Calculate the mass of 9.75x1021 atoms of helium.

How many moles of \ce{Fe2O3}FeX 2 OX 3 will be produced from 27.0 \text{ g}27.0 g27, point, 0, start text, space, g, end tex

ira [324]

Answer : The number of moles of $Fe_2O_3$ produced will be, 0.241 moles.

Solution : Given,

Mass of Fe = 27.0 g

Molar mass of $Fe$ = 56 g/mole

First we have to calculate the moles of $Fe$ .

$\text{ Moles of }Fe=\frac{\text{ Mass of }Fe}{\text{ Molar mass of }Fe}=\frac{27.0g}{56g/mole}=0.482moles$

Now we have to calculate the moles of $Fe_2O_3$

The balanced chemical reaction is,

$4Fe+3O_2\rightarrow 2Fe_2O_3$

From the reaction, we conclude that

As, 4 mole of $Fe$ react to give 2 mole of $Fe_2O_3$

So, 0.482 moles of $Fe$ react to give $\frac{0.482}{4}\times 2=0.241$ moles of $Fe_2O_3$

Thus, the number of moles of $Fe_2O_3$ produced will be, 0.241 moles.

6 0

2 years ago

Which statement accurately describes the illustration?

jasenka [17]

I'm not sure but I think it is A. image one has the most spread out particles like a gas, and b has closer together particles like a liquid or solid. since there are no choices that say A=gass and B=solid, so I am guessing it is answer A.

4 0

3 years ago

Read 2 more answers

In the space provided, for each element, type the ionic charge of an atom with a full set of valence electrons. Then, type the n

kobusy [5.1K]

11. ionic charge +1, helium.

12. ionic charge 2-, neon.

13. ionic charge 3+, neon.

3 0

3 years ago

The mole fraction of nitrogen in the air is 0.7808. this means that 78.08% of the molecules in the air are nitrogen. when the at

Pachacha [2.7K]

Answer:

$p = \boxed{\text{593 torr}}$

Explanation:

For this question, we must use Dalton's Law of Partial Pressures:

The partial pressure of a gas in a mixture of gases equals its mole fraction times the total pressure:

$p = \chi p_{\text{tot}}$

Data:

χ = 0.7808

$p_{\text{tot}} = \text{ 760 torr}$

Calculation:

$p = 0.7808 \times \text{ 760 torr}\\\\p= \boxed{\textbf{593 torr}}$

7 0

2 years ago

A sample of Manganese (II) chloride has a mass of 19.8 grams before heating, and 12.6 grams after heating until all the water is

IRINA_888 [86]

Answer:

no. of water molecules associated to each molecule of $MnCl_2$ = 4

Explanation:

Mass of $MnCl_2$ before heating = 19.8 g

Mass of $MnCl_2$ after heating = 12.6 g

Difference in mass of $MnCl_2$ before and after heating

= 19.8 - 12.6 = 7.2 g

Difference in mass corresponds to mass of water driven out.

Molar mass of water = 18 g/mol

No. of moles of water = $\frac{7.2}{18} = 0.4\ mol$

Mass of $MnCl_2$ obtained after heating is mass of anhydrous $MnCl_2$ .

Mass of anhydrous $MnCl_2$ = 12.6 g

Molar mass of $MnCl_2$ = 125.9 g/mol

No. of mol of anhydrous $MnCl_2$ = $\frac{125.9}{125.9} = 0.1\ mol$

so,

0.1 mol of $MnCl_2$ have 0.4 mol of water

1 mol of $MnCl_2$ will have = $\frac{0.4}{0.1} =4\ mol$

Hence, no. of water molecules associated to each molecule of $MnCl_2$ = 4

5 0

3 years ago