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Mariulka [41]
2 years ago
5

HELP PLSS!

Chemistry
1 answer:
Inessa05 [86]2 years ago
7 0
. The empirical formula is the simplest whole number ratio.

The correct answer is 1. HCI
You might be interested in
Does the temperature of the solvent (water) affect the rate of dissolving?​
gregori [183]
Yes. Heating up the solvent gives the molecules more kinetic energy. The more rapid motion means that the solvent molecules collide with the solute with greater frequency and the collisions occur with more force. Both factors increase the rate at which the solute dissolves.
3 0
3 years ago
How many molecules are contained in 0.800 mol o2?
Ksenya-84 [330]
Multiply .800 moles of O2 by Avagadro's number divided by 1 mole. This will get rid of the moles on the bottom and leave you with molecules. So technically .800 times 6.02x10^23.
6 0
2 years ago
To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 2.2-L bulb, then filled
Ber [7]

Answer:

N2

Explanation:

We use the ideal gas equation to calculate the number of moles of the diatomic gas. Then from the number of moles we can get

Given:

P = 2atm

1atm = 101,325pa

2atm = 202,650pa

T = 27 degrees Celsius = 27 + 273.15 = 300.15K

V = 2.2L

R = molar gas constant = 8314.46 L.Pa/molK

PV = nRT

Rearranging n = PV/RT

Substituting these values will yield:

n = (202,650 * 2.2)/(8314.46* 300.15)

n = 0.18 moles

To get the molar mass, we simply divide the mass by the number of moles.

5.1/0.18 = 28.5g/mol

This is the closest to the molar mass of diatomic nitrogen N2.

Hence, the gas is nitrogen gas

7 0
2 years ago
How much energy is required to vaporize 48.7 g of dichloromethane (CH2Cl2) at its boiling point, if its ΔHvap is 31.6 kJ/mol?
BartSMP [9]

Answer:

The answer is 18.12KJ is required to vaporise 48.7 g of dichloromethane at its boiling point

Explanation:

To solve the above question we have the given variable as follows

ΔHvap = heat of vaporisation of dichloromethane per mole = 31.6KJ/mole

However since the heat of vaporisation is the heat to vaporise one mole of dichloromethane, then, for 48.7 grams of dichloromethane, we have.

The number of moles of dichloromethane present = 48.7/84.93 = 0.573 moles

Therefore, the amount of heat required to vaporise 48.7 grams of dichloromethane at its boiling point is 31.6KJ/mole×0.573moles =18.12KJ

3 0
3 years ago
Based on the following molecular weight data of polypropylene, determine the degree of polymerization Molecular Weight Range (g/
Lady bird [3.3K]

Answer:

785

Explanation:

Molecular. X. W

Weight

8000-16000 0.05 0.03

16000-24000. 0.017. 0.08

24000-32000. 0.22. 0.18

32000-40000. 0.25. 0.35

40000-48000. 0.22. 0.27

48000-56000. 0.09. 0.09

Mean weight X*M. W*M

12000. 600. 240

20000. 3200. 2000

28000. 6720. 5600

36000. 10080. 10800

44000. 8800. 11880

52000. 3640 3640

Total=33040g\mol 36240

Note before repeat molecular weight m= 3*12.01+6*1.008=

42.08g/mol

Degree of polymerization= total W*M/w=33040/42.08 =785

8 0
3 years ago
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