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Zinaida [17]
3 years ago
12

HELPPPPPPPPPPPPPPPPPPPPPPPPPPPP!!!!!!!!!!

Chemistry
2 answers:
likoan [24]3 years ago
8 0

Answer:

BBB

Explanation:

solong [7]3 years ago
4 0

Answer:

ehn wet in be d problem which kind help u want biko?

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If I have 4 moles of a gas at a pressure of 5.6 atm and a volume of 12 liters, what is the temperature? (2005 K)
jolli1 [7]

Answer:

205K

Explanation:

The following were obtained from the question:

n = 4moles

P = 5.6atm

V = 12L

R = 0.082atm.L/Kmol

T =?

PV = nRT

T = PV/nR

T = (5.6 x 12)/ (4 x 0.082)

T = 205K

8 0
3 years ago
PLEASE HELP!!!!!! WILL AWARD BRAINLIEST!!!!!!
asambeis [7]

Answer:

A. Sedimentry rock

b. high pressure

Explanation:

6 0
3 years ago
A saline solution with a mass of 400 g has 30 g of NaCl dissolved in it. What is the mass/mass percent concentration of the solu
maria [59]
(g solute/g solution)*100 = % mass/mass

30 g / 400 * 100 

0,075 * 100

= 7,5% w/w

hope this helps!

3 0
3 years ago
2 Points
levacccp [35]
B aaaaaaaaaaassaaaaaaaaaaaaaaaaaaaaaaaaaaaaAaaaaaaaaaaaa
4 0
3 years ago
Read 2 more answers
The molar mass is determined by measuring the freezing point depression of an aqueous solution. A freezing point of -5.20°C is r
Dima020 [189]

Answer:

The empirical formula is C2H4O3

The molecular formula is C4H8O6

The molar mass is 152 g/mol

Explanation:

The complete question is: An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of 31.57% C and 5.30% H. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of -5.20°C is recorded for a solution made by dissolving 10.56 g of the compound in 25.0 g water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.

Step 1: Data given

Mass % of Carbon = 31.57 %

Mass % of H = 5.30 %

Freezing point = -5.20 °C

10.56 grams of the compound dissolved in 25.0 grams of water

Kf water = 1.86 °C kg/mol

Step 2: Calculate moles of Carbon

Suppose 31.57% = 31.57 grams

moles C = mass C / Molar mass C

moles C = 31.57 grams / 12.0 g/mol = 2.63 moles

Step 3: Calculate moles of Hydrogen:

Moles H = 5.30 grams / 1.01 g/mol

moles H = 5.25 moles

Step 4: Calculate moles of Oxygen

Moles O = ( 100 - 31.57 - 5.30) / 16 g/mol

Moles O = 3.95 moles

Step 5: We divide by the smallest number of moles

C: 2.63 / 2.63 = 1 → 2

H: 5.25/2.63 = 2 → 4

O: 3.95/ 2.63 = 1.5 → 3

The empirical formula is C2H4O3

The molar mass of the empirical formula = 76 g/mol

Step 6: Calculate moles solute

Freezing point depression = 5.20 °C = m * 1.86

m = 5.20 / 1.86

m = 2.80 molal = 2.80 moles / kg

2.80 molal * 0.025 kg = 0.07 moles

Step 7: Calculate molar mass

Molar mass = mass / moles

Molar mass = 10.56 grams / 0.07 moles

Molar mass = 151 g/mol

Step 8: Calculate molecular formula

151 / 76 ≈  2

We have to multiply the empirical formula by 2

2*(C2H4O3) = C4H8O6

The molecular formula is C4H8O6

The molar mass is 152 g/mol

6 0
3 years ago
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