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mario62 [17]
3 years ago
5

A gas with a volume of 100 mL at 80 °C is heated until its volume is 400 mL. What is the new temperature of the gas if the press

ure is unaltered?
Chemistry
1 answer:
xxMikexx [17]3 years ago
8 0

Charles law states a proportional relationship between volume and temperature.

\frac{V_{1} }{T_{1} } =\frac{V_{2} }{T_{2} }

V1= 100mL=0.1L

V2=400mL=0.4L

T1=80'c+273=353K

T2=?


T2=1412K-273=1139'C

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The gain of electrons by an element or ion in a reaction is called __.
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List the substances Ar, Cl2, CH4, and CH3COOH, in order of increasing strength of intermolecular attractions. List the substance
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Answer:

Order of increasing strength of intermolecular attraction:

CH_3COOH > Cl_2 > CH_4 > Ar

Explanation:

CH_3COOH can form hydrogen bond as H atom is attached with electronegative atom O.

Rest three, Cl_2,  CH_4,  Ar are non-polar molecules.

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Molecular size and mass of Cl_2 is high as compared to CH_4.

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Therefore,

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8 0
3 years ago
Read 2 more answers
How many grams of water are produced when 4.50 L of
MA_775_DIABLO [31]

The answer for the following problem has been mentioned below.

  • <em><u>Therefore the mass of the water is 5.802 grams.</u></em>

Explanation:

Given:

volume of oxygen (V) = 4.50 L

Temperature (T) = 425 K

pressure of oxygen (P) = 2.50 atm

Gram molecular mass of oxygen (M) = 16.0 grams

To calculate:

mass of water (m)

We know;

According to the ideal gas equation;

     P × V = n × R × T

As we know;

no of moles = \frac{m}{M}

m represents the mass of oxygen (m)

M represents the Gram molecular mass (M)

According to above mentioned equation;

           P × V = n × R × T

P represents the pressure of the oxygen

V represents the volume of the oxygen

n represents the no of moles of the oxygen

R represents the universal gas constant

where,

the value of R is 0.0821 L atm/K moles

Substituting the values in the above equation;

                  2.50 × 4.50 = \frac{m}{16.0} × 0.0821 × 425

                   11.25 =  \frac{m}{16.0} × 34.8925

                  180 = m × 34.8925

                  m = \frac{180}{34.8925}

                  m = 5.158 grams

Therefore the mass of the of oxygen is 5.158 grams

Now;

As we know;

           \frac{m_{1} }{M_{1} } = \frac{m_{2} }{M_{2} }

where;

m_{1} represents the mass of the oxygen

M_{1} represents the gram molecular mass of the oxygen

m_{2} represents the mass of the water

M_{2} represents the gram molecular mass of water

    From the above given formula,

      \frac{5.158}{16.0} = \frac{m_{2} }{18}

where;

Gram molecular weight of water = 18.0 u

    m_{2} = 5.802 grams

<em><u>Therefore the mass of the water is 5.802 grams.</u></em>

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