Ask question

Ask question

All categories

3 years ago

8

What is the rate constant of a first-order reaction that takes 563 secondsseconds for the reactant concentration to drop to half

of its initial value?

1 answer:

Stella [2.4K]3 years ago

4 0

Answer: The rate constant of a first-order reaction that takes 563 secondsseconds for the reactant concentration to drop to half of its initial value is $1.23\times 10^{-3}s^{-1}$

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

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

$k=\frac{0.693}{563s}=1.23\times 10^{-3}s^{-1}$

Thus rate constant of a first-order reaction that takes 563 secondsseconds for the reactant concentration to drop to half of its initial value is $1.23\times 10^{-3}s^{-1}$

You might be interested in

To interconvert the concentration units molality (m) and mass percent, you must also know the density of the solution. A) True B

uranmaximum [27]

Answer:

The correct answer is option false.

Explanation:

Molality of the solution defined as moles of substance present in 1 kilogram of solvent.

Moles = $\frac{Mass of solute}}{\text{molar mass of solute}}$

$Molality=\frac{Moles}{\text{Mass of solvent(kg)}}$

Mass of percent (w/w%) of the solution is defined as amount of solute present in 100 grams of solution.

$(w/w\%)_=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100$

So, if want to inter-convert molality into mass percent we can do that without knowing density of solution.

Mass of solution = Mass of solute + Mass of solvent

7 0

3 years ago

If a solution containing 45.101 g of mercury(II) acetate is allowed to react completely with a solution containing 12.026 g of s

AnnyKZ [126]

Answer:

14.533 grams of solid precipitate of mercury(II) dichromate will form.

Explanation:

$Hg(CH_3COO)_2(aq)+Na_2Cr_2O_7(aq)\rightarrow HgCr_2O_7(s)+2CH_3COONa(aq)$

Moles of mercury(II) acetate = $\frac{45.101 g}{318.70 g/mol}=0.14152 mol$

Moles of sodium dichromate = $\frac{12.026 g}{261.97 g/mol}=0.045906 mol$

According to reaction , 1 mole of sodium dichromate reacts with 1 mole of mercury(II) acetate , then 0.045906 moles of sodium dichromate will recat with :

$\frac{1}{1}\times 0.045906 mol=0.045906 mol$ of mercury(II) acetate

This means that mercury(II) acetate is present in an excess amount and sodium dichromate is present in limiting amount.So, amount of precipitate will depend upon moles of sodium dichromate.

According to reaction , 1 mole of sodium dichromate gives 1 mole of mercury(II) dichromate , then 0.045906 moles of sodium dichromate will give :

$\frac{1}{1}\times 0.045906 mol=0.045906 mol$ of mercury(II) dichromate

Mass of 0.045906 moles of mercury(II) dichromate:

0.045906 mol × 316.59 g/mol = 14.533 g

14.533 grams of solid precipitate of mercury(II) dichromate will form.

3 0

4 years ago

The area of a telescope lens is 7903 rnrn2 {a) What is the area in square feet (ft2)? (b) If it takes a technician 45 s to polis

LekaFEV [45]

Answer:

Area is 0.08504 $ft^{2}$ and $2.6\times 10^{3}s$ are needed for polishing

Explanation:

(a) ft is unit of F.P.S system whether mm is unit of M.K.S system.

According to unit conversion, $1mm=0.00328ft$

So, $1mm^{2}=(0.00328)^{2}ft^{2}=1.076\times 10^{-5}ft^{2}$

hence, $7903mm^{2}=7903\times 1.076\times 10^{-5}ft^{2}=0.08504ft^{2}$

So, area of telescope lens is 0.08504 $ft^{2}$

(b) $135mm^{2}$ are polished in 45s

So, $7903mm^{2}$ are polished in $\frac{45\times 7903}{135}s$ or $2.6\times 10^{3}s$

5 0

3 years ago

When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced. CaCO 3 ( s ) + 2

e-lub [12.9K]

<u>Answer:</u> The mass of calcium chloride formed is 15.21 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For calcium carbonate:</u>

Given mass of calcium carbonate = 32.0 g

Molar mass of calcium carbonate = 100 g/mol

Putting values in equation 1, we get:

$\text{Moles of calcium carbonate}=\frac{32.0g}{100g/mol}=0.32mol$

<u>For hydrochloric acid:</u>

Given mass of hydrochloric acid = 10.0 g

Molar mass of hydrochloric acid = 36.5 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrochloric acid}=\frac{10.0g}{36.5g/mol}=0.274mol$

The given chemical equation follows:

$CaCO_3(s)+2HCl(aq.)\rightarrow CaCl_2(aq.)+H_2O(l)+CO_2(g)$

By Stoichiometry of the reaction:

2 moles of hydrochloric acid reacts with 1 mole of calcium carbonate

So, 0.274 moles of hydrochloric acid will react with = $\frac{1}{2}\times 0.274=0.137mol$ of calcium carbonate

As, given amount of calcium carbonate is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrochloric acid is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of hydrochloric acid produces 1 mole of calcium chloride.

So, 0.274 moles of hydrochloric acid will produce = $\frac{1}{2}\times 0.274=0.137moles$ of calcium chloride.

Now, calculating the mass of calcium chloride from equation 1, we get:

Molar mass of calcium chloride = 111 g/mol

Moles of calcium chloride = 0.137 moles

Putting values in equation 1, we get:

$0.137mol=\frac{\text{Mass of calcium chloride}}{111g/mol}\\\\\text{Mass of calcium chloride}=(0.137mol\times 111g/mol)=15.21g$

Hence, the mass of calcium chloride formed is 15.21 grams.

4 0

3 years ago

Melting is considered a(n) ___ change because energy is gained by the substance as it changes state

Burka [1]

Melting is a physical change

4 0

3 years ago