What is the percent composition of nitrogen in sodium nitride

Answers

Answer 1

Answer:

The percent composition of nitrogen in sodium nitride is 16.86%

What is sodium nitride?

Sodium nitride is an inorganic compound that is used in preservatives and antidote for cyanide poisoning.

The molar mass of Na3N is 83 M

The mass of Na3N is 14 g

$% \;of \;N = (mass of N)/( molar mass of Na3N ) * 100$

$\% \;of \;N = (mass\; of \;N)/(molar\; mass\; of\; Na3N) * 100\n\n\n\% \;of \;N = (14)/(83) * 100 = 16.86\%$

Thus, the percent composition of nitrogen in sodium nitride is 16.86%.

Learn more about sodium nitride

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Answer 2

Answer:

16.86%

Explanation:

Na3N is sodium nitride.

% of N = mass of N/ molar mass of Na3N *100

% of N = 14/83*100= 16.86%

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The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment in a CSTR. When pure A is fed to a 10 dm 3 PFR at 300 K and a volumetric flow rate of 5 dm 3 /s, the conversion is 80%. When a mixture of 50% A and 50% inert (I) is fed to a 10 dm 3 CSTR at 320 K and a volumetric flow rate of 5 dm 3 /s, the conversion is also 80%. What is the activation energy in cal/mol

Answers

Answer:

The activation energy is $=8.1\,kcal\,mol^(-1)$

Explanation:

The gas phase reaction is as follows.

$A \rightarrow B+C$

The rate law of the reaction is as follows.

$-r_(A)=kC_(A)$

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = $10dm^(3)$

Temperature "T" = 300 K

Volumetric flow rate of the reaction $v_(o)=5dm^(3)s$

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

$V= (v_(0))/(k)[(1+\epsilon )ln((1)/(1-X)-\epsilon X)]$

Rearrange the formula is as follows.

$k= (v_(0))/(V)[(1+\epsilon )ln((1)/(1-X)-\epsilon X)]............(1)$

The feed has Pure A, mole fraction of A in feed $y_{A_(o)}$ is 1.

$\epsilon =y_{A_(o)}\delta$

$\delta$ = change in total number of moles per mole of A reacte.

$=1(2-1)=1$

Substitute the all given values in equation (1)

$k=(5m^(3)/s)/(10dm^(3))[(1+1)ln (1)/(1-0.8)-1 * 0.8] = 1.2s^(-1)$

Therefore, the rate constant in case of the plug flow reacor at 300K is $1.2s^(-1)$

The rate constant in case of the CSTR can be calculated by using the formula.

$(V)/(v_(0))= (X(1+\epsilon X))/(k(1-X)).............(2)$

The feed has 50% A and 50% inerts.

Hence, the mole fraction of A in feed $y_{A_(o)}$ is 0.5

$\epsilon =y_{A_(o)}\delta$

$\delta$ = change in total number of moles per mole of A reacted.

$=0.5(2-1)=0.5$

Substitute the all values in formula (2)

$(10dm^(3))/(5dm^(3))=(0.8(1+0.5(0.8)))/(k(1-0.8))=2.8s^(-1)$

Therefore, the rate constant in case of CSTR comes out to be $2.8s^(-1)$

The activation energy of the reaction can be calculated by using formula

$k(T_(2))=k(T_(1))exp[(E)/(R)((1)/(T_(1))-(1)/(T_(2)))]$

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

$E= R *((T_(1)T_(2))/(T_(1)-T_(2)))ln(k(T_(2)))/(k(T_(1)))$

Substitute the all values.

$=1.987cal/molK((300K *320K)/(320K *300K))ln (2.8)/(1.2)=8.081 *10^(3)cal\,mol^(-1)$

$=8.1\,kcal\,mol^(-1)$

Therefore, the activation energy is $=8.1\,kcal\,mol^(-1)$

How many grams of Cl are in 31.2g CF2Cl2

Answers

Answer:

Mass = 42.6 g

Explanation:

Given data:

Mass of CF₂Cl₂ = 31.2 g

Mass of Cl₂ = ?

Solution:

Number of moles of CF₂Cl₂ = mass/molar mass

Number of moles = 31.2 g/121 gmol

Number of moles = 0.3 mol

1 mole of CF₂Cl₂ contain 2 moles of Cl atom.

0.3 mol × 2 = 0.6 mol

Mass of Cl₂:

Mass = number of moles × molar mass

Mass = 0.6 mol × 71 g/mol

Mass = 42.6 g

What the first song come to your mind?

Answers

Answer: Hey Brother by Avicii

Let her go by Passenger

A potassium ion (K+) would most likely bond with _____.

Mg+
O
Cl-
Na+

Answers

Answer:

Chlorine

Explanation:

Answer:

That would be the negative ion: Cl-.

g Acetic acid is diluted with water to make a solution of vinegar. You have a sample of vinegar that contains 16.7 g of acetic acid. Determine the number of moles of acetic acid in the vinegar sample.

Answers

Answer:

0.278 mol

Explanation:

Step 1: Given and required data

Mass of acetic acid (m): 16.7 g

Chemical formula of acetic acid: CH₃COOH (C₂H₄O₂)

Step 2: Calculate the molar mass (M) of acetic acid

We will use the following expression.

M(C₂H₄O₂) = 2 × M(C) + 4 × M(H) + 2 × M(O)

M(C₂H₄O₂) = 2 × 12.01 g/mol + 4 × 1.01 g/mol + 2 × 16.00 g/mol = 60.06 g/mol

Step 3: Calculate the number of moles (n) of acetic acid

We will use the following expression.

n = m/M

n = 16.7 g/(60.06 g/mol) = 0.278 mol

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