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3 years ago

A solution of Cuso, is labelled 1.743 M. How much Cuso, in grams, must be used to make

Chemistry

2 answers:

zepelin [54]3 years ago

7 0

Answer:

Option A. 416.1 g.

Explanation:

Data obtained from the question include the following:

Molarity = 1.743 M

Volume = 1.4957 L

Mass of CuSO4 =..?

Next, we shall determine mole of CuSO4 present in the solution.

This is illustrated below:

Molarity is simply defined as the mole of solute per unit litre of the solution. Mathematically, it is expressed as:

Molarity = mole /Volume

With the above formula, we can obtain the mole CuSO4 present in the solution as follow:

Molarity = 1.743 M

Volume = 1.4957 L

Mole of CuSO4 =..?

Molarity = mole /Volume

1.743 = mole of CuSO4 /1.4957

Cross multiply

Mole of CuSO4 = 1.743 x 1.4957

Mole of CuSO4 = 2.607 moles

Finally, we shall convert 2.607 moles of CuSO4 to grams.

This can be obtained as follow:

Molar mass of CuSO4 = 63.55 + 32.06 + (16x4) = 159.61 g/m

Mole of CuSO4 = 2.607 moles

Mass of CuSO4 =..?

Mole = mass /molar mass

2.607 = mass of CuSO4 /159.61

Cross multiply

Mass of CuSO4 = 2.607 x 159.61

Mass of CuSO4 = 416.1 g

Therefore, 416.1 g of CuSO4 is needed to prepare the solution.

Aleks04 [339]3 years ago

3 0

A. 416.1 is your answer sir

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Animals breathe oxygen and release carbon dioxide during cellular respiration according to the equation below. How much oxygen i

kifflom [539]

Answer:

14.544 g of oxygen is needed to produce 120 grams of carbon dioxide.

Explanation:

Animals take in oxygen and breathe out carbon dioxide during cellular respiration. The reaction for the metabolism of the food in the animal body is:

$C_6H_{12}O_6 + O_2 \rightarrow 6CO_2 + 6H_2O + Energy$

As can be seen from the reaction stoichiometry that:

6 moles of carbon dioxide gas can be produced from 1 mole of oxygen gas in the process of metabolism of glucose.

Also,

Given :

Mass of carbon dioxide gas = 120 g

Molar mass of carbon dioxide gas = 44 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus, moles of carbon dioxide are:

$moles_{CO_2} = \frac{120 g}{44 g/mol}$

$moles_{CO_2} = 2.7273 mol$

As mentioned:

6 moles of carbon dioxide gas can be produced from 1 mole of oxygen gas in the process of metabolism of glucose.

1 mole of carbon dioxide gas can be produced from 1/6 mole of oxygen gas in the process of metabolism of glucose.

2.7273 mole of carbon dioxide gas can be produced from $\frac{1}{6} \times 2.7273$ moles of oxygen gas in the process of metabolism of glucose.

Thus, moles of oxygen gas needed = 0.4545 moles

Molar mass of oxygen gas = 32 g/mol

The mass of oxygen gas can be find out by using mole formula as:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Mass\ of\ oxygen\ gas = Moles \times Molar mass}$

$Mass\ of\ oxygen\ gas = 0.4545 \times 32}$

$Mass\ of\ oxygen\ gas = 14.544 g$

14.544 g of oxygen is needed to produce 120 grams of carbon dioxide.

3 0

3 years ago

Read 2 more answers

Three different samples were weighed using a different type of balance for each sample. The three were found to have masses of 0

Lena [83]

Answer:

6.096799125kg

Explanation:

According to the question, three different samples weighed using different types of balance had masses: 0.6160959 kg, 3.225 mg, and 5480.7 g.

Based on observation, the mass units in the three measurements are different but must be uniform in order to find the total mass. Hence, we need to convert to the standard unit (S.I unit of mass), which is kilograms (kg)

Since 1kg equals 1,000,000mg

Hence, 3.225mg will be 3.225/1000000

= 0.000003225kg

Also, 1kg equals 1000g

Hence, 5480.7g will be 5480.7/1000

= 5.4087kg

Hence, the total mass of the three samples (now in the same unit) are:

5.4807kg + 0.000003225kg + 0.6160959 kg

= 6.096799125kg

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3 years ago

Given a fixed amount of gas held at constant pressure, calculate the volume it would occupy if a 2.00 L sample were cooled from