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3 years ago

3 H2 (g) + N2 (g) 2 NH3 (g)

Chemistry

1 answer:

ale4655 [162]3 years ago

4 0

Answer:

Mass = 0.697 g

Explanation:

Given data:

Volume of hydrogen = 1.36 L

Mass of ammonia produced = ?

Temperature = standard = 273.15 K

Pressure = standard = 1 atm

Solution:

Chemical equation:

3H₂ + N₂ → 2NH₃

First of all we will calculate the number of moles of hydrogen:

PV = nRT

R = general gas constant = 0.0821 atm.L/mol.K

1atm ×1.36 L = n × 0.0821 atm.L/mol.K × 273.15 K

1.36 atm.L = n × 22.43 atm.L/mol

n = 1.36 atm.L / 22.43 atm.L/mol

n = 0.061 mol

Now we will compare the moles of hydrogen and ammonia:

H₂ : NH₃

3 : 2

0.061 : 2/3×0.061 = 0.041

Mass of ammonia:

Mass = number of moles × molar mass

Mass = 0.041 mol × 17 g/mol

Mass = 0.697 g

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3 years ago

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A) Calculate the average density of the following object (assume it is a perfect sphere). SHOW ALL YOUR WORK (formulas used, num

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Answers:

A) 2040 kg/m³; B) 58 600 km

Explanation:

A) Density

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$\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{1.3\times 10^{22} \text{ kg} }{6.37 \times 10^{9} \text{ km}^{3}}\times (\frac{\text{1 km}}{\text{1000 m}})^{3} = \text{2040 kg/m}^{3}$

B) Radius

$\text{Volume} = \frac{\text{mass}}{\text{density}} = \frac{5.68\times 10^{26} \text{ kg} }{687 \text{ kg/m}^{3} }= 8.268 \times 10^{23} \text{ m}^{3}$

$V = \frac{ 4}{3 }\pi r^{3}$

$r^{3} = \frac{3V }{4 \pi }\$

$r= \sqrt [3]{ \frac{3V }{4 \pi } }$

$r= \sqrt [3]{ \frac{3\times 8.268 \times 10^{23} \text{ m}^{3}}{4 \pi } }= \sqrt [3]{ 1.974 \times 10^{23} \text{ m}^{3}}= 5.82 \times 10^{7} \text{ m}=\text{58 200 km}$

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3 years ago

Please help on 44. /45./ 46./47,

Aleonysh [2.5K]

The only one that I can do without google is 47. Sorry that I can't answer the others. The answer to 47 is this: you know that the western side of the hill has the steepest slope because the ovals showing altitude are way closer together. The closer the circles/ovals are, the steeper the slope is.

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3 years ago

What is not matter in chemistry??

Igoryamba

Answer:

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Explanation:

Energy: Light, heat, kinetic and potential energy, and sound are non-matter because they are massless. Objects that have mass and are matter may emit energy.

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2 years ago

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What temperature will the water reach when 10.1 g CaO is dropped into a coffee cup containing 157 g H2O at 18.0°C if the followi

Zepler [3.9K]

Answer:

Final temperature attained by water = 34.6°C

Explanation:

The reaction of CaO and H₂O is an exothermic reaction. The equation of reaction is given below:

CaO + H₂O ----> Ca(OH)₂

The quantity of heat given off, ΔH°rxn = 64.8KJ/mol = 64800J/mol

Number of moles of CaO = mass/molar mass, where molar mass of Ca0 = 56g/mol, mass of CaO = 10.1g

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Quantity of heat given off by 0.179 moles = 64800 *0.179 = 11599.2J/mol

Using the formula, Quantity of heat, q = mass * specific heat capacity * temperature rise.

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Final temperature(T₂) - Initial temperature(T₁) = Temperature rise

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698.478T₂ = 24171.804

T₂ = 34.6°C

Therefore, final temperature attained by water = 34.6°C

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