When temperature drops, particle speed decreases.
True?
False?

Answer:

Colloid = Condensed milk, Paint

Solution = brine

Suspension = flour and water

Explanation:

Colloid:

Colloid is a mixture in which smaller particles of one substance are evenly distributed throughout into another substance.

These particles are smaller than the particles present in suspension but larger than the particles present in solution.

Condensed milk is example of colloid.

Paint is also example of colloid in which pigment particles dispersed.

Solution:

In solution one or more substances are dissolved into the another substance. The substance which is being dissolved is called solute while the substance which has tendency to dissolve the other substance is called solvent. It is homogenous mixture.

Salty water is example of solution in which salt are solute dissolved in solvent which is water.

Suspension:

In suspension solute particles are not dissolved they are dispersed into bulk of solvent. The solute particles are larger then the particles present in colloid.

flour and water is the example of suspension.

Answer:

2 Na⁺(aq) + S²⁻(aq) + Fe²⁺(aq) + 2 NO₃⁻(aq) ⇒ FeS(s) + 2 Na⁺(aq) + 2 NO₃⁻(aq)

Explanation:

Let's consider the molecular equation that occurs when aqueous solutions of sodium sulfide and iron(II) nitrate are mixed.

Na₂S(aq) + Fe(NO₃)₂ ⇒ FeS(s) + 2 NaNO₃(aq)

The complete ionic equation includes all the ions and the species that do not dissociate in water.

2 Na⁺(aq) + S²⁻(aq) + Fe²⁺(aq) + 2 NO₃⁻(aq) ⇒ FeS(s) + 2 Na⁺(aq) + 2 NO₃⁻(aq)

The complete ionic equation for the reaction between aqueous solution of sodium sulfide and iron(II) nitrate is

2Na⁺(aq) + S²⁻(aq) + Fe²⁺(aq) + 2NO₃⁻(aq) —> FeS(s) + 2Na⁺(aq) + 2NO₃⁻(aq)

The complete ionic equation for the reaction between aqueous solution of sodium sulfide and iron(II) nitrate can be written as follow:

In solution, Na₂S and Fe(NO₃)₂ will dissociate as follow:

Na₂S(aq) —> 2Na⁺(aq) + S²¯(aq)

Fe(NO₃)₂(aq) —> Fe²⁺(aq) + 2NO₃¯(aq)

The reaction will proceeds as illustrated below to give the complete ionic equation:

Na₂S(aq) + Fe(NO₃)₂(aq)—>

2Na⁺(aq) + S²¯(aq) + Fe²⁺(aq) + 2NO₃¯(aq) —> FeS(s) + 2Na⁺(aq) + 2NO₃¯(aq)

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Answer:

The balanced chemical equation for the reaction can be computed as:

[tex]\mathbf{Cr_2O_3_{(s)}+ 3H_2S_{(g)} \to Cr_2S_3_{(s)} + 3H_2O_{(l)}}[/tex]

2.103 moles of [tex]Cr_2O_3_{(s)}[/tex]  is being required.

Explanation:

The balanced chemical equation for the reaction can be computed as:

[tex]\mathbf{Cr_2O_3_{(s)}+ 3H_2S_{(g)} \to Cr_2S_3_{(s)} + 3H_2O_{(l)}}[/tex]

Recall that numbers of moles = [tex]\dfrac{mass}{molar \ mass}[/tex]

∴

number of moles of Cr2S3 = [tex]\dfrac{421}{200.19}[/tex]

number of moles of Cr2S3 = 2.103 moles

It  obvious from the balanced equation that:

1 mole of [tex]Cr_2S_3_{(s)}[/tex] requires 1 mole of  [tex]Cr_2O_3_{(s)}[/tex]  

Thus; 2.103 moles of  [tex]Cr_2S_3_{(s)}[/tex]  = 2.103 × (1/1) moles of [tex]Cr_2O_3_{(s)}[/tex]  

= 2.103 moles of [tex]Cr_2O_3_{(s)}[/tex]  

Answer:

Outcome Variable

Explanation:

Control group and Hypothesis are not what your measuring.

Test variable would be like- "amount of water added"

Outcome variable would be something like- "amount plants grew"

You would be measuring the amounts the plants grew, based on the water added.

Answer:

Its d

Explanation:

Got it correct

Answer:

d

Explanation:

did the test

Answer: A) Cold ice makes the liquid condense

On the flip side, warm temperatures make the liquid expand. Since there is only one direction to go, the liquid expands upward until it reaches a certain marker on the thermometer, to indicate the temperature.

Answer:

6.62 g of carbon.

