as representing a strong acid titrated with a strong base, a strong base titrated with a strong acid, a weak acid titrated with a strong base, a weak base titrated with a strong acid, or a polyprotic acid titrated with a strong base. A. Strong acid/ strong base.B. Strong base/ strong acid.C. Weak acid/ strong base.D. Weak base/ strong acid.E. Polyprotic acid/ strong base.

Answers

Answer:

Acid base titration curves shows the pH at equivalence point

Explanation:

Since the images were not shown, I will proceed to give a general description of the following acid-base titration curves:

In a strong acid-strong base titration, the acid and base will react to form a neutral solution. At the equivalence point of the reaction, hydronium (H+) and hydroxide (OH-) ions will react to form water, leading to a pH of 7.

The titration curve reflects the strengths of the corresponding acid and base. If one reagent is a weak acid or base and the other is a strong acid or base, the titration curve is irregular, and the pH shifts less with small additions of titrant near the equivalence point.

Polyprotic acids are able to donate more than one proton per acid molecule, in contrast to monoprotic acids that only donate one proton per molecule. In the titration curve of a polyptotic acid and a strong base, The curve starts at a higher pH than a titration curve of a strong base. There is always a steep climb in pH before the first midpoint. Gradually, the pH increases until it passes the midpoint; Right before the equivalence point there is a very sharp increase in pH.

Answer:

awzsexdrcftvgybhunimoknljbkhvjgchfx

Explanation:

Suppose the reaction between nitric oxide and bromine proceeds by the following mechanism: step elementary reaction rate constant (g) (g) (g) (g) (g) (g) Suppose also ≫. That is, the first step is much faster than the second.

Write the balanced chemical equation for the overall chemical reaction:
Write the experimentally-observable rate law for the overall chemical reaction.

Note: your answer should not contain the concentrations of any intermediates.

Answers

Answer:

Koverall [NO]^2 [Br2]

Balanced chemical reaction equation;

2NO + Br2 ⇄2NOBr

Explanation:

Consider the first step in the reaction;

NO(g) + Br2(g) ⇄ NOBr2(g) fast

The second step is the slower rate determining step

NOBr2(g) + NO(g) ⇄ 2NOBr(g)

Given that k1= [NOBr2]/[NO] [Br2]

k2= [NOBr2] [NO]

The concentration of the intermediate is now;

[NOBr2]= k1[NO][Br2]

It then follows that overall rate of reaction is

Rate= k1k2[NO]^2 [Br2]

Since k1k2=Koverall

Rate= Koverall [NO]^2 [Br2]

Answer:

NOBr_2(g)+NO(g)----> 2NOBr(g)

Explanation:

Answer retrieved from ALEKS

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