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4 years ago

Chem help please

Chemistry

1 answer:

Murrr4er [49]4 years ago

6 0

Considering the equation:

Li(s)+N₂(g)—->Li₃N(s)

Balancing this equation will give :

6Li(s)+N₂(g)—->2Li₃N(s)

It can be seen that 6 mol of Li reacts with 1 mol of N₂to produce 2 mol of Li₃N.

Here 33.6524g of nitrogen gas reacts with 58.7032g lithium to produce Li₃N.

Moles of Li = 58.7032 g * $\frac{1 mol }{6.94 g} = 8.46 mol Li$

Moles of $N_{2} = 33.6524 g * \frac{1 mol}{28 g} = 1.2 mol N_{2}$

Mass of $Li_{3} N from Li = 8.46 mol Li * \frac{2 mol Li_{3}N}{6 mol Li}* \frac{34.83 g Li_{3}N}{1 mol Li_{3}N} = 98.2 g$

Mass of $Li_{3} N from [tex] N_{2}$ = 1.2 mol $N_{2}$ * \frac{2 mol Li_{3}N}{1 mol N2}* \frac{34.83 g Li_{3}N}{1 mol Li_{3}N} = 98.2 g [/tex] = 83.6 g

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if you react 100g of ammonium chloride with excess calcium oxide, what Is the theoretical yeild (in grams) of ammonia? when you

Maurinko [17]

Answer:

31.78 grams

25.55%

Explanation:

The balanced reaction for ammonium chloride with calcium oxide will be:

2NH4Cl + Ca(OH)2 ---> CaCl2 + 2NH3 + 2H2O

The molecular weight for ammonium chloride(NH4Cl ) is 53.49g/mol, while the molecular weight for ammonium(NH3) is 17g/mol. The number of theoretical yield of ammonia from 100g of ammonium chloride will be:

100g / (53.49g/mol) * 2/2 * 17g/mol= 31.78 grams

If the actual yield is 8.12g, the percent yield will be: 8.12g/31.78g * 100% =25.55%

4 0

3 years ago

How many moles of propane gas would be present in 11 grams<br> of the gas at standard conditions?

Akimi4 [234]

The molar mass is usually referred to with
M
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m
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n
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=
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Since you have given (C3H8)=11 g
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(
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3
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8
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=
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g
and you already looked up (C3H8)=44.1 gmol−1
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(
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=
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g
m
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n
(
C
3
H
8
)
.

In this question it is quite hard to explain the use of significant figures. Those are used to imply a certain inaccuracy. Not enough information is given by the question, as of how accurate the measurement is. It is a mere exercise of converting one property into another. Here you should not worry about it.

5 0

3 years ago

How many moles are in 60.66 g CF2Cl2

mafiozo [28]

There are 0.501 moles in 60.66g of CF2Cl2

7 0

4 years ago

How much energy do individual photons of 470 nm light have

IceJOKER [234]

<h3>Answer:</h3>

4.227 × 10^-19 Joules

<h3>Explanation:</h3>

Energy of a photon of light is calculated by the formula;

E = hf, where h is the plank's constant, 6.626 × 10^-34 J-s and f is the frequency.

But, f = c/λ

Where, c is the speed of light (2.998 × 10⁸ m/s), and λ is the wavelength.

Given the wavelength is 470 nm or 4.7 × 10^-7 m

Therefore;

E = hc/λ

= (6.626 × 10^-34 J-s × 2.998 × 10^8 m/s) ÷ 4.7 × 10^-7 m

= 4.227 × 10^-19 Joules

Therefore, the energy of a photon with 470 nm is 4.227 × 10^-19 Joules

3 0

4 years ago

PLSSSSS HELP I DONT GET THIS PROBLEMMMM

Aleks [24]

Answer:

C. 7370 joules.

Explanation:

There is a mistake in the statement. Correct form is described below:

<em>Using the above data table and graph, calculate the total energy in Joules required to raise the temperature of 15 grams of ice at -5.00 °C to water at 35 °C. </em>

The total energy needed to raise the temperature is the combination of latent and sensible heats, all measured in joules, and represented by the following model:

$Q = m\cdot [c_{i} \cdot (T_{2}-T_{1})+L_{f} + c_{w}\cdot (T_{3}-T_{2})]$ (1)

Where:

$m$ - Mass of the sample, in grams.

$c_{i}$ - Specific heat of ice, in joules per gram-degree Celsius.

$c_{w}$ - Specific heat of water, in joules per gram-degree Celsius.

$L_{f}$ - Latent heat of fusion, in joules per gram.

$T_{1}$ - Initial temperature of the sample, in degrees Celsius.

$T_{2}$ - Melting point of water, in degrees Celsius.

$T_{3}$ - Final temperature of water, in degrees Celsius.

$Q$ - Total energy, in joules.

If we know that $m = 15\,g$ , $c_{i} = 2.06\,\frac{J}{g\cdot ^{\circ}C}$ , $c_{w} = 4.184\,\frac{J}{g\cdot ^{\circ}C}$ , $L_{f} = 334.72\,\frac{J}{g}$ , $T_{1} = -5\,^{\circ}C$ , $T_{2} = 0\,^{\circ}C$ and $T_{3} = 35\,^{\circ}C$ , then the final energy to raise the temperature of the sample is:

$Q = (15\,g)\cdot \left[\left(2.06\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (5\,^{\circ}C)+ 334.72\,\frac{J}{g} + \left(4.184\,\frac{J}{g\cdot ^{\circ}C}\right)\cdot (35\,^{\circ}C) \right]$

$Q = 7371.9\,J$

Hence, the correct answer is C.

8 0

3 years ago