What is a mol in chemistry

Magnesium reacts with hydrochloric acid to produce magnesium chloride and hydrogen gas. Write a balanced chemical equation for t

Alona [7]

Answer:

The balanced chemical equation is Mg + 2HCl ⇒ MgCl2 + H2

The molarity of the hydrochloric acid solution is HCl 0.04 M and the pH = 1.4.

The volume of hydrogen gas produced by the reaction of 0.510 g of Mg will be 0.482 L.

Explanation:

First, for the balanced equation you have to consider the oxidation state of the elements to find subscripts. Then you can find the correct coeficients. Mg= +2, Cl = -1.

Mg + HCL ⇒ MgCl2 + H2

For the molarity of the solution you have to notice tha if 0.510 grams of Mg reacts with 0,5 L of hydroclhoric acid, and from the previous equation 1 mol of Mg reacts with 2 mol HCl.

The atomic mass of Mg = 24.31 grs/mol

24.31 grs------------ 1 mol Mg

0.510 grs------------ x=0.02 mol Mg.

If 1 mol of Mg reacts with 2 mol HCl, then 0.02 mol of Mg will react with

0.04 mol HCl. So, the molarity of the solution is 0.04 M HCl.

Then to calculate the pH we use the formula pH = - log [H+]

⇒ pH = -log [0.04]⇒ pH=1.4.

Finally, from the balanced equation and the findings described, and considering that at 25°C and 1.00 atm 1 mol of gas has volume of 24.1 L.

1 mol H2----------- 24.1 L

0.02 mol H2----- x= 0.482L.

3 0

3 years ago

Express the concentration of a 0.0320 M aqueous solution of fluoride, F−, in mass percentage and in parts per million (ppm). Ass

denis23 [38]

Answer:

607 ppm

Explanation:

In this case we can start with the ppm formula:

$ppm=\frac{mg~of~solute}{Litters~of~solution}$

If we have a solution of 0.0320 M, we can say that in 1 L we have 0.032 mol of $F^-$ , because the molarity formula is:

$M=\frac{mol}{L}$

In other words:

$0.0320~M=\frac{mol}{1~L}$

$mol=0.032~M*1~L=0.032~mol$

$1~L~of~Solution=0.0320~mol~of~solute$

If we use the atomic mass of $F$ (19 g/mol) we can convert from mol to g:

$0.0320~mol~F^-\frac{19~g~F^-}{1~mol~F^-}~=~0.607~g$

Now we can convert from g to mg (1 g= 1000 mg), so:

$0.607~g\frac{1000~mg}{1~g}=607~mg$

Finally we can divide by 1 L to find the ppm:

$ppm=\frac{607~mg}{1~L}=~607~ppm$

We will have a concentration of 607 ppm.

I hope it helps!

4 0

3 years ago

How many moles are in 297 g of NH3?

BigorU [14]

Answer:

1. 17.4 moles.

2. 1.13 moles

3. 315.5 moles

4. 390.6g

5. 1.13x10⁶ moles

6. 14.8 moles

7. 337 moles

8. 2.15x10²⁴ molecules

9. 3.13x10²⁴ atoms

10. 1.38x10²⁴ particles

11. 1517g

12. 455g of CaF₂

Explanation:

We can convert formula units to moles or vice versa using Avogadro's number and moles to grams using molar mass of the substance:

1. Molar mass NH3: 17.031g/mol

297g * (1mol / 17.031g) = 17.4 moles

2. Molar mass MgCO3: 84.3g/mol

95g * (1mol / 84.3g) = 1.13 moles

3. Using Avogadro's number (6.022x10²³ formula units / mol):

1.9x10²⁶FU * (1mol / 6.022x10²³FU) = 315.5 moles

4. Molar mass H2O: 18g/mol

21.7mol * (18g / mol) = 390.6g

5. Using Avogadro's number (6.022x10²³ molecules / mol):

6.78x10²⁹molecules * (1mol / 6.022x10²³FU) = 1.13x10⁶ moles

6. 8.9x10²⁴FU * (1mol / 6.022x10²³FU) = 14.8 moles

7. Using Avogadro's number (6.022x10²³ atoms / mol):

2.03x10²⁶atoms* (1mol / 6.022x10²³FU) = 337 moles

8. 3.569mol * (6.022x10²³ molecules / 1mol) = 2.15x10²⁴ molecules

9. 5.2mol * (6.022x10²³ atoms / 1mol) = 3.13x10²⁴ atoms

10. Molar mass Li₂SO₄: 109.94g/mol:

36g * (1mol / 109.94g) * (6.022x10²³ molecules / 1mol) * (7 particles / 1molecule) = 1.38x10²⁴ particles

Assuming particles are atoms and in 1 molecule of Li₂SO₄ you have 7 atoms.

11. Molar mass Cl₂: 70.9g/mol:

21.4mol * (70.9g / mol) = 1517g

12. Molar mass CaF₂: 78.07g/mol:

3.51x10²⁴FU * (1mol / 6.022x10²³FU) * (78.07g / mol) = 455g of CaF₂

8 0

3 years ago

What will the pressure of H+ ions in a solution cause

suter [353]

ACIDIC BEHAVIOR OF SOLUTION

6 0

3 years ago

NItrogen in air reacts at high temperature to form NO2 according to the reaction:

Rina8888 [55]

Answer:

The correct answer is option E.

Explanation:

Structures for the reactants and products are given in an aimage ;

Number of double bonds in oxygen gas molecule = 1

Number of double bonds in nitro dioxide gas molecule = 1

Number of single bond in in nitro dioxide gas molecule = 1

Number of triple bonds in nitrogen gas molecule = 1

$N_2+2O_2\rightarrow 2NO_2,\Delta H=?$

$\Delta H=[2 mol\times \Delta H_{f,NO_2}]-[1 mol\times \Delta H_{f,N_2}-2 mol\times \Delta H_{f,O_2}]$

$\Delta H_{f,NO_2}=33.18 kJ/mol$

$\Delta H_{f,N_2}=0$ (pure element)

$\Delta H_{f,O_2}=0$ (pure element )

$\Delta H=2 mol\times 33.18 kJ/mol=66.36kJ=15.86 kcal$

The enthalpy of the given reaction is 15.86 kcal.

6 0

3 years ago