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2 years ago

What is the relative atomic mass of ,A of X?

Chemistry

1 answer:

Fed [463]2 years ago

7 0

11 A n 8 g sample of oxygen atoms contains the same number of atoms as 16 g of element x.
The question does not make sense. does it mean 11 g of A with 8 g Oxygen and 16 g X with 8 g Oxygen.

Does not make sense

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If you have 10 grams of Lithium Oxide what will the volume be? Show your work

gogolik [260]

Answer:

Volume occupied by 10 grams of Lithium Oxide at STP is 7.52 L

Explanation:

STP (Standard Temperature and Pressure) : It is used while performing calculation on gases and provide standards of measurements while performing experiment.

According to IUPAC (International Union of Pure And applied Chemistry) , standard temperature is 273.15 K and Pressure is 100 kPa (0.9869 atm). At STP condition 1 mole of substance occupies 22.4 L volume .

Molar mass of Lithium Oxide = 29.8 g/mol

$Li_{2}O$ = 2(6.9) + 15.99 = 29.8

Mass of 1 mole $Li_{2}O$ = 29.8 g

1 mole $Li_{2}O$ occupies, Volume = 22.4 L

29.8 g $Li_{2}O$ occupies, V = 22.4 L

1 g $Li_{2}O$ occupies ,V = $\frac{22.4}{29.8}$

1 g $Li_{2}O$ occupies ,V = 0.7516 L

10 g tex]Li_{2}O[/tex] occupies ,V = $0.7516 \times10$ L

V = 7.52 L

So, volume occupied by Lithium Oxide At STP is 7.52 L

4 0

3 years ago

if the temperature is constant what change in volume would cause the pressure of an enclosed gas to increase by a factor of 4

mezya [45]

P1V1=4P1V2
V2/V1= P1/4P1
V2/V1= 1/4
V2=1/4*V1
volume will decrease by factor 1/4

4 0

3 years ago

Read 2 more answers

In lab (write this down in your lab protocol), you will be given a stock solution that has a glucose concentration of 60 mg/dL.

Wittaler [7]

Answer:

1. The dilution factor for the serial dilution = 2

2. V2 = 1 mL

3. V1 = 0.5 mL

Explanation:

1. Dilution factor is the ratio of the initial concentration to the final concentration.

Dilution factor = initial concentration / final concentration

First dilution: initial concentration = 60 mg/dL

final concentration = 30 mg/dL

Dilution factor = 60 mg/dL / 30 mg/dL = 2

Second dilution: initial concentration = 30 mg/dL

final concentration = 15 mg/dL

Dilution factor = 30 mg/dL / 15 mg/dL = 2

Therefore, the dilution factor for the serial dilution = 2

2. From the dilution formula, C1V1 = C2V2; V2 = final volume to be prepared.

Since 1 mL of the various glucose solutions are to be prepared, the final concentration, V2 = 1 mL

3. From the dilution formula, C1V1 = C2V2; V1 = initial concentration of the solution to be prepared.

C1/C2 = V2/V1

Since the dilution factor, C1/C2 is 2, V2/V1 = 2

V1 = V2/2

V1 = 1 mL / 2

V1 = 0.5 mL

6 0

3 years ago

STEP 7: LEAD

GenaCL600 [577]

Answer:

answer is 76

Explanation:

pls mark me brainliest right

4 0

2 years ago

In this experiment, 0.070 g of caffeine is dissolved in 4.0mL of water. The caffeine is extracted from the aqueous solution thre

Len [333]

Answer:

0.068 g

Explanation:

The equation that is used to determine the fraction q remaining in water of volume V₁ after n extractions of volume V₂ is:

qⁿ = (V₁/(V₁ + KV₂))ⁿ

Substituting in the values gives the fraction of caffeine left in the water phase:

q³ = (4.0mL/(4.0mL + 4.6(2.0mL))³ = 0.0278

The fraction of caffeine extracted into the methylene chloride phase is:

1 - 0.0278 = 0.972

The amount of caffeine extracted into the methylene chloride is:

(0.070g)(0.972) = 0.068 g

6 0

2 years ago