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jolli1 [7]
3 years ago
5

Consider the balanced equation:

Chemistry
2 answers:
Mashutka [201]3 years ago
6 0

Answer:

A). 3 Mg, 2 P, 14 O, 12 H

Explanation:

Mamont248 [21]3 years ago
3 0
The answer is 6 moles of water will be produced.
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Calculate the molarity when 135 g of KCl is dissolved to make a 2.50 L solution. Round to two significant digits.
inysia [295]
<h2>0.72 mol/L</h2>

Explanation:

       \textrm{Molarity of a solution}=\frac{\textrm{Number of moles of solvent}}{\textrm{Volume of solution in Liters}}

       We are given a 2.50L solution which contains 135g of KCl.

       To calculate number of moles of KCl present, we need to find it's molar mass from charts. Molar Mass of KCl is known to be 74.55\frac{g}{Mol}.

       Number of moles of KCl present = \frac{\textrm{Given weight}}{\textrm{Molar Mass}}\textrm{ = }\frac{135g}{74.55\frac{g}{Mol}}\textrm{ = }1.81087mol

       Molarity = \frac{1.81087mol}{2.50L}=0.7243\frac{mol}{L}

∴ Molarity of given KCl solution = 0.72\frac{mol}{L}

7 0
3 years ago
Calcutale Grxn for the following equation at 25°C: <br><br> 4KClO3(s) → 3KClO4(s) KCl(s)
Step2247 [10]

Answer:

-133.2 kJ

Explanation:

Let's consider the following balanced equation.

4 KClO₃(s) → 3 KClO₄(s) + KCl(s)

We can calculate the standard Gibbs free energy of the reaction (ΔG°rxn) using the following expression.

ΔG°rxn = 3 mol × ΔG°f(KClO₄(s)) + 1 mol × ΔG°f(KCl(s)) - 4 mol × ΔG°f(KClO₃(s))

ΔG°rxn = 3 mol × (-303.1 kJ/mol) + 1 mol × (-409.1 kJ/mol) - 4 mol × (-296.3 kJ/mol)

ΔG°rxn = -133.2 kJ

5 0
3 years ago
A diver has 3,400 J of gravitational potential energy after climbing
Zanzabum

Answer:

57.8kg

Explanation:

Potential energy is given by:

U =mgh

Where U is potential energy, m is mass, g is acceleration due to gravity, and h is height. Using this equation:

3400=m(9.8)(6)\\3400=58.8m\\m=57.8kg

3 0
3 years ago
How good are you with chemistry
Ugo [173]

Answer:

<h2>━☆゚.*・。゚Fine, but not as good</h2>
7 0
3 years ago
Read 2 more answers
A student has a 2.19 L bottle that contains a mixture of O 2 , N 2 , and CO 2 with a total pressure of 5.57 bar at 298 K . She k
Sergeeva-Olga [200]

<u>Answer:</u> The partial pressure of oxygen gas is 2.76 bar

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 5.57 bar

V = Volume of the gas = 2.19 L

T = Temperature of the gas = 298 K

R = Gas constant = 0.0831\text{ L bar }mol^{-1}K^{-1}

n = Total number of moles = ?

Putting values in above equation, we get:

5.57bar\times 2.19L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{5.57\times 2.19}{0.0831\times 298}=0.493mol

To calculate the mole fraction of carbon dioxide, we use the equation given by Raoult's law, which is:

p_{A}=p_T\times \chi_{A}         ........(1)

where,

p_A = partial pressure of carbon dioxide = 0.318 bar

p_T = total pressure = 5.57 bar

\chi_A = mole fraction of carbon dioxide = ?

Putting values in above equation, we get:

0.318bar=5.57bar\times \chi_{CO_2}\\\\\chi_{CO_2}=\frac{0.381}{5.57}=0.0571

  • Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

We are given:

Moles of nitrogen gas = 0.221 moles

Mole fraction of nitrogen gas, \chi_{N_2}=\frac{0.221}{0.493}=0.448

Calculating the partial pressure of oxygen gas by using equation 1, we get:

Mole fraction of oxygen gas = (1 - 0.0571 - 0.448) = 0.4949

Total pressure of the system = 5.57 bar

Putting values in equation 1, we get:

p_{O_2}=5.57bar\times 0.4949\\\\p_{O_2}=2.76bar

Hence, the partial pressure of oxygen gas is 2.76 bar

6 0
3 years ago
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