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Oliga [24]
3 years ago
7

What is the molecular formula of the compound

Chemistry
1 answer:
Ipatiy [6.2K]3 years ago
3 0
Molecular formula: C10H15Cl5
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A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of 0.157g of the compound produced 0.213g of CO2 and
vladimir2022 [97]

Answer:

C_{7} H_{5}N_{3}O_{6}

Explanation:

First reaction gives you the number of moles or the mass from Carbon and hydrogen

for carbon:

0,213gCO_{2} .\frac{1molCO_{2}}{44gCO_{2}} .\frac{1molC}{1molCO_{2}} =0.005molC

0.213gCO_{2} .\frac{1molCO_{2}}{44gCO_{2}} .\frac{1molC}{1molCO_{2}} .\frac{12gC}{1molC} = 0.058gC\\

Analogously for hydrogen:

0.0310gH_{2}O have 0.0034gH or 0.0034mol of H

In the second reaction you can obtain the amount of nitrogen as a percentage and find the mass of N in the first sample.

0.023gNH_{3} .\frac{1molNH_{3}}{17gNH_{3}} .\frac{1molN}{1molNH_{3}} .\frac{14gN}{1molN} \frac{100}{0.103gsample} =18.4%N

now

\frac{18.4gN}{100gsample} .0.157gsample=0.0289gN in the first reaction

this is equivalet to 0.002mol of N

with this information you can find the mass of oxygen by matter conservation.

gO=total mass-(gN+gC+gH)=0.157-(0.0289+0.058+0.0034)=0.0666gO

this is equivalent to 0.004molO

finally you divide all moles obtained between the smaller number of mole (this is mol of H)

C\frac{0.0048}{0.0034} H\frac{0.0034}{0.0034} N\frac{0.002}{0.0034} O\frac{0.004}{0.0034} =C_{1.4} HN_{0.6} O_{1.2}

and you can multiply by  5   to obtain: C_{7} H_{5}N_{3}O_{6}

4 0
3 years ago
What is the best way to study for a chemistry SOL? Does anyone have any good study links?
Artist 52 [7]

There are a lot of good study sets on Quizlet. type in one of your questions and there is a good chance that a set has all of the required information to know for your Chemistry SOL. Good Luck!

5 0
3 years ago
What is the net ionic equation for the following? :)
I am Lyosha [343]

Answer:

1. Mg (s) + 2Na+(aq) → 2Na(s) + Mg²⁺(aq)

2. 2K(s) + Cd²⁺(aq) → 2K⁺(aq) + Cd(s)

Explanation:

The net ionic equation of a reaction express only the chemical species that are involved in the reaction:

1. Mg (s) + Na2CrO4 (aq) → 2Na + MgCrO4(aq)

The ionic equation:

Mg (s) + 2Na+(aq) + CrO4²⁻ (aq) → 2Na + Mg²⁺ + CrO4²⁻(aq)

Subtracting the ions that don't change:

<h3>Mg (s) + 2Na+(aq) → 2Na + Mg²⁺</h3>

2. 2K(s) + Cd(NO3)2(aq) → 2KNO3(aq) + Cd(s)

The ionic equation:

2K(s) + Cd²⁺(aq) + 2NO3⁻(aq) → 2K⁺(aq) + 2NO3⁻(aq) + Cd(s)

Subtracting the ions that don't change:

<h3>2K(s) + Cd²⁺(aq) → 2K⁺(aq) + Cd(s)</h3>

3 0
3 years ago
For a first-order reaction, A → B, the rate coefficient was found to be 3.4 × 10-4 s-1 at 23 °C. After 5.0 h, the concentration
Illusion [34]

Answer:

the original concentration of A = 0.0817092  M

Explanation:

A reaction is considered to be of first order it it strictly obeys the graphical equation method.

k_1 = \dfrac{2.303}{t}log \dfrac{a}{a-x}

where;

k = the specific rate coefficient  = 3.4 × 10⁻⁴ s⁻¹

t = time   = 5.0 h = 5.0 × 3600 = 18000 seconds

a = initial concentration = ???

a - x = remaining concentration of initial concentration at time t = 0.00018 mol L⁻¹

3.4 \times 10^{-4}= \dfrac{2.303}{18000}log \dfrac{a}{0.00018}

3.4 \times 10^{-4}= 1.27944 \times 10^{-4} \times log \dfrac{a}{0.00018}

\dfrac{3.4 \times 10^{-4}}{1.27944 \times 10^{-4}}=   log \dfrac{a}{0.00018}

2.657=   log \dfrac{a}{0.00018}

10^{2.657}= \dfrac{a}{0.00018}

453.94 = \dfrac{a}{0.00018}

a =453.94 \times 0.00018

a = 0.0817092  M

Thus , the original concentration of A = 0.0817092  M

8 0
3 years ago
What is the volume of 1.9 moles of chlorine gas (Cl2) at standard temperature and pressure (STP)?
masya89 [10]
Considering ideal gas behavior, the volume of 1 mol of gas at STP is 22.4 L; then the volume occupied by 1.9 moles is 1.9mol*22.4L/mol = 42. 6 L.

Answer: 43 L
5 0
3 years ago
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