Explanation:

We'll begin by calculating the molar mass of C6H12O6. This is illustrated below:

Molar mass of C6H12O6 = (12×6) + (12×1) + (16×6)

= 72 + 12 + 96

= 180 g/mol

Next, we shall determine the mass of carbon (C) in 1 mole (i.e 180 g) of C6H12O6. This is illustrated:

Mass of C in C6H12O6 = 6C = 12×6

Mass of C in C6H12O6 = 72 g

Thus, 180 g of glucose contains 72 g of carbon.

Finally, we shall determine the mass of carbon in 16.55 g of glucose. This can be obtained as follow:

180 g of glucose contains 72 g of carbon.

Therefore, 16.55 g of glucose will contain = (16.55 × 72)/180 = 6.62 g of carbon.

Thus, 16.55 g of glucose contains 6.62 g of carbon.

Answer:

easy

Explanation:

Divide the mass by the volume to calculate its density. If its density isn't the same as an equal amount of pure silver, the coin has some other metal in it.

Answer:

everything.

Explanation:

You need nice weather,good realation ships as well as convection

Answer:

A C D F

Explanation:

Answer:

0.167 mol Fe

Explanation:

4 Fe  +  3 O₂  ⇒  2 Fe₂O₃

Convert moles of oxygen to moles of iron using the mole ratio.  You can find the mole ratio using the chemical equation.  For every 4 moles of iron, you will need 3 moles of oxygen.

(0.125 mol O₂)  (4 mol Fe/3 mol O₂) = 0.167 mol Fe

Answer:

Your mom says to stop cheating

Explanation:

Answer:

c

Explanation:

Answer:

THE ANSWER IS CORRECT!!

Answer:

[tex]r_{N_2O_5}=-4.00mol/min\[/tex]

Explanation:

Hello.

In this case, since the chemical reaction is:

[tex]2N_2O_5\rightarrow 4NO_2+O_2[/tex]

We can write the rate ratio between N2O5 and NO2 as shown below:

[tex]\frac{1}{-2} r_{N_2O_5}= \frac{1}{4} r_{NO_2}[/tex]

Since N2O5 is consumed and NO2 produced at a rate of 8.00 mol/s, therefore, the rate of disappearance is N2O5 is:

[tex]r_{N_2O_5}= \frac{-2}{4} r_{NO_2}\\\\ r_{N_2O_5}=\frac{-1}{2} *8.00 mol/min\\\\ r_{N_2O_5}=-4.00mol/min[/tex]

Which is negative since N2O5 is a reactant.

Best regards!

Answer:

38.4%

Explanation:

Step 1 Obtain the concentration of Ca^2+ and SO4^2-

[Ca^2+] = 0.05 L (0.0125 mol/L) (1/0.1 L) = 6.25 * 10^-3 M

[SO4^2-] = 0.05 L (0.0152mol/L) (1/0.1L) = 7.60 * 10^-3 M

Step 2 Obtain the concentration of [SO4^2-]  left

         Ca^2+(aq)             +       SO4^2-(aq)------> CaSO4(s)

initial  6.25 * 10^-3 M             7.60 * 10^-3 M

final        0                               1.35 * 10^-3 M

Step 3 Determine the amount of Ca^2+ left from the common ion problem

                      CaSO4(s) --------> Ca^2+(aq)             +       SO4^2-(aq)

initial                                             0                                 1.35 * 10^-3 M

equilibrium                                    x                                (1.35 * 10^-3  + x) M

But Ksp of CaSO4 = 9.1 * 10^-6

Ksp = [Ca^2+] [SO4^2-]

9.1 * 10^-6= [x] [1.35 * 10^-3  + x]

9.1 * 10^-6= 1.35 * 10^-3 x + x^2

Rearranging:

x^2 + 1.35 * 10^-3 x + 9.1 * 10^-6 = 0

solving the quadratic yields x = 2.4 * 10^-3

Step 4 Calculate the amount of Ca^2+ remaining;

% Ca^2+ remaining = 2.4 * 10^-3/ 6.25 * 10^-3 M * 100

% Ca^2+ remaining = 38.4%

Answer:

0.41 M

Explanation:

A -> B

rate constant (k) = 0.039L/mol s

t = 23

Final concentration, [A]  = 0.30M

Initial concentration, [A]o = x

1 / [A]  = kt + 1 / [A]o

1 / [A]o = 1 / [A] - kt

1 / [A]o  = 1 / 0.30   - 0.039 (23)

1 / [A]  = 3.33 -  0.897 = 2.433

[A] = 0.41 M

Answer:

Molecular Formula

P4S6

Molecular Weight

316.3 g/mol

P[tex]_4[/tex]S[tex]_6[/tex] is the chemical formula of tetraphosphorus hexasulfide. These can only include one typographic line containing symbols.

