Help me please!! Is for my finals!!

Calculate electrons

Total electrons=10+2=12

The element is Magnesium with symbol Mg as atomic no is 12

Additional:-

M:-

Excited state of magnesium

A thermodynamically-controlled process describes a reaction that the most stable or more stable of two products is formed.

A  thermodynamically-controlled process is a reaction in which the ratio of the product formed is determined by the relative stability of the products.

Thermodynamic control ensures that the most stable or more stable of two products is formed.

Therefore, a thermodynamically-controlled process describes a reaction that the most stable or more stable of two products is formed.

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Heating up a reaction increases the speed of a reaction until the enzyme denatures.

Enzyme denaturation occurs when a biological protein catalyst does not work anymore due to a high temperature that alters its tridimensional conformation.

This cellular process (denaturation) is well known to be one of the main causes of enzymatic failure.

In conclusion, heating up a reaction increases the speed of a reaction until the enzyme denatures.

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The partial pressure of xenon is 571.5 KPa if the total pressure is 1436.4kPa.

The pressure exerted by a (specified) component in a mixture of gases.

According to Dalton's law of partial pressure, the net pressure of a gaseous mixture is always equal to the sum of the partial pressures of individual gases.

Now we are given that the total pressure is 1436.4kPa

The partial pressure of helium is 348.1kPa

The partial pressure of oxygen is 516.8kPa

Partial Pressure of xenon?

Using dalton's law;

1436.4kPa = 348.1kPa + 516.8kPa + partial pressure of xenon

The partial pressure of xenon is 571.5 KPa.

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Answer:

a.  V=44.9mL

Explanation:

For a gas undergoing all of these changes, it will be important to combine Boyle's Law and Charles' Law to form the following equation (if it isn't already known):

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

Where the Temperatures must be measured in Kelvin.

Recall that to convert Celsius to Kelvin, one must add 273 or use the equation [tex]T_C+273=T_K[/tex].

Thus, [tex]T_1=(18.1+273)[K]=291.1[K][/tex]  and [tex]T_2=(12.5+273)[K]=285.5[K][/tex]

To solve for the requested quantity, note that all of the other units match between beginning and end, so we substitute and solve:

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

[tex]\dfrac{(2.61[atm])V_1}{(291.1[K])}=\dfrac{(1.92[atm])(59.9[mL])}{(285.5[K])}[/tex]

[tex]\dfrac{(2.61[atm] \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----})\bold{V_1}}{291.1[K] \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----}}*\dfrac{291.1[K] \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----}}{2.61[atm]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----}}=\dfrac{(1.92[atm]\!\!\!\!\!\!\!\!\!\!\!{--})(59.9[mL])}{285.5[K\!\!\!\!\!{-}]}*\dfrac{291.1[K\!\!\!\!\!{-}]}{2.61[atm]\!\!\!\!\!\!\!\!\!\!\!{--}}[/tex]

[tex]V_1=44.928677576[mL][/tex]

Accounting for significant digits, [tex]V_1=44.9[mL][/tex]

The missing components of the neutralisation reaction include the following:

Neutralisation reaction is defined as the type of reaction that leads to the formation of salt and water when an acid and a base reacts.

From the reactions given the missing components are replaced as follows:

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Answer:

[tex]\fbox {C}[/tex]

Explanation:

By repeating the experiment with new amounts of each reactant, this conflict can be resolved as we can see the effect of a reactant on the final amount of product formed.

The Molar Concentration Of A =0.099 .

The Relative Molar Mass Of A = 35.0129 gm

Given,

Mass of HX = 3.5 g

Moles of solution B ([tex]Na_{2} CO_{3}[/tex]) = 0.05 moles

Volume of HX = 26.10 mL

Volume of Solution B = 25 mL

Molecular weight of solution B = 2(atomic weight of Na )+ atomic weight of C + 3(atomic weight of O)

                 = 2(23) + 12 + 3(16)

                 =106 gm

Equivalent weight of  [tex]Na_{2} CO_{3}[/tex] = molecular weight / 2 = 106 /2  =53 g

Mole = mass / molecular weight

∴0.05 = mass / 106

∴ mass = 5.3 gm [tex]Na_{2} CO_{3}[/tex]

Normality = mass ÷ (equivalent weight × volume of the solution in liter)

                = 5.3 ÷( 53 × 0.025)

                =4 N

So, by using formula ,

[tex]N_{1} V_{1} =N_{2} V_{2}[/tex]

[tex]N_{1}[/tex] = normality of solution B = 4 N

[tex]V_{1}[/tex] = volume of solution B = 25 mL

[tex]N_{2}[/tex] = normality of solution A = ? N

[tex]V_{2}[/tex] =Volume of solution A = 26.1 mL

∴ 4×25 = [tex]N_{2}[/tex] × 26.1

∴[tex]N_{2}[/tex] = 3.83 N

∴ normality of solution A = 3.83 N

from Formula of the normality we can find the equivalent weight of the A

Normality = mass of HX ÷ (equivalent weight × volume of the solution in liter)

3.83 = 3.5 ÷( equivalent weight × 0.0261)

∴equivalent weight = 35.0129 g

In case of HX the electron transfer is 1 ,so equivalent weight = molecular weight ; which is also termed as relative molar mass in given case.

