Anyone good at chemistry ? Maybe you can help me out?

A 5.00-cm cube of magnesium has a mass of 217.501 g. What is the density of magnesiummetal?

Doss [256]

Answer:

d = 43.5 g/cm³

Explanation:

Given data:

Mass of magnesium cube = 217.501 g

Volume of magnesium cube = 5.00 cm³

Density of magnesium cube = ?

Solution:

Formula:

d = m/v

d = density

m = mass

v = volume

by putting values,

d = 217.501 g/ 5.00 cm³

d = 43.5 g/cm³

8 0

3 years ago

Suppose you are working with a NaOH stock solution but you need a solution with a lower concentration for your experiment. Calcu

Monica [59]

Answer: The volume of the 1.224 M NaOH solution needed is 26.16 mL

Explanation:

In order to prepare the dilute NaOH solution, solvent is added to a given amount of the NaOH stock solution up to a final volume of 250.0 mL.

Since only solvent is added, the amount of the solute, NaOH, in the dilute solution is the same as in the volume taken from the stock solution.

Molarity (M) is calculated from the following equation:

M = n ÷ V

where n is the number of moles of the solute in the solution, and V is the volume of the solution.

Accordingly, the number of moles of the solute is given by

n = M x V

Now, let's designate the stock NaOH solution and the dilute solution as (1) and (2), respectively . The number of moles of NaOH in each of these solutions is:

n (1) = M (1) x V (1)

n (2) = M (2) x V (2)

As the amount of NaOH in the dilute solution is the same as in the volume taken from the stock solution,

n (1) = n (2)

and

M (1) x V (1) = M (2) x V (2)

For the stock solution, M (1) = 1.244 M, and V (1) is the volume needed. For the dilute solution, M (2) = 0,1281 M, and V (2) = 250.0 mL.

The volume of the stock solution needed, V (1), is calculated as follows:

V (1) = M (2) x V (2) ÷ M (1)

V (1) = 0.1281 M x 250.0 mL ÷ 1.224 M

V (1) = 26.16 mL

The volume of the 1.224 M NaOH solution needed is 26.16 mL.

7 0

3 years ago

What is propsed as evidence that supports the Big Bang Theory?

Stels [109]

The answer to your question is Hubble’s law

4 0

3 years ago

Which statement describes the phase change that occurs when dry ice is placed in an open container at room temperature?

zloy xaker [14]

Answer:

Dry ice undergoes sublimation, an endothermic change at room temperature.

7 0

3 years ago

Read 2 more answers

A chemist uses hot hydrogen gas to convert chromium(III) oxide to

Darya [45]

Answer: 3.024 g grams of hydrogen are needed to convert 76 grams of chromium(III) oxide, $Cr_{2}O_{3}$

Explanation:

The reaction equation for given reaction is as follows.

$Cr_{2}O_{3} + 3H_{2} \rightarrow 2Cr + 3H_{2}O$

Here, 1 mole of $Cr_{2}O_{3}$ reacts with 3 moles of $H_{2}$ .

As mass of chromium (III) oxide is given as 76 g and molar mass of chromium (III) oxide $(Cr_{2}O_{3})$ is 152 g/mol.

Number of moles is the mass of substance divided by its molar mass. So, moles of $Cr_{2}O_{3}$ is calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{76 g}{152 g/mol}\\= 0.5 mol$

Now, moles of $H_{2}$ .given by 0.5 mol of $Cr_{2}O_{3}$ is calculated as follows.

$0.5 mol Cr_{2}O_{3} \times \frac{3 mol H_{2}}{1 mol Cr_{2}O_{3}}\\= 1.5 mol H_{2}$

As molar mass of $H_{2}$ is 2.016 g/mol. Therefore, mass of $H_{2}$ is calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\1.5 mol = \frac{mass}{2.016 g/mol}\\mass = 3.024 g$

Thus, we can conclude that 3.024 g grams of hydrogen are needed to convert 76 grams of chromium(III) oxide, $Cr_{2}O_{3}$ .

7 0

3 years ago