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3 years ago

What is the percent of Li in Li3N ? Please!!

Chemistry

2 answers:

lisabon 2012 [21]3 years ago

6 0

I think it maybe 59.7852%

rjkz [21]3 years ago

3 0

Answer: Li percentage in $Li_3N$ is 59.8%.

Explanation: The question asks to calculate the mass percentage of Li in $Li_3N$ .

The formula used for mass percentage of an element in a compound is:

mass percentage = $(\frac{element mass}{compound mass})*100$

Atomic mass of Li is 6.94 and atomic mass of N is 14.0.

There are three Li in the given compound. So, the mass of Li in the compound is = 3(6.94) = 20.82

Mass of compound = 20.82 + 14.0 = 34.82

Mass percentage of Li = $(\frac{20.82}{34.82})*100$ = 59.8%

Hence, the percentage of Li in $Li_3N$ is 59.8%.

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Answer:

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3 years ago

If a sample of air initially occupies 240L at 2 atm how much pressure is required to compress it to 20L at constant temperature

IceJOKER [234]

Answer:

24 atm.

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 240 L

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P₁V₁ = P₂V₂

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3 years ago

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How many moles of water are formed when 22 mol of methane combusts?

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3 years ago

First-order reaction that results in the destruction of a pollutant has a rate constant of 0.l/day. (a) how many days will it ta

Klio2033 [76]

Rate equation for first order reaction is as follows:

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}$

Here, k is rate constant of the reaction, t is time of the reaction, $A_{0}$ is initial concentration and $A_{t}$ is concentration at time t.

The rate constant of the reaction is $0.1 day^{-1}$ .

(a) Let the initial concentration be 100, If 90% of the chemical is destroyed, the chemical present at time t will be 100-90=10, on putting the values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{10}=23.03 days$

Thus, time required to destroy 90% of the chemical is 23.03 days.

(b) Let the initial concentration be 100, If 99% of the chemical is destroyed, the chemical present at time t will be 100-99=1, on putting the values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{1}=46.06 days$

Thus, time required to destroy 99% of the chemical is 46.06 days.

(c) Let the initial concentration be 100, If 99.9% of the chemical is destroyed, the chemical present at time t will be 100-99.9=0.1, on putting the values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{0.1}=69.09 days$

Thus, time required to destroy 99.9% of the chemical is 69.09 days.

5 0

3 years ago