∴The Relative Molar Mass Of A = 35.0129 g

Molar concentration = mass / molar mass

                               = 3.5 / 35.0129

                               = 0.099 mole

∴ The Molar Concentration of A  is 0.099 .

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9.06 L  is the volume of 0.24 mol oxygen (02) gas at 345 k and a pressure of 0.75.

The ideal gas law (PV = nRT) relates to the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

We know that the ideal gas equation relates the pressure, volume, temperature, and moles of gases as:

PV =nRT

This is a combined gas law based on other gas law like Boyle's law, Charle's law, Avagadro's law, Gay Lussac's law

P =  pressure = 0.75 atm

V = volume = ?

n = moles = 0.24 mol

R = 0.0821 L atm / molK  [gas constant]

T = temperature = 345 K

Putting the values

Volume =[tex]\frac{nRT}{ P}[/tex]

Volume = 0.24 X 0.0821 X 345 / 0.75

= 9.06 L

Hence, 9.06 L is the volume of 0.24 mol oxygen (02) gas at 345 k and a pressure of 0.75.

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Answer: it would be a 1 to 1 ratio

Explanation: originally it would be 2 to 2 but you have to reduce

[tex](40.0)(7.426)=\boxed{297.0 \text{ g (to 4 sf)}}[/tex]

Answer:

A.

Explanation:

Well, Nicola Tesla was from Croatia.

Albert Einstein was from Germany.

Thomas Edison was from the U.S.

So I think all over.

3 moles of NaOH reacts with 1 mole of phosphoric acid.

The moles ratio of the reaction between NaOH and Phosphoric acid is given by the equation of the reaction below:

[tex]3\:NaOH + H_{3}PO_{4} \rightarrow Na_{3}PO_{4} + 3\:H_{2}O \\ [/tex]

Based on the equation of the reaction, 3 moles of NaOH reacts with 1 mole of phosphoric acid.

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Answer:

Explanation:

Answer: The pH of the solution is 5.65

Explanation:

The relationship between the pH and the pOH is that [tex]pH+pOH=14[/tex].

Given this, we can plug in the pOH and subtract that from 14.

[tex]14-pOH=pH\\14-8.35=5.65[/tex]

I hope this helped! Pls give brainliest!! :)

Answer:

Give the man above me brainliest

Answer:

31536000

Explanation:

There are 31536000 seconds in a day.

60s x 60min x 24hrs x 365 days = 31536000sec

Answer:

C

Explanation:

with combustion you must know that your products must be carbon dioxide and water. I think you copied your question incorrectly.

Answer:

5.0 mol

Explanation:

From the coefficients of the equation, we know that for every 2 moles of Mn consumed, 2 moles of MnO are produced.

Thus, the number of moles of Mn consumed is equal to the number of moles of MnO produced, and thus the answer is 5.0 mol.

The moles of MnO formed with the reaction of 5 moles of Mn is 3 moles.

In a chemical reaction, the reagent that is present in a lesser quantity and governs the rate of the reaction is termed as the limiting reagent.

In the reaction of the formation of MnO, according to the stoichiometric law 2 moles of Mn reacts with 1 moles of Oxygen.

Thus, the moles of oxygen consumed by 7 moles of Mn is:

2 moles Mn = 1 mole O₂

7 moles Mn = 3.5 moles O₂

The available moles of O₂ = 5 mol.

The remaining moles of O₂ = 5-3.5 moles

The remaining moles of O₂ = 1.5 moles

The reaction of 5 moles Mn requires 2.5 moles of O₂, whereas the available moles of oxygen is 1.5 moles. Thus, oxygen serves as the limiting reagent.

The moles of MnO formed with the reaction of 5 moles Mn and 1.5 moles O₂

1 mole O₂ = 2 moles MnO

1.5 moles O₂ = 2 * 1.5 moles MnO

1.5 moles O₂ = 3 moles MnO

Thus, the moles of MnO formed with the reaction of 5 moles of Mn with the available oxygen is 3 moles.

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Calcium nitrate is made up of 6 oxygen atoms, 2 nitrogen atoms and one calcium atom.

Ca(NO₃)₂

Thus, calcium nitrate is made up of 6 oxygen atoms, 2 nitrogen atoms and one calcium atom.

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Answer:

B)There are six atoms of oxygen.

C)There are two nitrogen atoms.

Explanation:

edge 2023

Answer:

=> 119.4597 g

Explanation:

From our question, we have been provided with the molarity of the resultant solution.

We know that, molarity of a solution is contained in 1 L or 1000 cm³ of solution.

Hence we have 1.50 moles in 1000 cm³.

