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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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If 20 grams of Zinc phosphate reacts with excess hydrochloric acid and produces 18 grams of Zinc chloride what is the percent yi

fenix001 [56]

Answer:

$Y=85\%$

Explanation:

Hello!

In this case, since we know the balanced chemical reaction, we are first able to realize there is a 1:3 mole ratio between zinc phosphate and zinc chloride; it means that we can first compute the moles of the desired product via stoichiometry:

$n_{ZnCl_2}=20gZn_3(PO_4)_2*\frac{1molZn_3(PO_4)_2}{386.11gZn_3(PO_4)_2}*\frac{3molZnCl_2}{1molZn_3(PO_4)_2}=0.16gZnCl_2$

Next, since those moles are associated with the theoretical yield of zinc chloride, we obtain the corresponding mass:

$m_{ZnCl_2}^{theoretical}=0.16molZnCl_2*\frac{136.29gZnCl_2}{1molZnCl_2} =21gZnCl_2$

Finally, we compute the percent yield by diving the actual yield (18 g) by the theoretical yield:

$Y=\frac{18g}{21g}*100\%\\\\Y=85\%$

Best regards!

4 0

2 years ago

When balance the following chemical equations N2+O2= NO2 the coefficient for O2 is

Papessa [141]

Answer:

Explanation:

N₂ + O₂ ---> NO₂

balance

N₂ + 2O₂ ---> 2NO₂

4 0

2 years ago

Use the following half-reactions to construct a voltaic cell:

velikii [3]

<u>Answer:</u> The correct answer is $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V$

<u>Explanation:</u>

We are given:

$Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o=-0.73V\\\\Ag^+(aq.)+e^-\rightarrow Ag(s);E^o=+0.80V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.

Chromium will undergo oxidation reaction and will get oxidized.

The half reactions for the above cell is:

Oxidation half reaction: $Cr(s)\rightarrow Cr^{3+}+3e^-;E^o_{Cr^{3+}/Cr}=-0.73V$

Reduction half reaction: $Ag^{+}+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V$ ( × 3)

Net equation: $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.)$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.80-(-0.73)=1.53V$

Hence, the correct answer is $3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V$

4 0

2 years ago

Caculate the mass of oxygen needed to burn 175.0grams of octane

IgorC [24]

A water wave traveling in a straight line on a lake is described by the equation
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8 0

3 years ago

a sample of an unknown gas has a mass of 0.582g. it's volume is 21.3 mL a a temp of 100 degrees Celsius and a pressure of 754 mm

Zielflug [23.3K]

Answer:

Molar mass = 0.09 × 10⁴ g/mol

Explanation:

Given data:

Mass = 0.582 g

Volume = 21.3 mL

Temperature = 100°C

Pressure = 754 mmHg

Molar mass = ?

Solution:

(21.3 /1000 = 0.0213 L)

(100+273= 373 K)

(754/760 = 0.99 atm)

PV = nRT

n = PV/RT

n = 0.99 atm × 0.0213 L / 0.0821 atm. L. mol⁻¹. k⁻¹ × 373 K

n =0.02 mol/ 30.6

n = 6.5 × 10⁻⁴ mol

Molar mass = Mass/ number of moles

Molar mass = 0.582 g / 6.5 × 10⁻⁴ mol

Molar mass = 0.09 × 10⁴ g/mol

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3 years ago