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elena-14-01-66 [18.8K]
3 years ago
11

294 g of potassium dichromate contains 52 g of chromium and 39 g of potassium. What is the mass percent of oxygen in the compoun

d?
Chemistry
1 answer:
xenn [34]3 years ago
5 0
The answer is 69.048%.

Mass of potassium dichromate (K₂Cr₂O₇) is 294 g, which is a 100%.

Given mass of chromium is 52 g. So, the mass percent of chromium in the compound is 17.687%:
52 g : x% = 294 g : 100%
x = 52 ÷ 294 × 100 %
x = 17.687%

Given mass of potassium is 39 g. <span>So, the mass percent of potassium in the compound is 13.265%:
39 g : x% = 294 g : 100%
x = </span>39 ÷ <span>294 × 100 %
x = 13.265%

Therefore, </span>the mass percent of oxygen in the compound is <span>69.048%:
100% - </span>17.687% - <span>13.265% = 69.048%.</span>
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A compound is found to be 30.45% n and 69.55 % o by mass. if 1.63 g of this compound occupy 389 ml at 0.00°c and 775 mm hg, what
Charra [1.4K]
1) mass composition

N: 30.45%
O: 69.55%
   -----------
   100.00%

2) molar composition

Divide each element by its atomic mass

N: 30.45 / 14.00 = 2.175 mol

O: 69.55 / 16.00 = 4.346875

4) Find the smallest molar proportion

Divide both by the smaller number

N: 2.175 / 2.175 = 1

O: 4.346875 / 2.175 = 1.999 = 2

5) Empirical formula: NO2

6) mass of the empirical formula

14.00 + 2 * 16.00 = 46.00 g

7) Find the number of moles of the gas using the equation pV = nRT

=> n = pV / RT = (775/760) atm * 0.389 l / (0.0821 atm*l /K*mol * 273.15K)

=> n = 0.01769 moles

8) Find molar mass

molar mass = mass in grams / number of moles = 1.63 g / 0.01769 mol = 92.14 g / mol

9) Find how many times the mass of the empirical formula is contained in the molar mass

92.14 / 46.00 = 2.00

10) Multiply the subscripts of the empirical formula by the number found in the previous step

=> N2O4

Answer: N2O4
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3 years ago
True or false all countries are legally bound to survey and monitor for disease within their bodies
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Answer:

False..........

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Glycerol (molar mass 92.09 g/mol) has been suggested for use as an alternative fuel. The enthalpy of combustion, Δ
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The mass of glycerol to that would need to be combusted to heat 500.0g of water from 20.0°C to 100.0°C is; 9.32 grams.

We must establish the fact that energy is neither created nor destroyed.

Therefore, the amount of heat absorbed by water is equal to the amount of heat released by the combustion of glycerol.

Total heat absorbed by water, H(water) is;

Q(water) = m C (T2 - T1)

Q(water) = 500 × 4.184 × (100-20)

Q(water) = 167.36 kJ

Consequently, the quantity of heat evolved by the combustion of glycerol is;

Q(glycerol) = 167,360 J = n × ΔH°comb

where, n = no. of moles of glycerol.

167.36 kJ= n × 1654 kJ/mole

n = 167.36/1654

n = 0.1012 moles of glycerol.

Therefore, mass of glycerol combusted, m is;

m = n × Molar mass

m = 0.1012 × 92.09

m = 9.32 g.

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brainly.com/question/20709115

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Carbon dioxide in the atmosphere dissolves in raindrops to produce carbonic acid (H2CO3), causing the pH of clean, unpolluted ra
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Answer:

The range of [H⁺] is from 2.51 x 10⁻⁶ M to 6.31 x 10⁻⁶ M,

Explanation:

To answer this problem we need to keep in mind the <u>definition of pH</u>:

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-5.2 = log [H⁺]

10^{-5.2} = [H⁺]

6.31 x 10⁻⁶ M = [H⁺]

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-5.6 = log [H⁺]

10^{-5.6} = [H⁺]

2.51 x 10⁻⁶ M = [H⁺]

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Atomic radii increase when going down a group and decreases when going towards the anion periods. So A and D.
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