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soldi70 [24.7K]

3 years ago

8

A slow joggernruns a mile in 13min calculate the speed in inch/second

1 answer:

levacccp [35]3 years ago

7 0

1mile/13min * 1min/60sec * 63360in/1mil = 63360in/780sec = 81.23 in/sec

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drek231 [11]

Answer:

Explanation:

The acid level has changed

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3 years ago

O<br> Plz answer this plz plz plz

saul85 [17]

Answer:

Q1. C

Q2 and Q3 are correct.

Explanation:

Since F=ma, and the force is a constant,

for the greatest acceleration, the mass of the ball must be the least.

Thus ball C has the greatest acceleration.

Let's check:

A) F=ma

a=F/m

a= F/68

B) a=F/72

C) a= F/64 (✓)

The smaller the denominator, the larger the value of a.

(Think: 1/2 >1/3)

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3 years ago

Uv gel enhancements rely on ingredients from the monomer liquid and ________ chemical family.

pishuonlain [190]

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3 years ago

If you made 6 moles of NO2 How many grams of N2 did you use N2+2O2> 2NO2

denis23 [38]

Answer: 1:2

Explanation:

Believe me its correct.

7 0

3 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

3 years ago