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Elodia [21]
2 years ago
7

Calculate the mass (g) of each product formed when 43.82 g of diborane (B₂H₆)

Chemistry
1 answer:
guajiro [1.7K]2 years ago
6 0

Stoichiometry:

The area of study in chemistry concerned with the quantities of chemical species produced or required for a given chemical reaction is stoichiometry. The stoichiometric coefficients of the balanced chemical equation are often used to relate the amounts of substances to one another.

Evaluation :

The balanced chemical equation for the reaction is given as follows:

B_{2} H_{6} + 6H_{2} O  → 2H_{3} BO_{3}  + 3H_{2}

By using the molar mass of the given chemical species, along with the stoichiometric coefficients of the balanced chemical equation, we determine the mass of each product as:

43.82 g B_{2} H_{6} ×\frac{1 Mol B_{2} H_{6} }{27.66 g} ×\frac{3molH_{2} }{1 molB_{2}H_{6}  } ×\frac{2g}{1 mol H_{2} } = 9.51 g H_{2}

43.82 g B_{2} H_{6} × \frac{1 mol B_{2} H_{6} }{27.66 g}    ×   \frac{1 mol H_{3}BO_{3}  }{1 mol B_{2} H_{6} } × \frac{61.8g}{1 mol H_{3} BO_{3} }  = 196 gH_{3} BO_{3}

Learn more about chemical reaction :

brainly.com/question/1893305

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Irina-Kira [14]

Answer:

t=147.24\ min

Explanation:

Given that:

Half life = 30 min

t_{1/2}=\frac{\ln2}{k}

Where, k is rate constant

So,  

k=\frac{\ln2}{t_{1/2}}

k=\frac{\ln2}{30}\ min^{-1}

The rate constant, k = 0.0231 min⁻¹

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given that:

The rate constant, k = 0.0231 min⁻¹

Initial concentration [A_0] = 7.50 mg

Final concentration [A_t] = 0.25 mg

Time = ?

Applying in the above equation, we get that:-

0.25=7.50e^{-0.0231\times t}

750e^{-0.0231t}=25

750e^{-0.0231t}=25

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3 years ago
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Answer:

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I hope it's helpful!

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