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3 years ago

Please help!

Chemistry

1 answer:

lord [1]3 years ago

8 0

Answer:

$\large \boxed{\text{6 L}}$

Explanation:

We can use Gay-Lussac's Law of Combining Volumes to solve this problem.

Gases at the same temperature and pressure react in the same ratios as their coefficients in the balanced equation.

1. Write the chemical equation.

Ratio: 1 L 3 L

N₂ + 3H₂ ⟶ 2NH₃

V/L: 2

2. Calculate the volume of H₂.

According to Gay-Lussac, 3 L of H₂ react with 1 L of N₂.

Then, the conversion factor is (3 L H₂/1 L N₂).

$\text{Volume of H}_{2} = \text{2 L N}_{2} \times \dfrac{ \text{3 L H}_{2} }{\text{1 L N}_{2}}= \textbf{6 L H}_{2}\\\text{You need $\large \boxed{\textbf{6 L}}$ of hydrogen,}$

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Which term names the concentration of salts dissolved in a liquid?

Viefleur [7K]

Hey there!
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4 years ago

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Calculate the standard cell potential given the following standard reduction potentials: Al3++3e−→Al;E∘=−1.66 V Cu2++2e−→Cu;E∘=0

Maru [420]

Answer:

Explanation:

To calculate the cell potential we use the relation:

Eº cell = Eº oxidation + Eº reduction

Now in order to determine which of the species is going to be oxidized, we have to remember that the more the value of the reduction potential is negative, the greater its tendency to be oxidized is. In electrochemistry we use the values of the reductions potential in the tables for simplicity because the only thing we need to do is change the sign of the reduction potential for the oxized species .

So the species that is going to be oxidized is the Aluminium, and therefore:

Eº cell = -( -1.66 V ) + 0.340 V = 5.06 V

Equally valid is to write the equation as:

Eº cell = Eº reduction for the reduced species - Eº reduction for the oxidized species

These two expressions are equivalent, choose the one you fell more comfortable but be careful with the signs.

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3 years ago

What is the volume of an oxygen tank if it contains 12 moles of oxygen at 273 K under 75 kPa?

Zina [86]

Given:
n = 12 moles of oxygen
T = 273 K, temperature
p = 75 kPa, pressure

Use the ideal gas law, given by
$pV=nRT \\ or \\ V= \frac{nRT}{p}$
where
V = volume
R = 8.3145 J/(mol-K), the gas constant

Therefore,
$V= \frac{(12\,mol)(8.3145\, \frac{J}{mol-K} )(273\,K)}{75 \times 10^{3} \, Pa}= 0.3632\,m^{3}$

Answer: 0.363 m³

5 0

3 years ago

Hydrogen gas was collected by water displacement. what was pressure of the h2 collected if the temperature was 26°c?

Vedmedyk [2.9K]

The ideal gas law may be written as
$p= \frac{\rho R T}{M}$
where
p = pressure
ρ =density
T = temperature
M = molar mass
R = 8.314 J/(mol-K)

For the given problem,
ρ = 0.09 g/L = 0.09 kg/m³
T = 26°C = 26+273 K = 299 K
M = 1.008 g/mol = 1.008 x 10⁻³ kg/mol

Therefore
$p= \frac{(0.09 \, kg/m^{3})*(8.314 \, J/(mol-K))*(299 \, K)}{1.008 \times 10^{-3} \, kg/mol} =2.2195 \times 10^{5} \, Pa$

Note that 1 atm = 101325 Pa
Therefore
p = 2.2195 x 10⁵ Pa
= 221.95 kPa
= (2.295 x 10⁵)/101325 atm
= 2.19 atm

Answer:
2.2195 x 10⁵ Pa (or 221.95 kPa or 2.19 atm)

4 0

3 years ago

The first-order rate constant for the reaction of methyl chloride (CH3Cl) with water to produce methanol (CH3OH) and hydrochlori

Dvinal [7]

Answer:

K(48.5°C) = 1.017 E-8 s-1

Explanation:

CH3Cl + H2O → CH3OH + HCl

at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1

at T2 = 48.5°C (321.5 K) ⇒ K2 = ?

Arrhenius eq:

K(T) = A e∧(-Ea/RT)
Ln K = Ln(A) - [(Ea/R)(1/T)]

∴ A: frecuency factor

∴ R = 8.314 E-3 KJ/K.mol

⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)

⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)

(1)/(2):

⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)

⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)

⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)

⇒ Ln (K1/K2) = - 3.422

⇒ K1/K2 = e∧(-3.422)

⇒ (3.32 E-10 s-1)/K2 = 0.0326

⇒ K2 = (3.32 E-10 s-1)/0.0326

⇒ K2 = 1.017 E-8 s-1

7 0

3 years ago