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JulijaS [17]
2 years ago
11

How many g are in 3 mol of Iron II Carbonate FeCO3

Chemistry
1 answer:
kozerog [31]2 years ago
4 0

Answer:

348 g FeCO3

Explanation:Use your 3 mol's and find the atomic number of Iron II and CO3 which i think is 115.854 so you multiply it by the 3 and that should give you your grams in FeCO3.

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Suppose the half-life is 9.0 s for a first order reaction and the reactant concentration is 0.0741 M 50.7 s after the reaction s
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<u>Explanation:</u>

The equation used to calculate half life for first order kinetics:

k=\frac{0.693}{t_{1/2}}

where,

t_{1/2} = half-life of the reaction = 9.0 s

k = rate constant = ?

Putting values in above equation, we get:

k=\frac{0.693}{9}=0.077s^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}     ......(1)

where,

k = rate constant  = 0.077s^{-1}

t = time taken for decay process = 50.7 sec

[A_o] = initial amount of the reactant = ?

[A] = amount left after decay process =  0.0741 M

Putting values in equation 1, we get:

0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}

[A_o]=3.67M

Now, calculating the time taken by using equation 1:

[A]=0.0055M

k=0.077s^{-1}

[A_o]=3.67M

Putting values in equation 1, we get:

0.077=\frac{2.303}{t}\log\frac{3.67}{0.0055}\\\\t=84.5s

Hence, the time taken by the reaction is 84.5 seconds

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3 years ago
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