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VARVARA [1.3K]
3 years ago
11

If 21 mL of gas is subjected to a temperature change from 10.0°C to 120°C and a pressure change from 1.0 atm to 15 atm, the new

volume is: (Be sure to use the correct number of significant figures.) 1.9 mL 1.0 mL 0.53 mL 440 mL
Chemistry
1 answer:
poizon [28]3 years ago
3 0

 The  new volume is  1.9  ml


<u><em> calculation</em></u>

The new   volume  is  calculated using the   combined gas  equation

that is  P1V1/T1 = P2V2/T2  where

P1= 1.0 atm

V1= 21   ml

T1 = 10.0 c  into kelvin = 10 + 273 = 283 k

P2 = 15 atm

V2=? Ml

T2 = 120 c  into kelvin = 120 +273 = 393 K


make V2  the subject of the   formula by  multiplying  both side  by T2/P2

V2=T2P1v1 / P2T1

V2=[(393 k  x 1.0 atm  x 21  ml) / (15 atm x 283 K)]  = 1.9  Ml


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When of a certain molecular compound X are dissolved in of dibenzyl ether , the freezing point of the solution is measured to be
kondaur [170]

This question is incomplete, the complete question is;

When 4.28 g of a certain molecular compound X are dissolved in 60.0 g of dibenzyl ether [(C₆H₅CH₂)₂0] , the freezing point of the solution is measured to be -3.2°C . Calculate the molar mass of X.

If you need any additional information on dibenzyl ether, use only what you find in the ALEKS Data resource. Also, be sure your answer has a unit symbol, and is rounded to significant digit.

Answer: molar mass of solute (X) is 88.03 g/mol

Explanation:

Given that;

mass of solute = 4.28 g

mass of solvent = 60.0 g = 0.060 kg        (Dibenzyl ether)

depression constant kf = 6.17 °CKg/mol

Freezing Point of solvent T₀ = 1.80°C       (Dibenzyl ether)

freezing point of solution Tsol = -3.20°C

Now we know that

Depression in freezing point ΔTf = depression constant kf × molaity m

and (ΔTf = T₀-Tsol)

so T₀ - Tsol = kf × m

we substitute

1.80 - (-3.20) = 6.17  × m

5 = 6.17 × m

m = 5 / 6.17

m = 0.8103 kg/mol

so molaity m = 0.8103 kg/mol

we know that

Molaity of solute m = (mass of solute / M.wt of solute) × ( 1 / mass of solvent in Kg)

solve for molar mass of solute

molar mass of solute =  (mass of solute / molaity) × ( 1 / mass of solvent in Kg)

now we substitute

molar mass = (4.28g / 0.8103 kg/mol) × (1 / 0.060kg)

molar mass = ( 5.2839 × 16.66 ) g/mol

molar mass = 88.0297 g/mol ≈ 88.03 g/mol

Therefore molar mass of solute (X) is 88.03 g/mol

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How many times higher is the concentration of H+ in the Hubbard Brook sample than in unpolluted rainwater?
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Answer:

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Answer: 10-7= 0.0000001 moles per liter

2. 5.6 (unpolluted rainwater)

Answer: 10-5.6 = 0.0000025 moles per liter

3. 3.7 (first acid rain sample in North America)

Answer: 10-3.7 = 0.00020 moles per liter

The concentration of H+ in the Hubbard Brook sample is 0.00020/0.0000025, which is 80 times higher than the H+ concentration in unpolluted rainwater.

Explanation:

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Mass defect for oxygen-16 = 0. 13261 amu How many kilograms does the mass defect represent? 1. 66 × 10-27 kg 2. 20 × 10 -28 kg 3
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Mass defect for oxygen-16 = 0. 13261 amu, in the kilograms the mass defect equals to 2.20 × 10⁻²⁸ kg.

<h3>What is mass defect?</h3>

Mass defect is the difference between the mass of of an whole atom and the combined mass of its individual particles present in that atom.

We know that, 1 amu = 1.6 × 10⁻²⁷ kg

Given that, mass defect for oxygen-16 = 0.13261 amu

To calculate this defect in terms of kilograms, we have to convert into kg unit as:

0.13261 amu = 0.13261 amu × 1.6 × 10⁻²⁷ kg/amu

0.13261 amu = 2.20 × 10⁻²⁸ kg

Hence option (2) is correct.

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