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Step2247 [10]
2 years ago
10

Find the percent composition of OXYGEN in Manganese (III) nitrate, Mn(NO3)3.

Chemistry
1 answer:
Minchanka [31]2 years ago
6 0

Answer:

Mn(NO3)3 Molar mass = (54.93) + (3 x 14) + (9 x 16) = 240.93g/mol

Total number of Oxygen atoms in formula = 9

Total mass of oxygen atoms theoretically in the formula = 9x16 = 144g/mol

Oxygen's percentage composition:

(144/240.93) x 100 = 59.7% of the compound is oxygen

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Cual es la contribucion importante de los alquimistas
Lelechka [254]

Answer:

Hello

Explanation:

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3 0
3 years ago
How many different bases of DNA​
const2013 [10]

Answer:

four

Explanation:

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4 0
3 years ago
The element californium (cf) sells for $1000 per µg. assuming 6.02 x 1023 atoms of cf have a mass of 251 grams, how many atoms o
saveliy_v [14]

Answer:- 2.40*10^1^0atoms

Solution:- It is a simple unit conversion problem. We could solve this using dimensional analysis.

We know that, 1 US dollar = 100 cents

1 cent  = 1 US penny

So, 1 US dollar = 100 US pennies

1g=10^6\mu g

Let's make the set up starting with 1 penny as:

1penny(\frac{$1}{100pennies})(\frac{1\mu g}{$1000})(\frac{1g}{10^6\mu g})(\frac{6.02*10^2^3atoms}{251g})

= 2.40*10^1^0atoms

Therefore, we can bye 2.40*10^1^0atoms of Cf in one US penny.

3 0
3 years ago
1. A gas sample at a pressure of 5.00 atm has a volume of 3.00 L. If the gas pressure is changed to 760 mm Hg, what volume will
tatyana61 [14]

Answer:

V₂ =  15.00 atm

Explanation:

Given data:

Initial pressure = 5.00 atm

Initial volume = 3.00 L

Final pressure = 760 mmHg ( 760/760 = 1 atm)

Final volume = ?

Solution:

P₁V₁ = P₂V₂

V₂ = P₁V₁ /  P₂

V₂ =  5.00 atm × 3.00 L / 1 atm

V₂ =  15.00 atm

8 0
3 years ago
Calculate the solubility of carbon dioxide at 400 kPa.
BaLLatris [955]

The solubility of carbon dioxide at 400 kPa  at room temperature is ;

( B ) 0.61 CO2/L

<u>Given data </u>

pressure of CO₂ = 400 Kpa = 3.95 atm

Kh of CO₂ = 3.3 * 10⁻² mol/L.atm

<h3>Calculate the solubility of carbon dioxide </h3>

Solubility = pressure * Kh value of CO₂

                = 3.95 atm * 3.3 * 10⁻² mol / L.atm

                = 0.13 mol/l  CO₂

                = 0.61 CO₂ / L

Hence we can conclude that the solubility of CO₂ at 400 kPa is 0.13 mol/l  CO₂.

Learn more about solubility : brainly.com/question/23946616

From the options the closest answer is ( B ) 0.61 CO₂ / L

7 0
2 years ago
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