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erastovalidia [21]
3 years ago
10

A pump contains 0.5 L of air at 203kPa. You draw back on the piston of the pump until the pressure reads 25.4kPa. What is the vo

lume of air in the pump?
a. 0.2L
b. 1.0L
c. 4.0L
d. 2.0L
Chemistry
1 answer:
densk [106]3 years ago
6 0
<h3><u> Answer</u>;</h3>

= 4.0 L

<h3><u>Explanation;</u></h3>

Boyle's law states that the volume of a fixed mass of a gas is inversely proportional to pressure at a constant temperature.

Therefore; <em>Volume α 1/pressure</em>

<em>Mathematically; V α 1/P</em>

<em>V = kP, where k is a constant;</em>

<em>P1V1 = P2V2</em>

<em>V1 = 0.5 l, P1 =203 kPa, P2 = 25.4 kPa</em>

<em>V2 = (0.5 × 203 )/25.4 </em>

<em>     = 3.996 </em>

<em>    ≈ </em><em><u>4.0 L</u></em>

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Which of the following reagents is/are likely to be used to form the Britton–Robinson buffer solution used in the experiment?
Ierofanga [76]

Answer:

I. A polyprotic, weak acid

II. Na2HPO4

Explanation:

Buffer solutions are those that, upon the addition of an acid or base, are capable of reacting by opposing the part of the basic or acid component to keep the pH fixed.

Buffers consist of hydrolytically active salts that dissolve in water. The ions of these salts are combined with acids and alkalis. These hydrolytically active salts are the products that result from the reaction between weak acids and strong alkalis such as calcium carbonate (from carbonic acid and calcium hydroxide) or between strong acids and weak alkalis such as ammonium chloride (a from hydrochloric acid and ammonium hydroxide).

A buffer acid reacts when a weak acid or weak base is combined with its corresponding hydrolytic salt in a water solution, a buffer system called a buffer is formed. As in this case a weak polyrotic acid with Na2HPO4, which allows the solution to be maintained at a pH of 3.8 against small aggregate amounts of both acid and base, thus favoring the reaction at a pH of 3.8

A buffer system is not always appropriate, because the ions of some hydrolytic salts can, for example, damage organisms that come into contact with it.

4 0
3 years ago
For a particular isomer of C 8 H 18 , the combustion reaction produces 5113.3 kJ of heat per mole of C 8 H 18 ( g ) consumed, un
ale4655 [162]

Answer:

ΔH°f(C₈H₁₈(g)) = -210.9 kJ/mol

Explanation:

Let's consider the combustion of C₈H₁₈.

C₈H₁₈(g) + 25/2 O₂(g) ⟶ 8 CO₂(g) + 9 H₂O(g) ΔH°rxn = − 5113.3 kJ

We can calculate the standard enthalpy of formation of C₈H₁₈(g) using the following expression.

ΔH°rxn = 8 mol × ΔH°f(CO₂(g)) + 9 mol × ΔH°f(H₂O(g)) - 1 mol × ΔH°f(C₈H₁₈(g)) - 25/2 mol × ΔH°f(O₂(g))

1 mol × ΔH°f(C₈H₁₈(g)) = 8 mol × ΔH°f(CO₂(g)) + 9 mol × ΔH°f(H₂O(g)) - 25/2 mol × ΔH°f(O₂(g)) - ΔH°rxn

1 mol × ΔH°f(C₈H₁₈(g)) = 8 mol × (-393.5 kJ/mol) + 9 mol × (-241.8 kJ/mol) - 25/2 mol × 0 kJ/mol - (− 5113.3 kJ)

ΔH°f(C₈H₁₈(g)) = -210.9 kJ/mol

5 0
3 years ago
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galben [10]
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3 years ago
Equal volumes of hydrogen and helium gas are at the same pressure. The atomic mass of helium is four times that of hydrogen. If
OverLord2011 [107]

Answer:

The ratio of the temperature of helium to that of hydrogen gas is 2:1.

Explanation:

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Temperature of hydrogen gas =T

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Mass of the hydrogen gas = m

Moles of the hydrogen gas = n=\frac{m}{2M}

Volume of the hydrogen gas = V

Using an ideal gas equation:

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Mass of the helium gas = m' =m

Moles of the helium  gas = n'=\frac{m}{M'}=\frac{m}{4M}

Volume of the helium gas = V' = V

Using an ideal gas equation:

P'V'=n'RT'=\frac{mRT'}{4M}...(2)

Divide (2) by (1)

\frac{P'V'}{PV}=\frac{\frac{mRT'}{4M}}{\frac{mRT}{2M}}

\frac{T'}{T}=\frac{2}{1}

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Answer:

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