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3 years ago

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In a first-order decomposition reaction. 58.6% of a compound decomposes in 11.7 min. How long (in min) does it take for 80.2% of

the compound to decompose

1 answer:

gavmur [86]3 years ago

6 0

Answer:

t = 21.5 min.

Explanation:

Hello!

In this case, since the kinetics of a first-order reaction is:

$\frac{[A]}{[A]_0}=exp(-kt)$

Thus, since we are given the 11.7 min for a 58.6-% consumption, we can compute the rate constant, k:

$ln(1-0.586)=-kt\\\\k=\frac{ln(0.414)}{-t}=\frac{-0.882}{11.7min}=0.0754min^{-1}$

Now, for the second problem, as the new consumption is 80.2%, we can compute the required time as shown below:

$ln(1-0.802)=-kt\\\\t=\frac{ln(198)}{k} \\\\t=\frac{-1.62}{0.0754min^{-1}}\\\\t=21.5min$

Best regards!

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How many grams sodium bromide can be formed from 51 grams of sodium hydroxide?

raketka [301]

Explanation:

When working with moles only, you will start by applying stoichiometry to determine how the reactants will affect your amount of products in this reaction. For this question, we will assume that other reactants are in infinite qualities, so therefore, it is the amount of aluminum that we will be concerned with. You need to figure out how much aluminum is in the specified amount of aluminum bromide, and then how much aluminum hydroxide that will be able to create. Make sure all your units cancel out!

9.24 mol AlBr3 x (1 mol Al / 1 mol AlBr3) x (1 mol Al(OH)3 / 1 mol Al) = 9.24 mol AlBr3

When you're working with mole ratios that involve grams to moles conversions, the first thing you want to do is calculate the molecular weight of each component you are being asked about. Because the question was given to you as words instead of chemical formulas, you will want to figure out the chemical formulas. For example, aluminum hydroxide is Al(OH)3 and aluminum bromide is AlBr3. To calculate molecular weight, you will want to consult a periodic table, find the molecular weight for each atom, and then calculate the correct sum of each molecular weight. Make sure you keep track of the number of each atom you have, i.e. 3 oxygen and 3 hydrogen for aluminum hydroxide.

Na = 22.990 g/mol

O = 15.999 g/mol

H = 1.008 g/mol

NaOH = 22.990 g/mol + 15.999 g/mol + 1.008 g/mol = 39.997 g/mol

Al = 26.982 g/mol

O = 15.999 g/mol

H = 1.008 g/mol

Al(OH)3 = 26.982 g/mol + (3 x 15.999 g/mol) + (3 x 1.008 g/mol) = 78.003 g/mol

Now, if you begin with an amount of NaOH in grams, you will first have to convert that to moles in order to use the mole ratio.

24 g NaOH x (1 mol NaOH / 39.997 g NaOH) = 0.600 mol NaOH

Now, you will have to account for the part of the sodium hydroxide that will be present in the aluminum hydroxide. In this case, it is the hydroxide (OH) portion of the formula. There is one mole of OH in each mole of NaOH, but there are 3 moles of OH in each mole of Al(OH)3. You will start with the 0.600 mol NaOH you know you have and then use the mole ratio.

0.600 mol NaOH x (1 mol OH / 1 mol NaOH) x (1 mol Al(OH)3 / 3 mol OH) = 0.200 mol Al(OH)3

Finally, when you are converting from grams to grams, you will have to find the molecular weight of both the reactant and the product, convert reactants in grams to reactants in moles, then use the mole ratio, then convert the moles of product back to grams of product. This time, you are concerned about the mole ratio of sodium, as that is the element that is in both chemical formulas.

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How many moles are in 25 grams of HF

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$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

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$m(\text{Hg}) = m(\text{residure})$
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First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.

Similarly, for the second compound

$m(\text{O}) = m(\text{loss}) = 0.016 \; g$
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$n = m/M$ ; $0.4172 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol$
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