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galben [10]
3 years ago
7

2000 2×10 What is the answer?

Chemistry
2 answers:
Zepler [3.9K]3 years ago
6 0

Answer:

40000

Explanation:

harina [27]3 years ago
4 0

Answer: 40000

Explanation: I really hope it works for you

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A gas effuses 1.55 times faster than propane (C3H8)at the
stepladder [879]

Answer:  The mass of the gas is 18.3 g/mol.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows:

\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

\frac{Rate_{X}}{Rate_{C_3H_8}}=1.55

\frac{Rate_{X}}{Rate_{C_3H_8}}=\sqrt{\frac{M_{C_3H_8}}{M_{X}}}

1.55=\sqrt{\frac{44}{M_{X}}

Squaring both sides and solving for M_{X}

M_{X}=18.3g/mol

Hence, the molar mas of unknown gas is 18.3 g/mol.

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4 years ago
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dybincka [34]

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7 0
3 years ago
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The half-life of a positron is very short. It reacts with an electron, and the masses of both are converted to two gamma-ray pho
sp2606 [1]

Explanation:

(a)   It is known that relation between energy and mass is as follows.

            E = 2 \times mc^{2}

where,    E = energy

              m = mass

              c = speed of light = 3 \times 10^{8} m/s

As it is given that mass is 9.109 \times 10^{-31} kg. So, putting the given values into the above formula as follows.

             E = 2 \times mc^{2}

                       = 2 \times 9.109 \times 10^{-31} kg \times 3 \times 10^{8}m/s

                       = 1.638 \times 10^{-13} J

Therefore, we can conclude that the energy produced by the reaction between one electron and one positron is 1.638 \times 10^{-13} J.

(b) When gamma ray photons are produced then they will have the same frequency. Relation between energy and frequency is as follows.

                    E = h \times \nu   ..... (1)

where,     h = plank's constant = 6.626 \times 10^{-34} J.s

              \nu = frequency

Also,     E = 2 \times mc^{2} ........ (2)

Hence, equating equations (1) and (2) as follows.

                    h \times \nu = 2 \times mc^{2}        

So,    

6.626 \times 10^{-34} Js \times \nu = 1.638 \times 10^{-13} J

                           \nu = 1.236 \times 10^{20} Hz

Thus, we can conclude that the frequency is 1.236 \times 10^{20} Hz.

5 0
4 years ago
2HgCl2(aq) + C2O42–(aq) → 2Cl–(aq) +2CO2(g) + Hg2Cl2(s)
alexgriva [62]

Explanation:

Expression for rate of the given reaction is as follows.

             Rate = k[HgCl_{2}]x [C_{2}O^{2-}_{4}]y[/tex]

Therefore, the reaction equations by putting the given values will be as follows.

       1.8 \times 10^{-5} = k[0.105]x [0.15]y ............. (1)

       7.2 \times 10^{-5} = k [0.105]x [0.30]y ........... (2)

       3.6 \times 10^{-5} = k [0.0525]x [0.30]y ............ (3)

Now, solving equations (1) and (2) we get the value of y = 2. Therefore, by solving equation (2) and (3)  we get the value of x = 1.

Therefore, expression for rate of the reaction is as follows.

     Rate = k[HgCl_{2}]x [C_{2}O^{2-}_{4}]y

          Rate = k [HgCl2]1 [C_{2}O^{-2}_{4}]2

Hence, total order = 1 + 2 = 3

According to equation (1),

               1.8 \times 10^{-5} = k[0.105]x [0.15]y  

            1.8 \times 10^{-5} = k [0.105]1 [0.15]2  

                      k = 7.6 \times 10^{-3} M^{-2} min^{-1}  

Thus, we can conclude that rate constant for the given reaction is 7.6 \times 10^{-3} M^{-2} min^{-1}.

8 0
3 years ago
The amount of energy required to raise the temperature of a swimming pool depends on the mass of water present.
cluponka [151]
True. Refer to the heat equation of water
3 0
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