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Leviafan [203]
3 years ago
9

A piece of rubber has a mass of 495 kg and a volume of 355 liters. Will it float in water, which has a density of 1.0 g/mL?

Chemistry
1 answer:
hoa [83]3 years ago
6 0
Compare the density of the rubber to water. If it is less that 1 g/mL then it floats in the water.

You must compare density in the same units. So convert to g/mL 

( \frac{495kg}{355L} )*( \frac{1000g}{1kg} )*( \frac{1L}{1000mL} ) = 1.39 g/mL

The rubber does not float in water.
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Part 1. determine the molar mass of a 0.458-gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm. show your work.
d1i1m1o1n [39]
  • The molar mass of 0.458-gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm is 9.15g/mol.
  • If this sample was placed under extreme pressure, the volume of the sample will decrease.

<h3>How to calculate molar mass?</h3>

The molar mass of a substance can be calculated by first calculating the number of moles using ideal gas law equation:

PV = nRT

Where;

  • P = pressure
  • V = volume
  • T = temperature
  • R = gas law constant
  • n = no of moles

0.98 × 1.2 = n × 0.0821 × 287

1.18 = 23.56n

n = 1.18/23.56

n = 0.05moles

mole = mass/molar mass

0.05 = 0.458/mm

molar mass = 0.458/0.05

molar mass = 9.15g/mol

  • Therefore, the molar mass of 0.458-gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm is 9.15g/mol
  • If this sample was placed under extreme pressure, the volume of the sample will decrease.

Learn more about gas law at: brainly.com/question/12667831

8 0
2 years ago
Hii pls help me to balance the equation thanksss​
tiny-mole [99]

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\boxed{\pmb{\color{gold}{\sf{2SO_{2}(g) + O_{2}(g)\dashrightarrow 2SO_{3}(g)}}}}

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7 0
2 years ago
Read 2 more answers
How many moles are in 22 grams of Argon
kakasveta [241]
The answer would be 0.55 moles! good luck!
4 0
3 years ago
Read 2 more answers
A reaction that occurs in the internal combustion engine is n2(g) + o2(g) ⇌ 2 no(g) (a) calculate δh o and δs o for the reaction
jekas [21]
1) ΔrH = 2mol·ΔfH(NO) - (ΔfH(O₂) + ΔfH(N₂)).
ΔrH = 2 mol · 90.3 kJ/mol - (0 kJ/mol + 0 kJ/mol).
ΔrH = 180.6 kJ.
2) ΔS = 2mol·ΔS(NO) - (ΔS(O₂) + ΔS(N₂)).
ΔS = 2mol · 210.65 J/mol·K - (1mol · 205 J/mol·K + 1 mol · 191.5 J/K·mol).
ΔS = 24.8 J/K.
3) ΔG = ΔH - TΔS.
55°C: ΔG = 180.6 kJ - 328.15 K · 24.8 J/K = 172.46 kJ.
2570°C: ΔG = 180.6 kJ - 2843.15 K · 24.8 J/K = 110.09 kJ.
3610°C: ΔG = 180.6 kJ - 3883.15 K · 24.8 J/K = 84.29 kJ.
7 0
3 years ago
If a balloon is filled with a mixture of helium and oxygen, which gas will escape faster? Why?
Serjik [45]

Answer:

The rate of leakage will be higher for helium; its molecules move about 3 times faster than oxygen’s

Explanation:

Step 1: Data given

Molar mass helium = 4.0 g/mol

Molar mass O2 = 32 g/mol

Step 2: Graham's law

Graham's Law of Effusion states that the rate of effusion of a gas is inversely proportional to the square root of the molecular mass : 1/(Mr)^0.5

Rate of escape for He = 1/(4.0)^0.5 = 0.5

Rate of escape for O2 = 1/(32)^0.5 = 0.177

The rate of leakage will be higher for helium; its molecules move about 3 times faster than oxygen’s

4 0
3 years ago
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