Also, we know that, 1 cm³ equivalent to 1 g for water.

Thus, from our question we have 995 cm³ of water.

Lets now solve our problem;

1.5 moles are contained in 1000 cm³

x moles are contained in 995 cm³

Cross multiply

[tex]x \: moles \: = \frac{(995 {cm}^{3} \times 1.5 \: moles )}{(1000 {cm}^{3} )} \\ = \frac{1492.5 \: moles}{1000} \\ = 1.4925 \: moles[/tex]

1.4925 moles of NH4NO3 are required to be dissolved in water.

To find the mass of solute (NH4NO3) we use the formula;

Mass = RFM (Relative Formula Mass) x Number of moles

[tex] = 80.04 \: \frac{g}{mol} \times1.4925 \: mol \\ = 119.4597 \: g [/tex]

Therefore the mass of NH4NO3 needed to be dissolved in 995 g of water is 119.4597 g

Answer:

119.4597

Explanation:

Acellus

Al₂(SO₄)₃

= 2.Al+3.S+12.O

= 2.27 + 3.32+12.16

= 54+96+192

=342 g/mol

The temperature is always the same.

Assume we have something in solid phase. As we increase the temperature, the particles on the solid increase their kinetic energy, thus, the particles move more.

This causes that the volume of the object increases (for example when we heat up a metal and it dilates) and this keeps happening until we reach a critical point, when we are near a change of phase.

At this point the energy given is not used to increase the temperature of the object, but is used to "break" bonds in such a way that the particles are more free than before.

When all these bonds are "broken" the change of phase is completed, and in the case of the solid, we go from solid phase to liquid phase.

So, the temperature is always the same at the beginning of a change of state compare with the temperature at the end of the change

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Answer:

(b)  19.9 moles

Explanation:

To convert from moles Al to moles O₂, you need to multiply the given value by the mole-to-mole ratio of each component. This ratio is constructed via the coefficients in the balanced reaction. The desired unit (moles O₂) should be placed in the numerator to allow for the cancellation of units.

2 Al₂O₃ ----> 4 Al + 3 O₂
                    ^         ^            

26.5 moles Al           3 moles O₂
---------------------  x  ----------------------  =  19.9 moles O₂
                                 4 moles Al

Answer:

T₂ = 26.7 °C

Explanation:

Because you are dealing with pressure and temperature, you can use Gay-Lussac's Law to find the final answer. The equation looks like this:

P₁ / T₁ = P₂ / T₂

In this form, "P₁" and "T₁" represent the initial pressure and temperature. "P₂" and "T₂" represent the final pressure and temperature. You have been given the values for all of the variables but "T₂". Therefore, by plugging these values into the equation, you can simplify to find "T₂".

P₁ = 2.11 atm                   T₁ = 25 °C

P₂ = 2.25 atm                 T₂ = ?

P₁ / T₁ = P₂ / T₂                                            <--- Gay-Lussac's Law equation
(2.11 atm) / (25 °C) = (2.25 atm) / T₂           <--- Plug variables into equation

0.0844 = (2.25 atm) / T₂                            <--- Divide 2.11 by 25

(0.0844) x T₂ = 2.25 atm                           <--- Multiply both sides by T₂

T₂ = 26.7                                                    <--- Divide both sides by 0.0844

Answer: 3

Explanation:

Density is mass/volume divide the mass 9.6 by the volume of 3.2

[tex]\frac{9.6}{3.2} =3[/tex]

The density of the object displaced in water is equal to 3 g/ml.

The density of a substance can be described as the substance's mass per unit of its volume. The symbol is generally used to represent density ρ and the letter D can be used.

The mathematical formula of the density of a material can be represented as shown below:

Density = Mass/volume

ρ = m/V

The density of a material varies with pressure as well as temperature. There is a small variation in the density of the solids and liquids but much larger for gases. Increasing the pressure of the substance decreases the volume and thus increases its density.

As the object displaced water of volume of 3.2 ml.  

Therefore, the volume of the object, V = 3.2 ml

The mass of the object, m = 9.6 g

The density of the object placed in water can be calculated as:

Density of object = 9.6/3.2 = 3 g/ml

Therefore, the density of the object is 3 g/ml.

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Specialized periodicals in which scientists publish the results of their works are called scientific journals.

In educational publishing, a scientific journal is a periodical book intended to similarly the progress of technology, typically by way of reporting new studies.

Journal articles may include original research, re-analyses of studies, opinions of literature in a selected place, proposals of new but untested theories, or opinion pieces.

These scientific journals include the following.

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The 18o-labeled methanol (CH3O*H) will appear in the products side at position b.

The 18O label will appear at position b in the product as indicated in the image.

This methoxy group in the product formed in position b comes from the 18O-labeled methanol (CH3OH).

While the oxygens at positions a and c in the product come from the unlabeled hemiacetal.

Thus, the 18o-labeled methanol (CH3O*H) will appear in the products side at position b.

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