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vaieri [72.5K]
3 years ago
9

When the volume of a gas is

Chemistry
1 answer:
Montano1993 [528]3 years ago
3 0

Answer:

\boxed {\boxed {\sf 232.9 \textdegree C}}

Explanation:

This question asks us to find the temperature change given a volume change. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula is:

\frac {V_1}{T_1}= \frac{V_2}{T_2}

The volume of the gas starts at 250 milliliters and the temperature is 137 °C.

\frac{250 \ mL}{137 \textdegree C}= \frac{V_2}{T_2}

The volume of the gas is increased to 425 milliliters, but the temperature is unknown.

\frac{250 \ mL}{137 \textdegree C}= \frac{425 \ mL}{T_2}

We are solving for the new temperature, so we must isolate the variable T₂. First, cross multiply. Multiply the first numerator and second denominator, then multiply the first denominator and second numerator.

250 \ mL * T_2 = 137 \textdegree C * 425 \ mL

Now the variable is being multiplied by 250 milliliters. The inverse of multiplication is division. Divide both sides of the equation by 250 mL.

\frac{250 \ mL * T_2}{250 \ mL}=\frac{ 137 \textdegree C * 425 \ mL}{250 \ mL}

T_2=\frac{ 137 \textdegree C * 425 \ mL}{250 \ mL}

The units of milliliters (mL) cancel.

T_2=\frac{ 137 \textdegree C * 425 }{250 }

T_2= \frac{58225}{250} \textdegree C

T_2=232.9 \textdegree C

The temperature changes to <u>232.9 degrees Celsius.</u>

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The molarity of the acid given the data from the question is 0.30 M

<h3>Balanced equation </h3>

2HNO₃ + Ba(OH)₂ —> Ba(NO₃)₂ + 2H₂O

From the balanced equation above,

  • The mole ratio of the acid, HNO₃ (nA) = 2
  • The mole ratio of the base, Ba(NO₃)₂ (nB) = 1

<h3>How to determine the molarity of the acid</h3>

From the question given above, the following data were obtained:

  • Volume of acid, HNO₃ (Va) = 39.7 mL
  • Volume of base, Ba(NO₃)₂ (Vb) = 24 mL
  • Molarity of base, Ba(NO₃)₂ (Cb) = 0.250 M
  • Molarity of acid, HNO₃ (Ma) =?

MaVa / MbVb = nA / nB

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Divide both side by 39.7

Ma = 12 / 39.7

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