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2 years ago

The equation below shows the decomposition of lead nitrate. How many molecules of PbO are produced when 11.5 g NO2 is formed?

Chemistry

1 answer:

Nadusha1986 [10]2 years ago

8 0

Answer is in the file below

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Determine the molarity for each of the following solution solutions:

____ [38]

Answer :

(a)The molarity of KCl solution is, 0.9713 mole/L

(b)The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

(c)The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

(d)The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

(e)The molarity of $Br_2$ solution is, 0.0565 mole/L

(f)The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

Explanation :

(a) 1.457 mol of KCl in 1.500 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Solute is KCl.

$\text{Molarity of the solution}=\frac{1.457mole}{1.500L}=0.9713mole/L$

The molarity of KCl solution is, 0.9713 mole/L

(b) 0.515 gram of $H_2SO_4$ , in 1.00 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Solute is $H_2SO_4$

Molar mass of $H_2SO_4$ = 98 g/mole

$\text{Molarity of the solution}=\frac{0.515g}{98g/mole\times 1.00L}=0.00525mole/L$

The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

(c) 20.54 g of $Al(NO_3)_3$ in 1575 mL of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Solute is $Al(NO_3)_3$

Molar mass of $Al(NO_3)_3$ = 213 g/mole

$\text{Molarity of the solution}=\frac{20.54g\times 1000}{213g/mole\times 1575L}=0.0612mole/L$

The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

(d) 2.76 kg of $CuSO_4.5H_2O$ in 1.45 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Solute is $CuSO_4.5H_2O$

Molar mass of $CuSO_4.5H_2O$ = 250 g/mole

$\text{Molarity of the solution}=\frac{2760g}{250g/mole\times 1.45L}=7.61mole/L$

The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

(e) 0.005653 mol of $Br_2$ in 10.00 ml of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Solute is $Br_2$ .

$\text{Molarity of the solution}=\frac{0.005653mole\times 1000}{10.00L}=0.0565mole/L$

The molarity of $Br_2$ solution is, 0.0565 mole/L

(f) 0.000889 g of glycine, $C_2H_5NO_2$ , in 1.05 mL of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Solute is $C_2H_5NO_2$

Molar mass of $C_2H_5NO_2$ = 75 g/mole

$\text{Molarity of the solution}=\frac{0.000889g\times 1000}{75g/mole\times 1.05L}=0.0113mole/L$

The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

5 0

3 years ago

¿Cuál es la molaridad de 275 ml de solución, que se preparó disolviendo 42 g de hidroxido de sodio NaOH en agua?

Dovator [93]

CEYSDINO IS A FILE SPAMMER SHE OR HE USES DIS FILE TO GET YOUR INFORMATION...

8 0

3 years ago

In clinical applications, the unit parts per million (ppm) is used to express very small concentrations of solute, where 1 ppm i

Norma-Jean [14]

Answer: 25 μg of calcium in a total volume of 57 mL to ppm is 0.4386ppm

Explanation:

Please see the attachments below

3 0

2 years ago

g What is the molarity of hydrochloric acid if 40.95 mL of HCl is required to neutralize 0.550 g of sodium oxalate, Na2C2O4

kykrilka [37]

Answer:

0.0002 M

Explanation:

The molarity of the HCl required would be 0.0002 M.

First, let us consider the balanced equation of the reaction:

$Na_2C_2O_4 + 2HCl = 2NaCl + H_2 + 2CO_2$

Stoichiometrically, 1 mole of $Na_2C_2O_4$ reacts with 2 moles of $HCl$ for a complete neutralization reaction.

Recall that: mole = $\frac{mass}{molar mass}$

Mole of 0.550 g sodium oxalate = 0.550/134 = 0.0041 mole

If 1 mole $Na_2C_2O_4$ requires 2 moles HCl, then 0.0041 mole will require:

0.0041 x 2 = 0.0082 mole HCl

Volume of the HCl = 40.95 L

Molarity = mole/volume

Hence, molarity of the HCl = 0.0082/40.95 = 0.0002 M

7 0

3 years ago

If 3.45 grams of Mg are burned how many grams of MgO are produced

coldgirl [10]

The number of grams of Mgo that are produced is calculated as follows
calculate the moles of Mg used

that is moles =mass/molar mass
= 3.45 g/ 24.305g/mol = 0.142 moles

write the equation for reaction
that is 2 Mg + O2 = 2 MgO

by use of mole ratio between Mg to Mgo which is 2:2 this implies that the moles of Mgo is also 0.142 moles

mass of Mgo= moles x molar mass of MgO
molar mass of Mgo = 24.305 + ( 3x15.999) = 72.302 g/mol

mass= 0.142 mol x 72.302 g/mol= 10.27 grams

4 0

2 years ago

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