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Tems11 [23]
3 years ago
15

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:

los metales alcalinotérreos: berilio (Be), magnesio (Mg), calcio (Ca), estroncio (Sr), bario (Ba) y radio (Ra).

o simplemente llamado grupo 2A

Explanation:

8 0
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

Initial pressure (P₁) = 2 atm

Final volume (V₂) = 20 L

Temperature = constant

Final pressure (P₂) =?

The final pressure required, can be obtained by using the Boyle's law equation as shown below:

P₁V₁ = P₂V₂

2 × 240 = P₂ × 20

480 = P₂ × 20

Divide both side by 20

P₂ = 480 / 20

P₂ = 24 atm

Thus, the final pressure required is 24 atm.

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3 years ago
WHAT IS A ROCK?
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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
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