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2 years ago

N2 + 3H2 - 2NH3

Chemistry

1 answer:

vladimir1956 [14]2 years ago

4 0

The number of mole of nitrogen gas, N₂, needed to produce 150 g of ammonia, NH₃ is 4.41 moles

<h3>How to determine the mole of NH₃ produced </h3>

Mass of NH₃ = 150 g
Molar mass of NH₃ = 14 + (3×1) = 17 g/mol

Mole of NH₃ =?

Mole = mass /molar mass

Mole of NH₃ = 150 / 17

Mole of NH₃ = 8.82 moles

<h3>How to determine the mole of N₂ needed </h3>

Balanced equation

N₂ + 3H₂ —> 2NH₃

From the balanced equation above,

2 moles of NH₃ were produced by 1 mole of N₂.

Therefore,

8.82 moles of NH₃ will be produced by = 8.82 / 2 = 4.41 moles of N₂.

Thus, 4.41 moles of N₂ is needed for the reaction.

Learn more about stoichiometry:

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The net ionic equation for the precipitation reaction that occurs when aqueous solutions of silver(i) nitrate and chromium(iii)

eduard

The net ionic equation formed is

Ag^+(aq)+Cl^−(aq)→AgCl(s)

Chromium(III) nitrate and silver(I) chloride are the products of the balanced molecular equation for the reaction between chromium(III) chloride and silver(I) nitrate. An (s) next to the chemical formula for silver(I) chloride designates it as an insoluble salt.

CrCl3(aq)+3AgNO3(aq)→Cr(NO3)3(aq)+3AgCl(s)

Silver and the chloride ions are the two ions that must interact to create silver(I) chloride. By designating ions as the reactants and silver(I) chloride as the product, the net ionic equation is formed.

Ag^+(aq)+Cl^−(aq)→AgCl(s)

Ionic Equation:

In general, anions and cations react to generate a compound in a dissolved media, which is known as an ionic reaction. Water-insoluble salts are created when the ions of water-soluble salts interact with one another in an aqueous media.

To learn more about Ionic equaion click the given link

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3 0

1 year ago

Does adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent, a lesser extent, or a greater e

igor_vitrenko [27]

Answer:

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em> by adding 1 mol of $C_06H_{12}O_6$ to 1 kg of water.

Explanation:

1) Moles of NaCl , $n_1=1 mol$

Mass of water = m= 1 kg = 1000 g

Moles of water = $n_2=\frac{1000 g}{18 g/mol}=55.55 mol$

Vapor pressure of the solution = $p$

Vapor pressure of the pure solvent that is water = $p_o=17.5 Torr$

Mole fraction of solute(NaCl)= $\chi_1=\frac{n_1}{n_1+n_2}$

$\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}$

$\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}$

$p=17.19 Torr$

The vapor pressure for the NaCl solution at 17.19 Torr.

2) Moles of sucrose , $n_1=1mol$

Mass of water = m = 1 kg = 1000 g

Moles of water = $n_2=\frac{1000 g}{18 g/mol}=55.55 mol$

Vapor pressure of the solution = $p'$

Vapor pressure of the pure solvent that is water = $p_o=17.5 Torr$

Mole fraction of solute ( glucose)= $\chi_1=\frac{n_1}{n_1+n_2}$

$\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}$

$\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}$

$p'=17.19 Torr$

The vapor pressure for the glucose solution at 17.19 Torr.

p = p' = 17.19 Torr

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent by adding 1 mol of $C_06H_{12}O_6$ to 1 kg of water.

3 0

3 years ago

5.11 g of MgSO₄ is placed into 100.0 mL of water. The water's temperature increases by 6.70°C. Calculate ∆H, in kJ/mol, for the

cupoosta [38]

Answer: Thus ∆H, in kJ/mol, for the dissolution of MgSO₄ is -66.7 kJ

Explanation:

To calculate the entalpy, we use the equation:

$q=mc\Delta T$

where,

q = heat absorbed by water = ?

m = mass of water = ${\text {volume of water}}\times {\text {density of water}}=100.0ml\times 1.00g/ml=100.0g$

c = heat capacity of water = 4.186 J/g°C

$\Delta T$ = change in temperature = $6.70^0C$

$q=100.0g\times 4.184J/g^0C\times 6.70^0C=2803.3J=2.8033kJ$

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

The heat absorbed by water will be equal to heat released by $MgSO_4$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass = 5.11 g

Molar mass = 120 g/mol

Putting values in above equation, we get:

$\text{Moles of }MgSO_4=\frac{5.11g}{120g/mol}=0.042mol$

0.042 moles of $MgSO_4$ releases = 2.8033 kJ

1 mole of $MgSO_4$ releases = $\frac{2.8033 kJ}{0.042}\times 1=66.7kJ$

Thus ∆H, in kJ/mol, for the dissolution of MgSO₄ is -66.7 kJ

3 0

3 years ago

An acetylene tank has a volume of 390.0 L. It is stored at a temperature of 23.5 °C and has a

xeze [42]

Considering the ideal gas law, there are 279.42 moles of acetylene in the tank.

<h3>Definition of ideal gas</h3>

Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.

<h3>Ideal gas law</h3>

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of gases:

P×V = n×R×T

<h3>Moles of acetylene</h3>

In this case, you know:

P= 1765 kPa= 17.4192 atm (being 101.325 kPa= 1 atm)
V= 390 L
n= ?
R= 0.082 $\frac{atmL}{molK}$
T= 23.5 °C= 296.5 K (being 0 °C= 273 K)

Replacing in the ideal gas law:

17.4192 atm× 390 L = n×0.082 $\frac{atmL}{molK}$ × 296.5 K

Solving:

$n=\frac{17.4192 atmx 390 L}{0.082 \frac{atmL}{molK}x296.5 L}$

<u><em>n= 279.42 moles</em></u>

Finally, there are 279.42 moles of acetylene in the tank.

Learn more about ideal gas law:

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4 0

2 years ago

What would the product look like if ROOR weren't used?

S_A_V [24]

Answer:

The least substituted product (anti-Markovnikov)

Explanation:

The ROOR is used in the addition reaction of HBr to an organic substance (an alkene for example).

In normal conditions (with no ROOR) the adition of the halogen will be performed in the most substituted C (following the rule of Markovnikov that says that the stability increases with the more substituted is the C).

But in presence of ROOR, the reaction takes other mechanism (free radicals), and the product in this case is the one with the Br added in the least substituted C.

4 0

3 years ago