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Valentin [98]
3 years ago
9

A reaction yield 6.26 g of a product what is the percent yield if the theoretical yield is 18.81 g

Chemistry
1 answer:
marta [7]3 years ago
7 0

Explanation:

remember the equation percentage yield = actual/theoretical yield

so 6.26/18.81 X 100 gives u 33.28017012 so to 3 SF ot would be 33.3%

hope this helps:)

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When Ca⁺ undergoes alpha decay, it will lose an alpha particle, which has mass 4 and 2 protons.

⁴¹₂₀Ca⁺ → ³⁷₁₈X⁺ + ⁴₂α

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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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