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Snowcat [4.5K]
3 years ago
6

Question 3

Chemistry
2 answers:
Arte-miy333 [17]3 years ago
6 0

Answer:

80%

Explanation:

dicarbon hexahydride is C2H6

So mass percent = (24/30) x 100 = 80%

Fofino [41]3 years ago
5 0
80% carbon and 20.% is hydrogen
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They are highly lustrous metals which are resistant to corrosion. They are highly malleable and ductile so can be transformed into any shape or design.

Explanation:

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Pls help ASAP will give brianlist
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If the temperature of air in gabor's lungs is 37∘c (98.6∘f), and the volume is 6l, how many moles of air n must be released by t
umka21 [38]
Missing question:
Suppose Gabor, a scuba diver, is at a depth of 15 m. Assume that:
1. The air pressure in his air tract is the same as the net water pressure at this depth. This prevents water from coming in through his nose.
2. The temperature of the air is constant (body temperature).
3. The air acts as an ideal gas.
4. Salt water has an average density of around 1.03 g/cm^3, which translates to an increase in pressure of 1.00 atm for every 10.0 m of depth below the surface. Therefore, for example, at 10.0 m, the net pressure is 2.00 atm.

T = 37°C = 310 K.
p₁ = 2,5 atm = 253,313 kPa.
p₂ = 1 atm = 101,325 kPa.
Ideal gas law: p·V = n·R·T.
n₁ = 253,313 kPa · 6 L ÷ 8,31 J/mol·K · 310 K.
n₁ = 0,589 mol.
n₂ = 101,325 kPa · 6 L ÷ 8,31 J/mol·K · 310 K.
n₂ = 0,2356 mol.
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4 years ago
1. What is the mitochondra in a cell? What is its primary funtion?
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3 years ago
Read 2 more answers
A eudiometer contains a 65.0 ml sample of a gas collected
SCORPION-xisa [38]

Answer:

53.1 mL

Explanation:

Let's assume an ideal gas, and at the Standard Temperature and Pressure are equal to 273 K and 101.325 kPa.

For the ideal gas law:

P1*V1/T1 = P2*V2/T2

Where P is the pressure, V is the volume, T is temperature, 1 is the initial state and 2 the final state.

At the eudiometer, there is a mixture between the gas and the water vapor, thus, the total pressure is the sum of the partial pressure of the components. The pressure of the gas is:

P1 = 92.5 - 2.8 = 89.7 kPa

T1 = 23°C + 273 = 296 K

89.7*65/296 = 101.325*V2/273

101.325V2 = 5377.45

V2 = 53.1 mL

6 0
4 years ago
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