A chemical formula is a way to describe the chemical ratios of the atoms that make up a specific chemical compound or molecule in chemistry. Chemical element numbers, symbols, and on occasion other symbols, including parentheses, dashes, brackets, commas, and plus (+) and minus () signs, are used to represent the chemical elements. Which may also include subscripts or superscripts. Since a chemical formula inherently wordless, it cannot be considered a chemical name. A chemical formula does not equal a complete chemical structural formula, despite the fact that it may suggest certain basic chemical structures. P[tex]_4[/tex]S[tex]_6[/tex] is the chemical formula of tetraphosphorus hexasulfide.

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Answer:

Binding properties

Explanation:

Answer:

87.0

Explanation:

Hopes this helps

Answer:

[tex]P_2=2.04atm[/tex]

Explanation:

Hello.

In this case, by using the combined ideal gas law, we can identify the change in pressure, temperature an volume as shown below:

[tex]\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}[/tex]

In such a way, since we are asked to compute the final pressure, P2, we proceed as follows:

[tex]P_2=\frac{P_1V_1T_2}{T_1V_2} \\\\P_2=\frac{1.12atm*2.29L*302K}{285K*1.33L}\\ \\P_2=2.04atm[/tex]

Best regards!

Answer:

What ae the options?  

Explanation:

Answer:

[tex]\Delta G=8015\frac{J}{mol}[/tex]

Explanation:

Hello.

In this case, for the following chemical reaction in gaseous phase:

[tex]C_2H_4+H_2O\rightleftharpoons C_2H_5OH[/tex]

We can write the equilibrium expression as:

[tex]Kp=\frac{p_{C_2H_5OH}}{p_{H_2O}*p_{C_2H_4}}[/tex]

it means that given the pressures, we can compute Kp as shown below:

[tex]Kp=\frac{0.100atm}{0.275atm*0.350atm}=1.04[/tex]

Next, we compute Kc at 298 K:

[tex]Kc=\frac{Kp}{(RT)^{1-1-1}}=\frac{Kp}{(RT)^{-1}}[/tex]

Whereas 1-1-1 comes from the stoichiometric coefficients of products minus reactants. Therefore, Kc is:

[tex]Kc=\frac{1.04}{(0.08206*298)^{-1}}=25.4[/tex]

Thus, the Gibbs free energy for this reaction is:

[tex]\Delta G=RTln(kC)=8.314\frac{J}{mol*K}*298K*ln(25.4)\\ \\\Delta G=8015\frac{J}{mol}[/tex]

Best regards!

Answer: The answer is C. 0.1MCaCl2

Explanation:

We have AgCl whose least solubility is to be determined at 25degree Celsius

so upon looking at the options given we observe C 0.1MCaCl2 and D 0.1MHCl both have chlorine ions present and we know that here common ion effect will happen that causes decreases the solubility of the AgCl here

but when Calcium chloride dissociates it gives 2 moles of chlorine ions

[tex]CaCl2[/tex]→[tex]Ca^+2 + 2Cl^-[/tex]

so chlorine ion will be present in more amount [tex]2*0.1\\[/tex]=[tex]0.2[/tex] as compared to the chlorine produced by HCl

[tex]HCl[/tex]→[tex]H^+ +Cl^-[/tex] only 1 mole of chlorine is produced.

Hence AgCl will be least soluble in 0.1M CaCl2

Explanation:

A. 3 groups bound to it with no lone pairs

Answer: 1.79 x 10^24 molecules H2O

Explanation:

Use the following conversion factors to convert from molecules of oxygen to molecules of water: 1 mole O2/ 6.02 x 10^23 molecules O2, 2 mole H2O/ 1 mole O2, 6.02 x 10^23 molecules H2O/ 1 moles H2O

The question is incomplete; the complete question is;

Liquid octane CH3CH26CH3 reacts with gaseous oxygen gas O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. If 2.80g of carbon dioxide is produced from the reaction of 3.4g of octane and 5.9g of oxygen gas, calculate the percent yield of carbon dioxide. Be sure your answer has the correct number of significant digits in it.

Answer:

55%

Explanation:

The balanced equation is:

C8H18(g) + 25/2O2(g) ----------> 8CO2(g) + 9H2O(g)

The limiting reactant IS first determined. This is the reactant that yields the least amount of product.

3.4 g octane / 114 g/mol = 0.030 mol octane

From the reaction equation;

1 mole of octane yields 8 moles of CO2

0.030 mol octane yields 0.030 * 8/1 = 0.24 moles of CO2

5.9 g O2 / 32 g/mol = 0.18 mol O2

25/2 moles of O2 yields 8 moles of CO2

0.18 moles of O2 yields 8 * 0.18/25/2 = 0.1152 moles of CO2

Oxygen is the limiting reactant.

Number of moles of carbon dioxide produced (actual yield) = 2.8 g/44g/mol = 0.0636 moles of CO2

% yield = actual yield/ theoretical yield * 100

% yield =   0.0636/0.1152 * 100= 55%