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Mama L [17]
3 years ago
15

4. Look at the following chemical reaction and determine what is true about Lithium (Li) and

Chemistry
1 answer:
stich3 [128]3 years ago
5 0
All shown on photo including oxidation states

You might be interested in
Determina el pH del café cuya concentración de H+ es de 0.00001M
lara31 [8.8K]

Answer:

5

Explanation:

Given parameters:

Hydrogen ion concentration  =  0.00001M

Unknown:

pH of the solution =?

Solution:

The pH is used to estimate the degree of acidity or alkalinity of a solution. To solve for pH of any solution, we use the expression below;

          pH  = -log [H⁺]

[H⁺] is the hydrogen ion concentration

        pH  = -log (1 x 10⁻⁵)

      pH = -(-5) = 5

4 0
3 years ago
5. If 1 g of a gas occupies a volume of 300 mL at STP, what is the molecular weight of the gas?
ikadub [295]

Answer:

74,67 gr/mol

Explanation:

At STP 1 mole of an ideal gas has volume of 22,4 L. Since we know the volume of the gas we can find the number of moles of the gas. (300 mL=0,3 L)

n=0,3L/22,4 L=0,01339 mol

Since we know weight of the gas as 1 g, we can find the molecular weight as;

MW=1 g/0,01339 mol =74,67 gr/mol

3 0
3 years ago
Oxygen gas generated in the thermal decomposition of potassium chlorate is collected over water. At 24ºC and an atmospheric pres
Yanka [14]

Answer:

its 0.163 g

Explanation:

From the total pressure and the vapour pressure of water we can calculate the partial pressure of O2​

PO 2 =P t −P H 2 O

​= 760 − 22.4

=  737.6 mmHg

From the ideal gas equation we write.

W= RT/PVM = (0.0821Latm/Kmol)(273+24)K(0.974atm)(0.128L)(32.0g/mol/) =0.163g

6 0
2 years ago
a 25 ml volume of a sodium hydroxide solution requires 19.6 mL of a 0.189 M HCl acid for neutralization. A 10 mL volume of phosp
Drupady [299]

Answer: 0.172 M

Explanation:

a) To calculate theconcentration of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=0.189M\\V_1=19.6mL\\n_2=1\\M_2=?\\V_2=25mL

Putting values in above equation, we get:

1\times 0.189\times 19.6=1\times M_2\times 25\\\\M_2=0.148

b) To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_3PO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=3\\M_1=?\\V_1=10mL\\n_2=1\\M_2=0.148\\V_2=34.9mL

Putting values in above equation, we get:

3\times M_1\times 10=1\times 0.148\times 34.9\\\\M_1=0.172M

The concentration of the phosphoric acid solution is 1.172 M

3 0
3 years ago
What is the pH of a solution that has a [OH-] of 5.08x10^-5 M
sattari [20]

Answer:

The pH of a solution that has a [OH-] of 5.08x10^-5 M is 5

Explanation:

just took the test

onedg2020

3 0
2 years ago
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