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kirill [66]
4 years ago
6

If temperature is kept constant and the volume of a gas is doubled, what will happen to the pressure?

Chemistry
1 answer:
Mrrafil [7]4 years ago
5 0

Answer:

pressure = P1/2

Explanation:

using boyles law

P1V1=P2V2

V2 = 2V1

P1V1=P2 x 2V1

P1 = p2 x 2

P2 = P1/2

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OHC-CH2-CH2-CH2-CHO<br><br> What is the IUPAC name of this compound ?
Rufina [12.5K]

Answer:

Pentan_1,5_di-al

Explanation:

OHC-CH₂-CH₂-CH₂-CHO

This is Pentan_1,5_di-al

If we break this compound, we will observe that there is presence aldehyde group and hence the functional group "al". This aldehyde is bonded to carbon 1 and carbon 5 respectively.

Also the pentan is due to presence of 5 carbon atoms.

Therefore, the IUPAC name of this compound (OHC-CH₂-CH₂-CH₂-CHO) is  Pentan_1,5_di-al

5 0
3 years ago
H2O has a mc021-1.jpgHvap = 40.7 kJ/mol. What is the quantity of heat that is released when 27.9 g of H2O condenses?
babymother [125]

Energy released from changing the phase of a substance from the gas phase to liquid phase can be calculated by using the specific latent heat of vaporization. The heat of fusion of water at 0 degrees Celsius is 40.7 kJ/mol. Calculation are as follows:<span> </span>


Energy = 27.9 g (1 mol / 18.02 g) x 40.7 kJ/mol


Energy = 63.09 kJ


5 0
3 years ago
Read 2 more answers
Identify fossil types match each each type of fossil to its description
dem82 [27]
Well we need to see the fossil
4 0
3 years ago
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A bar of gold is 5.0mm thick, 10.0cm long and 2.0cm wide. It has a mass of exactly 193.0g. What is the desity of gold?
Tanzania [10]
<h3>Answer:</h3>

19.3 g/cm³

<h3>Explanation:</h3>

Density of a substance refers to the mass of the substance per unit volume.

Therefore, Density = Mass ÷ Volume

In this case, we are given;

Mass of the gold bar = 193.0 g

Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm

We are required to get the density of the gold bar

Step 1: Volume of the gold bar

Volume is given by, Length × width × height

Volume =  0.50 cm × 10.0 cm × 2.0 cm

             = 10 cm³

Step 2: Density of the gold bar

Density = Mass ÷ volume

Density of the gold bar = 193.0 g ÷ 10 cm³

                                      = 19.3 g/cm³

Thus, the density of the gold bar is 19.3 g/cm³

3 0
3 years ago
A sample of an unknown compound is vaporized at 150.°C . The gas produced has a volume of 960.mL at a pressure of 1.00atm , and
IrinaK [193]

Answer:

34.02 g.

Explanation:

Hello!

In this case, since the gas behaves ideally, we can use the following equation to compute the moles at the specified conditions:

PV=nRT\\\\n=\frac{1.00atm*0.960L}{0.08206\frac{atm*L}{mol*K}*(150+273)K} =0.0277mol\\\\

Now, since the molar mass of a compound is computed by dividing the mass over mass, we obtain the following molar mass:

MM=\frac{0.941g}{0.0277mol} \\\\MM=34.02g/mol

So probably, the gas may be H₂S.

Best regards!

6 0
3 years ago
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