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3 years ago

Find the density of a substance that’s had a volume of 24.7mL and a mass of 49.1g. Give the answer with two decimals.

Chemistry

2 answers:

Fudgin [204]3 years ago

8 0

Answer:

1.99 g/mL

Explanation:

Use the formula d = m/v, where m is the mass and v is the volume.

Plug in the values:

d = 49.1/24.7

d = 1.99 g/mL

Gnesinka [82]3 years ago

8 0

Answer:

About 1.99 g/mL

Explanation:

The density of a substance can be found using the following formula:

d= m/v

where m is the mass and v is the volume.

The mass of this substance is 49.1 grams and the volume is 24.7 milliliters.

m= 49.1 g

v= 24.7 mL

Substitute these values into the formula.

d= 49.1 g / 24.7 mL

Divide 49.1 g by 24.7 mL

d= 1.98785425 g/mL

We are asked to round to two decimal places, or the hundredth place. The 8 in the thousandth place tells us to round the 8 to a 9 in the hundredth place.

d ≈ 1.99 g/mL

The density of the this substance is approximately 1.99 grams per milliliter.

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The enthalpy of solution (∆H) of KOH is -57.6 kJ/mol. If 3.66 g KOH is dissolved in enough water to make a 150.0 mL solution, wh

Wewaii [24]

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

$3.66g \times \frac{1mol}{56.11g} \times \frac{(-57.6kJ)}{mol} = -3.76 kJ$

According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.

$Qr+Qa = 0\\\\Qa = -Qr = 3.76 kJ$

150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

$150.0 mL \times \frac{1.02g}{mL} = 153 g$

Given the specific heat capacity of the solution (c) is 4.184 J/g･°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.

$Qa = c \times m \times \Delta T\\\\\Delta T = \frac{Qa}{c \times m} = \frac{3.76 \times 10^{3}J }{\frac{4.184J}{g.\° C } \times 153g} = 5.87 \° C$