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3 years ago

How many grams of H2SO4 are needed to prepare 500. mL of a .250M solution?

Chemistry

1 answer:

zavuch27 [327]3 years ago

7 0

Answer:

We need 12.26 grams H2SO4

Explanation:

Step 1: Data given

Volume of a H2SO4 solution = 500 mL = 0.500 L

Concentration of the H2SO4 solution = 0.250 M

Molar mass of H2SO4 = 98.08 g/mol

Step 2: Calculate moles H2SO4

Moles H2SO4 = concentration * volume

Moles H2SO4 = 0.250 M * 0.500 L

Moles H2SO4 = 0.125 moles

Step 3: Calculate mass of H2SO4

Mass of H2SO4 = moles * molar mass

Mass of H2SO4 = 0.125 moles * 98.08 g/mol

Mass of H2SO4 = 12.26 grams

We need 12.26 grams H2SO4

You might be interested in

How many molecules are in 100 g of C6H120,?*

Grace [21]

Answer:

3.37 × 10²³ molecules

Explanation:

Given data:

Mass of C₆H₁₂O₆ = 100 g

Number of molecules = ?

Solution:

Number of moles of C₆H₁₂O₆:

Number of moles = mass/molar mass

Number of moles = 100 g/ 180.16 g/mol

Number of moles = 0.56 mol

Number of molecules:

1 mole contain 6.022 × 10²³ molecules

0.56 mol × 6.022 × 10²³ molecules /1 mol

3.37 × 10²³ molecules

6 0

3 years ago

Liquid nitrogen boils at –195.8°C. Express the boiling point of liquid nitrogen in kelvin

inessss [21]

Answer:

$\boxed {\boxed {\sf 77.35 \ K}}$

Explanation:

The Celsius and Kelvin scales are used to measure the temperature of matter. Their scales and unit differences are the same (1 K increase = 1 °C increase), but they have different starting points.

So, the conversion is quite simple and only requires addition because of the different starting points. The formula is:

$T_K = T_C+ 273.15$

The boiling point of liquid nitrogen is -195.8 °C. We can substitute this value into the formula.

$T_K= -195.8 + 273.15$

$T_K= 77.35 K$

The boiling point of liquid nitrogen is 77.35 Kelvin.

4 0

2 years ago

Please help !!

ipn [44]

Answer:

For 2: The % yield of the product is 92.34 %

For 3: 12.208 L of carbon dioxide will be formed.

Explanation:

For 2:

The percent yield of a reaction is calculated by using an equation:

$\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100$ ......(1)

Given values:

Actual value of the product = 78.4 g

Theoretical value of the product = 84.9 g

Plugging values in equation 1:

$\% \text{yield}=\frac{78.4 g}{84.9g}\times 100\\\\\% \text{yield}=92.34\%$

Hence, the % yield of the product is 92.34 %

For 3:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(2)

Given mass of carbon dioxide = 24 g

Molar mass of carbon dioxide = 44 g/mol

Plugging values in equation 1:

$\text{Moles of carbon dioxide}=\frac{24g}{44g/mol}=0.545 mol$

At STP conditions:

1 mole of a gas occupies 22.4 L of volume

So, 0.545 moles of carbon dioxide will occupy = $\frac{22.4L}{1mol}\times 0.545mol=12.208L$ of volume

Hence, 12.208 L of carbon dioxide will be formed.

5 0

3 years ago

Identify the true statements about colloids.

Mashutka [201]

Answer : The true statements are:

(a) Emulsions are a type of colloid

(b) The particles of a colloid are larger than the particles of a solution

(d) Many colloids scatter light (tyndal effect)

Explanation :

Colloid : It is defined as the solution in which the one substance is insoluble in another solution that means the insoluble substance rotating in the solution.

The particles of a colloid are larger than the particles of a solution.

Colloid do not separate on standing.

Cannot be separated by filtration.

Scatter light (Tyndall effect).

For example :

Milk is considered as a colloid because various substances (fats, proteins etc..) are present in milk which are suspended in a solution.

Suspension : It is a heterogeneous mixture in which some of the particles are settle down in the mixture on standing or over time.

The particles in a suspension are far larger than those of a solution.

Emulsion : It is a mixture of two or more liquids that are normally immiscible.

Emulsions are a type of colloid.

6 0

3 years ago

Calculate the payback time if she insulates the loft with 300 mm insulation when: area of loft space = 100m^2 cost of roll of 10

ycow [4]

Answer: Payback time = 0.0075

Explanation: Since payback time is calculated as:

payback time = $\frac{installation cost}{annual savings}$

First determine the installation cost:

100 mm thick insulation covers 8.3 m². Then 300 mm covers 24.9 m².

To cover 8.3m² costs £20. Then, the cost to cover 24.9 m² is:

cost = $\frac{20*24.9}{8.3}$

cost = £60

The cost of putting the insulation is £120, so the total cost is:

total cost = £60 + £120

total cost = £180

Savings per year per 100 mm thick is £80. For 300 mm, the value of annual savings is:

savings = $\frac{300*80}{100}$

savings = 240

payback time = $\frac{installation cost}{annual savings}$

payback time = $\frac{180}{240}$

payback time = 0.75

8 0

3 years ago