How can you increase the energy of a glass of water?

7. How many formula units are equal to a 0.25 g sample of Chromium (III) sulfate,

ryzh [129]

Answer:

38.541 × 10¹⁹ formula units

Explanation:

Given data:

Mass of chromium sulfate = 0.25 g

Formula units in 0.25 g = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ formula units of water

Number of moles of chromium sulfate = Mass / molar mass

Number of moles of chromium sulfate = 0.25 g/ 392.16 g/ mol

Number of moles of chromium sulfate = 6.4 × 10⁻⁴ moles

Number of formula units:

1 mole = 6.022 × 10²³ formula units

6.4 × 10⁻⁴ moles × 6.022 × 10²³ formula units / 1 mol

38.541 × 10¹⁹ formula units

7 0

3 years ago

Some elements are liquids or gases.<br><br><br> True <br> False

gregori [183]

The answer to this ? is true

4 0

3 years ago

Which of these agreements has goals of peace and shared scientific research?

Jet001 [13]

Option C, Antarctic Treaty System, is the right answer.

The Antarctic Treaty System, control global connections with regard to Antarctica; the only continent of the earth without the native population of human beings. In other words, the ATS is the entire system of adjustments developed for the intention of managing associations among states in the Antarctic. The main goal of the ATS is to guarantee "in the affair of all humankind that Antarctica shall proceed always to be practised completely for peaceful objectives and shall not become the view or gadget of universal disharmony.

8 0

3 years ago

Read 2 more answers

Is particle size one of the factors that affect solubility? How?

Over [174]

Hello!

The reduction in particle size results in an increased rate of solution. It occur because is harder for a a solvent to surround bigger molecules.

Hugs!

4 0

4 years ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0

4 years ago