All categories

3 years ago

6 How many moles are in a 321 g sample of magnesium?

Chemistry

1 answer:

Tatiana [17]3 years ago

3 0

Answer:

13.4mol of Mg

Explanation:

Given parameters:

Mass of magnesium = 321g

Unknown:

Number of moles = ?

Solution:

The number of moles of a substance is given as;

Number of moles = $\frac{mass}{molar mass}$

Molar mass of Mg = 24g/mol

Insert the parameters and solve;

Number of moles = $\frac{321}{24}$ = 13.4mol of Mg

You might be interested in

What is the mole ratio of NH3 to N2?

Ksenya-84 [330]

To answer this question a balanced chemical equation is necessary. The correct equation is: N2 + 3H2 = 2NH3
From this equation, one mole of nitrogen react with 3 moles of hydrogen to give 2 moles of ammonia.
Therefore, the mole ratio of NH3 to N2 is 2:1

8 0

3 years ago

If you put hydrogen (H1+) and Hydroxide (OH1) together what it’s it called?

Tamiku [17]

Answer:

The hydroxyl ions (OH-) released will combine with any hydrogen ions (H+) in the solution to form water molecules (OH- + H+ = H2O).

Explanation:

3 0

3 years ago

Iridium has two naturally occurring isotopes, Iridium-191 and Iridium-193. If the average atomic mass of orodum s 192.217, what

Lina20 [59]

Answer:

The percent isotopic abundance of Ir-193 is 60.85 %

The percent isotopic abundance of Ir-191 is 39.15 %

Explanation:

we know there are two naturally occurring isotopes of iridium, Ir-191 and Ir-193

First of all we will set the fraction for both isotopes

X for the isotopes having mass 193

1-x for isotopes having mass 191

The average atomic mass is 192.217

we will use the following equation,

193x + 191(1-x) = 192.217

193x + 191 - 191x = 192.217

193x- 191x = 192.217 - 191

2x = 1.217

x= 1.217/2

x= 0.6085

0.6085 × 100 = 60.85 %

60.85% is abundance of Ir-193 because we solve the fraction x.

now we will calculate the abundance of Ir-191.

(1-x)

1-0.6085 =0.3915

0.3915× 100= 39.15 %

6 0

3 years ago

A 50.0 mL sample of a 1.00 M solution of CuSO4 is mixed with 50.0 mL of 2.00 M KOH in a calorimeter. The temperature of both sol

Reika [66]

Answer : The enthalpy change for the process is 52.5 kJ/mole.

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the solution

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the reaction

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the solution

$c_1$ = specific heat of calorimeter = $12.1J/^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_2$ = mass of water or solution = $Density\times Volume=1/mL\times 100.0mL=100.0g$

$\Delta T$ = change in temperature = $T_2-T_1=(26.3-20.2)^oC=6.1^oC$

Now put all the given values in the above formula, we get:

$q=[(12.1J/^oC\times 6.1^oC)+(100.0g\times 4.18J/g^oC\times 6.1^oC)]$

$q=2623.61J$

Now we have to calculate the enthalpy change for the process.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat released = 2626.61 J

n = number of moles of copper sulfate used = $Concentration\times Volume=1M\times 0.050L=0.050mole$

$\Delta H=\frac{2623.61J}{0.050mole}=52472.2J/mole=52.5kJ/mole$

Therefore, the enthalpy change for the process is 52.5 kJ/mole.

8 0

3 years ago

What is Potassium Oxalate name formula?

melisa1 [442]

Answer:

formula pottasium Oxalate C2k204

Explanation:

C2k204

6 0

3 years ago