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pentagon [3]
3 years ago
13

The rate constant for the decomposition reaction of H2O2 is 3.66 × 10−3 s−1 at a particular temperature. What is the concentrati

on of H2O2 in a solution that was initially 10.0 M H2O2 after 15.0 minutes have passed?
Chemistry
1 answer:
Pepsi [2]3 years ago
7 0

Answer:  3.72 M

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 = 3.66\times 10^{-3}s^{-1}

t = age of sample = 15.0 minutes

a = let initial amount of the reactant  = 10.0 M

a - x = amount left after decay process = ?

15.0\times 60s=\frac{2.303}{3.66\times 10^{-3}}\log\frac{10.0}{(a-x)}

\log\frac{100}{(a-x)}=1.43

\frac{100}{(a-x)}=26.9

(a-x)=3.72M

The concentration of H_2O_2 in a solution after 15.0 minutes have passed is 3.72 M

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f the Ksp for HgBr2 is 2.8×10−14, and the mercury ion concentration in solution is 0.085 M, what does the bromide concentration
goldfiish [28.3K]

Answer:

0.057 M

Explanation:

Step 1: Given data

Solubility product constant (Ksp) for HgBr₂: 2.8 × 10⁻⁴

Concentration of mercury (II) ion: 0.085 M

Step 2: Write the reaction for the solution of HgBr₂

HgBr₂(s) ⇄ Hg²⁺(aq) + 2 Br⁻

Step 3: Calculate the bromide concentration needed for a precipitate to occur

The Ksp is:

Ksp = 2.8 × 10⁻⁴ = [Hg²⁺] × [Br⁻]²

[Br⁻] = √(2.8 × 10⁻⁴/0.085) = 0.057 M

7 0
3 years ago
Which of the following statements regarding glucose is FALSE?a) Glucose is the main component of starch and glycogen.b) Glucose
marishachu [46]

Answer:

The correct option is C.

Explanation:

Carbohydrates are one of the macro molecules that are consumed by living organisms. The end product of carbohydrate is glucose. Glucose is a very important fuel that the body cells used to produce energy, which they use to carry out their daily activities. Glucose is also known as blood sugar and it is the only fuel that living cells can use for the production of ATP. Other food macro molecules such as lipids and proteins can also be converted to glucose if there is a need for that. Glucose is always stored in the body in form of glycogen.

The statement given in option C about glucose is wrong because glucose is a monosaccharide and not a disaccharide.

8 0
4 years ago
To find the Ce4+ content in a solid sample, 4.3718 g of the solid sample were dissolved and treated with excess iodate to precip
gulaghasi [49]

Answer:

3.43 %

Explanation:

We need  to calculate first the number of moles of CeO2 produced in the combustion. Given its formula we know how many moles of Ce atom are present. From there calculate the mass this number of moles this represent and then one can calculate the percentage.

0.1848 g CeO2 x 1 mol CeO2/172.114g = 0.00107 mol CeO2

0.00107 mol CeO2 x 1 mol Ce/ 1 mol CeO2 = 0.00107 mol Ce

.00107 mol Ce x 140.116 g Ce/ mol  =  0.150 g Ce

0.150 g Ce/ 4.3718 g sample  x 100 = 3.43 %

5 0
3 years ago
Read 2 more answers
What is spectrophotometry? How can this be useful in identifying drugs?
liq [111]

Answer:

What is spectrophotometry? It is an analytic method that it use substances light absorbance property

How can this be useful in identifying drugs? It possible compare spectra and identify problem substance

Explanation:

The way substances absorb light is unique and because of that it is possible to use spectrophotometry for substances identifying. the model spectra are measured in ideal conditions, so, it is difficult ensure the same conditions and to achieve identical spectra.

I hope I have been helpful

7 0
4 years ago
Must all species reproduce in order for the species to survive? *
xz_007 [3.2K]

Answer:

NOPE

Explanation:

7 0
3 years ago
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