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horsena [70]
2 years ago
13

Someone please do this!!!

Chemistry
1 answer:
Lilit [14]2 years ago
4 0

Answer:

1) Fe = 69.9%

O = 31.1%

2) H = 5.19%

O = 16.5%

N = 28.9%

C = 49.5%

Explanation:

One easy way to do percent compositions is to assume you have 100g of a substance.

1) Lets say we have 100g of Fe2O3.

The total molar mass would be:

= 55.845*2+15.999*3 = 159.687

The molar mass of the Fe2 alone is:

=55.845*2 = 111.69

Thus, the grams of Fe2(out of a 100) could be calculated by multiplying 100g * the molar mass ratio of Fe2 to the whole:

= 100g *\frac{111.69}{159.687} = 69.9431

Which is approximately 69.9%.

We can find the amount of O3 by simply subtracting, as the rest of the compound is made of O3. Thus, the % composition of O3 is 31.1%

You can then do this same process to the next question, getting us the following:

H = 5.19%

O = 16.5%

N = 28.9%

C = 49.5%

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lina2011 [118]

Answer:

The warmer, lighter air rises, bringing cooler, heavier air to low altitudes.

Air at higher altitudes doesn't have as much air weighing down on it from above.

Explanation:

In short - air pressure is the result of the cumulative force that air molecules act on objects below them due to Earth's gravity. The higher the altitude, the less air molecules there are to act a force below them, and therefore, there's less air pressure at higher altitudes.

7 0
2 years ago
Read 2 more answers
The second-order rate constants for the reaction of oxygen atoms witharomatic hydrocarbons have been measured (R. Atkinson and J
Pepsi [2]

Answer:

A = 1,13x10¹⁰

Ea = 16,7 kJ/mol

Explanation:

Using Arrhenius law:

ln k = -Ea/R × 1/T + ln(A)

You can graph ln rate constant in x vs 1/T in y to obtain slope: -Ea/R and intercept is ln(A).

Using the values you will obtain:

y = -2006,9 x +23,147

As R = 8,314472x10⁻³ kJ/molK:

-Ea/8,314472x10⁻³ kJ/molK = -2006,9 K⁻¹

<em>Ea = 16,7 kJ/mol</em>

Pre-exponential factor is:

ln A = 23,147

A = e^23,147

<em>A = 1,13x10¹⁰</em>

<em></em>

I hope it helps!

6 0
2 years ago
What mass of CH3COOH is present in a 250 mL cup of 1.25 mol/L solution of vinegar?
aleksley [76]
If    1000 ml (1 L) of CH₃COOH contain 1.25 mol
let  250 ml  of CH₃COOH contain x

⇒  x =  \frac{250 ml * 1.25mol}{1000 ml}
        
        =  0.3125 mol

∴ moles of CH₃COOH in 250ml is 0.3125 mol

Now, Mass = mole  ×  molar mass
        
                   = 0.3125 mol  × [(12 × 2)+(16 × 2)+(1 × 4)] g/mol
 
                   = 18.75 g

∴ Mass of CH₃COOH present in a 250 mL cup of 1.25 mol/L solution of vinegar is <span>18.75 g</span>
4 0
3 years ago
Please help it’s due today
Romashka [77]
U literally put all yo information on their
8 0
3 years ago
Onsider the reaction below.
mamaluj [8]

Answer:

3.14 moles of hydrogen are produced

Explanation:

This is because for every 1 molesof hydrogen are produced 2 moles of oxygen are produced. So we take 6.28 divide it by 2 and we wend up with 3.14.

8 0
2 years ago
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