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2 years ago

2 points

Chemistry

1 answer:

quester [9]2 years ago

6 0

(b) 12.7 L

Explanation:

Given:

Volume, V₁ = 12 L

Pressure, P₁ = 760 mm Hg (at STP)

Temperature, T₁ = 273 K (at STP)

Pressure, P₂ = 720 mm Hg

Temperature, T₂ = 0°C = 273 K

Volume, V₂ = ?

We know:

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

Substituting the value we get:

$\frac{760 X 12}{273} = \frac{720 X V_2}{273} \\\\V_2 = \frac{760 X 12}{720} \\\\V_2 = 12.7 L$

Therefore, V₂ is 12.7 L

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3 years ago

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A certain gas is present in a 10.0 LL cylinder at 4.0 atmatm pressure. If the pressure is increased to 8.0 atmatm the volume of

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Answer:

The gas obeys Boyle’s law and the value of $k_i\&k_f$ both are equal to 40.0 atm L.

Explanation:

Initial volume of the gas = $V_1=10.0 L$

Initial pressure of the gas = $P_1=4.0 atm$

Final volume of the gas = $V_2=5.0 L$

Final pressure of the gas = $P_2=8.0 atm$

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$PV=k$

The equation given by this law is:

$P_1V_1=P_2V_2$

$P_1\propto \frac{1}{V_1}$

$P_1V_1=k_i$

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The gas in the cylinder is obeying Boyle's law.

The gas obeys Boyle’s law and the value of $k_i\&k_f$ both are equal to 40.0 atm L.

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3 years ago

A 16.0 mL sample of a 1.04 M potassium sulfate solution is mixed with 14.3 mL of a 0.880 M barium nitrate solution and this prec

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Number of moles in the K2SO4 sample
= (16/1000)*1.04= 0.01664 mol

Number of moles in the Ba(NO3)2 sample
= (14.3/1000*0.880)= 0.01258 mol

Since the reaction is a 1:1 ratio between the two reactants, the limiting reagent is the one containing a smaller number of moles, namely Ba(NO3)2.

The molecular mass of BaSO4 is 137.3+(32.06+4*16.00)=233.4
Therefore the theoretical yield of Barium Sulphate is
233.4*0.01258=2.937 g
Actual yield = 2.60 g (given)
Therefore the percentage yield = 2.60/2.937=88.54%

Answer:
1. the limiting reagent is Barium Nitrate (Ba(NO3)2)
2. the theoretical yield is 2.94 g
3. the percentage yield is 88.5%

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2 years ago

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Answer: The particle released in the given reaction is one alpha particle

Explanation:

In a nuclear reaction, the total mass and total atomic number remains the same.

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To calculate A:

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