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olganol [36]
3 years ago
9

The dimensions of a cupboard are 1.31 m wide

Chemistry
2 answers:
Mnenie [13.5K]3 years ago
4 0

Answer:

The volume of cupboard is 2.0043 m³.

Explanation:

Given data:

width of cupboard = 1.31 m

length of cupboard = 0.9 m

height of cupboard = 1.70 m

Volume = ?

Solution:

Volume = length × width × height

Volume = 0.9 m × 1.31 m × 1.70 m

Volume = 2.0043 m³

The volume of cupboard is 2.0043 m³.

melomori [17]3 years ago
3 0

Answer: 2.0

Explanation:

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7 0
3 years ago
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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.

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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

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.

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7 0
2 years ago
In a 100 g sample of iron(III) oxide, how many grams of iron are present?
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In order to calculate the mass of iron in a given sample of iron(III) oxide, we must first know the mass percentage of iron in iron(III) oxide. This is calculated by:
[mass of iron in one mole of iron(III) oxide/ mass of one mole of iron(III) oxide] * 100 
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= [(2 * 56) / (2 * 56 + 3 * 16)] * 100
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