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Sergeu [11.5K]
3 years ago
6

SOMEONE HELP ME

Chemistry
1 answer:
AnnZ [28]3 years ago
6 0

Answer:

1. 1.33 gram of carbon

2. 2.38g of carbon dioxide

Explanation:

From the given information:

Total amount of CO₂ = 4.86 grams

Atomic mass of C = 12 g/mole

molar mass of CO₂ = 44 g/mole

∴

The mass of the Carbon (C) in grams is:

= Total\ amount \ of \ CO_2 \times \dfrac{12 \ g/mol}{44 \ g/mol}

= 4.86 \ g  \times \dfrac{12 \ g/mol}{44 \ g/mol}

= 1.33 gram of carbon

2.

Here, the total amount of CO₂ = unknown

Atomic mass of O₂ = 32 g/mole

molar mass of CO₂ = 44 g/mole

amount of oxygen = 1.73 g

∴

The mass of CO₂ = Total \ amount \ of \ O_2 \times \dfrac{44 \ g/mol}{32\ g/mol}

=1.73  \times \dfrac{44 \ g/mol}{32\ g/mol}

= 2.38 g of carbon dioxide

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At 1000°C, cyclobutane (C4H8) decomposes in a first-order reaction, with the very high rate constant of 79 1/s, to two molecules
leva [86]

Explanation:

It is known that for first order reaction, the equation is as follows.

           t = \frac{2.303}{K} log \frac{[C_{4}H_{8}]_{o}}{[C_{4}H_{8}]_{t}}

    t = ?,        K = rate constant = 79 1/s

Initial conc. of C_{4}H_{8} = 1.68

Decompose amount of C_{4}H_{8} = 52% of 1.68

                                                   = \frac{52}{100} \times 1.68

                                                   = 0.8736

                                                   = 0.87

Now, [C_{4}H_{8}]_{t} = (1.68 - 0.87)

                         = 0.81

Therefore, calculate the value of t as follows.

               t = \frac{2.303}{K} log \frac{[C_{4}H_{8}]_{o}}{[C_{4}H_{8}]_{t}}

                  = \frac{2.303}{79 s^{-1}} log \frac{1.68}{0.81}

                  = \frac{2.303}{79 s^{-1}} \times 0.316

                  = 9.212 \times 10^{-3} s

                  = 0.00921 s

Thus, we can conclude that 0.00921 s will be taken for 52% of the cyclobutane to decompose.

7 0
3 years ago
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