Which of the following is a product when Na2O and H2O react together?

What is the percent lithium in lithium nitrate?

seropon [69]

Answer:

Around 60%

Explanation:

4 0

3 years ago

What is the total pressure of air in lungs of an individual with oxygen at 100. mmHg, nitrogen at 573 mmHg, carbon dioxide at 0.

ryzh [129]

Answer: The total pressure of air in lungs of an individual is 760.28 mm Hg

Explanation:

According to Dalton's law, the total pressure is the sum of individual pressures.

$p_{total}=p_A+p_B+p_C...$

Given : $p_{total}$ =total pressure of gases = ?

$p_{O_2}$ = partial pressure of oxygen = 100 mm Hg

$p_{N_2}$ = partial pressure of nitrogen = 573 mm Hg

$p_{CO_2}$ = partial pressure of Carbon dioxide = 0.053 atm = 40.28 mm Hg(1 atm = 760 mmHg)

$p_{H_2O}$ = partial pressure of water vapor = 47 torr = 47 mm Hg (1torr=1 mm Hg)

putting in the values we get:

$p_{total}=(100+573+40.28+47)mmHg$

$p_{total}=760.28mmHg$

Thus the total pressure of air in lungs of an individual is 760.28 mm Hg

5 0

3 years ago

if a steel spoon were to be plated with silver, state what would be suitable as the anode, cathode, electrolyte

Tresset [83]

Answer:

The anode will be silver

The cathode will be the steel spoon

The electrolyte will be silver nitrate

Explanation:

Pls mark BRAINLIEST

7 0

3 years ago

Read 2 more answers

3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15

shutvik [7]

Answer:

C₅H₁₀O₅

Explanation:

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

$\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}$

(b) Mass of H

$\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}$

(c) Mass of O

Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

$\text{Moles of C = 1400 mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}$