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sleet_krkn [62]

2 years ago

13

What type of chemical reaction is C3H8 + O2 ---> CO2 + H2O ?

1 answer:

yuradex [85]2 years ago

7 0

E - Combustion
You can tell because the products are CO2 and H2O :)

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A sample of gas is held at constant pressure. Increasing the Kelvin temperature of this gas sample causes the average kinetic en

Maksim231197 [3]

Increase.
I don't even know the point of the first sentence. Average Kinetic Energy *is* temperature. The higher AVE the higher the temp. The higher the temp the faster the AVE. They're directly correlated.

5 0

2 years ago

A 100g sample of Carbon-14 has a half-life of 5 years. How much Carbon-14 is left after 10 years?

attashe74 [19]

Answer: The amount of carbon-14 left after 10 years is 25 g

Explanation:

Formula used :

$a=\frac{a_o}{2^n}$

where,

a = amount of reactant left after n-half lives = ?

$a_o$ = Initial amount of the reactant = 100 g

n = number of half lives = $\frac{\text {given time}}{\text {half life}}=\frac{10}{5}=2$

Putting values in above equation, we get:

$a=\frac{100g}{2^2}$

$a=\frac{100g}{4}=25g$

Therefore, the amount of carbon-14 left after 10 years is 25 g

6 0

2 years ago

Given the balanced equation below, calculate the moles of aluminum that are needed to react completely with 13.2 moles of FeO?

irina [24]

The number of moles of aluminium that are needed to react completely with 13.2 moles of FeO is 8.8 moles

calculation

2Al + 3FeO → 3aFe +Al2O3

by use of of mole ratio of Al: FeO from equation above = 2:3 the moles of Al is therefore

= 13.2 x 2/3=8.8 moles of Al

3 0

2 years ago

Iron(III) oxide and hydrogen react to form iron and water, like this: Fe_2O_3(s) + 3H_2(g) rightarrow 2Fe(s) + 3H_2O(g) At a cer

Sergeeva-Olga [200]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The equilibrium constant is $K_c= 2.8*10^{-4}$

Explanation:

From the question we are told that

The chemical reaction equation is

$Fe_{2} O_{3}_{(s)} + 3H_{2}_{(g)} -----> 2Fe_{(s)} + 3H_{2} O_{(g)}$

The voume of the misture is $V_m = 5.4L$

The molar mass of $Fe_{2} O_{3}_{(s)}$ is a constant with value of $M_{Fe_{2} O_{3}_{(s)} } = 160g/mol$

The molar mass of $H_{2}_{(g)}$ is a constant with value of $H_2 = 2g/mol$

The molar mass of $H_{2}O$ is a constant with value of $H_2O = 18g/mol$

Generally the number of moles is mathematically given as

$No \ of \ moles \ = \frac{mass}{molar\ mass}$

For $Fe_{2} O_{3}_{(s)}$

$No \ of\ moles = \frac{3.54}{160}$

$= 0.022125 \ mols$

For $H_{2}$

$No \ of\ moles = \frac{3.63}{2}$

$= 1.815 \ mols$

For $H_{2}O$

$No \ of\ moles = \frac{2.13}{18}$

$= 0.12 \ mols$

Generally the concentration of a compound is mathematicallyrepresented as

$Concentration = \frac{No \ of \ moles }{Volume }$

For $Fe_{2} O_{3}_{(s)}$

$Concentration[Fe_2 O_3] = \frac{0.222125}{5.4}$

$= 4.10*10^{-3}M$

For $H_{2}$

$Concentration[H_2] = \frac{1.815}{5.4}$

$= 0.336M$

For $H_{2}O$

$Concentration [H_2O] = \frac{0.12}{5.4}$

$= 0.022M$

The equilibrium constant is mathematically represented as

$K_c = \frac{[concentration \ of \ product]}{[concentration \ of \ reactant ]}$

Considering $H_2O \ for \ product$

And $H_2 \ for \ reactant$

At equilibrium the

$K_c = \frac{0.022}{0.336}$

$K_c= 2.8*10^{-4}$

3 0

2 years ago

Please help balance this equation <br> _B2Br6 + _HNO3= _B(NO3)3+_HBr

Trava [24]

Answer: B2Br6 + 6HNO3 → 2B(NO3)3 + 6BrH

Equation

B2Br6+HNO3=B(NO3)3+HBr

B=2 B=1

BR=6 BR=1

H=1 H=1

N=1 N=3

O=3 O=9

ANSWER

B2Br6 + 6HNO3 → 2B(NO3)3 + 6BrH

B=2 B=2

BR=6 BR=6

H=6 H=6

N=6 N=6

O=18 O=18

HOPE THIS HELPS

4 0

2 years ago