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3 years ago

Calculate the percent composition of Mg(NO3)2.

Chemistry

1 answer:

dimaraw [331]3 years ago

3 0

Answer:

%M = 16.2%

%N = 18.9%

%O = 64.9%

Explanation:

Mg(NO3)2

Atomic mass of Magnesium M = 24

Atomic mass of Nitrogen N=14

Atomic mass of Oxygen O=16

Molar mass of Mg(NO3)2 = 24 + (14+[16*3})*2

Mg(NO3)2 = 24 + (14+48)*2

Mg(NO3)2 = 24 + 124

Mg(NO3)2 = 148

Percentage composition;

Magnesium M= (24/148) × 100

Magnesium M=0.162 × 100 =16.2%

Oxygen O=(16*6/148) × 100

Oxygen O= 0.649 × 100 = 64.9%

Nitrogen N= (14*2/148) × 100

Nitrogen N=0.189 × 100=18.9%

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What is the molarity of a solution of 17.0 g of nh4br in enough h2o to make 158 ml of solution? answer in units of m?

vivado [14]

Unit of M is also mole/L, where mole is the moles of solute and L is the volume of the solution. The latter is given: 158 mL or 0.158 L. So we need to find out the moles of NH4Br.

Moles of NH4Br = Mass of NH4Br/molar mass of NH4Br = 17.0g/(14+1*4+79.9)g/mol = 0.1736 mole.

So, the molarity of the solution = 0.1736mole/0.158L = 1.10 mole/L = 1.10 M

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3 years ago

Consider the second-order reaction:

kirza4 [7]

Answer:

Initial concentration of HI is 5 mol/L.

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

Explanation:

$2HI(g)\rightarrow H_2(g)+I_2(g)
$

Rate Law: $k[HI]^2
$

Rate constant of the reaction = k = $6.4\times 10^{-9} L/mol s$

Order of the reaction = 2

Initial rate of reaction = $R=1.6\times 10^{-7} Mol/L s$

Initial concentration of HI = $[A_o]$

$1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2$

$[A_o]=5 mol/L$

Final concentration of HI after t = [A]

t = $4.53\times 10^{10} s$

Integrated rate law for second order kinetics is given by:

$\frac{1}{[A]}=kt+\frac{1}{[A_o]}$

$\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}$

$[A]=0.00345 mol/L$

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

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3 years ago

Based on the following molecular weight data of polypropylene, determine the degree of polymerization Molecular Weight Range (g/

Lady bird [3.3K]

Answer:

785

Explanation:

Molecular. X. W

Weight

8000-16000 0.05 0.03

16000-24000. 0.017. 0.08

24000-32000. 0.22. 0.18

32000-40000. 0.25. 0.35

40000-48000. 0.22. 0.27

48000-56000. 0.09. 0.09

Mean weight X*M. W*M

12000. 600. 240

20000. 3200. 2000

28000. 6720. 5600

36000. 10080. 10800

44000. 8800. 11880

52000. 3640 3640

Total=33040g\mol 36240

Note before repeat molecular weight m= 3*12.01+6*1.008=

42.08g/mol

Degree of polymerization= total W*M/w=33040/42.08 =785

8 0

3 years ago

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shtirl [24]

Answer : The balanced equation for the complete combustion of cyclopentane will be :

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Explanation :

As we know that the hydrocarbons can undergo complete or incomplete combustion that depends on the amount of oxygen available.

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The general reaction will be :

$\text{Hydrocarbon}+O_2\rightarrow CO_2+H_2O$

Incomplete combustion : It takes place when there is a poor supply of oxygen. In this, hydrocarbon react with the oxygen to give carbon monoxide, carbon dioxide and water as a product.

The general reaction will be :

$\text{Hydrocarbon}+O_2\text{(limited)}\rightarrow CO+CO_2+H_2O$

As per question, the complete combustion of cyclopentane react with oxygen to gives carbon dioxide and water as a product.

Therefore, the balanced chemical reaction will be :

$2C_5H_{10}+15O_2\rightarrow 10CO_2+10H_2O$

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3 years ago

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kupik [55]

Answer:

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4 years ago