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siniylev [52]
2 years ago
15

4) The initial rate of the reaction between substances P and Q was measured in a series of

Chemistry
1 answer:
ASHA 777 [7]2 years ago
6 0

Answer:

The initial rate of the reaction between substances P and Q was measured in a series of

experiments and the following rate equation was deduced.

rate = k[P]^{2} [Q]

Complete the table of data below for the reaction between P and Q

Explanation:

Given rate of the reaction is:

rate= k[P]^{2} [Q]\\=>[Q]=\frac{rate}{k.[P]^{2} } \\and \\\\\\\ [P]=\sqrt{\frac{rate}{k.[Q]} }

Substitute the given values in this formulae to get the [P], [Q] and rate values.

From the first row,

the value of k can be calulated:

k=\frac{rate}{[P]^{2}[Q] } \\  =\frac{4.8*10^-3}{(0.2)^{2} 2. (0.30)} \\ =0.4

Second row:

2. Rate value:

rate =0.4* (0.10)^{2} * (0.10)\\\\        =4.0*10^-3mol.dm^-3.s^-1

3.Third row:

[Q]=\frac{rate}{k.[P]^{2} } \\     =9.6*10^-3 / (0.4 *(0.40)^{2} \\    =0.15mol.dm^{-3}

4. Fourth row:

[P]=\sqrt{\frac{rate}{k.[Q]} }\\=>[P]=\sqrt{\frac{19.2*10^-3}{0.60*0.4} } \\=>[P]=0.283mol.dm^{-3}

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2 years ago
Calculate the mass of a 0.297-mL sample of a liquid with a density of 0.930 g/mL.
tekilochka [14]

Answer: 0.27621 g

Explanation:

0.297 ml *0.930 g/ml=0.27621 g

8 0
1 year ago
Calculate the energy (in kJ) required to heat 10.1 g of liquid water from 55 oC to 100 oC and change it to steam at 100 oC. The
Maksim231197 [3]

Answer:

           \large\boxed{\large\boxed{24.6kJ}}

Explanation:

<u>1. Energy to heat the liquid water from 55ºC to 100ºC</u>

     Q=m\times C\times \Delta T

  • m = 10.1g
  • C = 4.18g/JºC
  • ΔT = 100ºC - 55ºC = 45ºC

     Q=10.1g\times 4.18J/g\ºC\times 45\ºC=1,899.81J

<u>2. Energy to change the liquid to steam at 100ºC</u>

      L=\lambda \times n

  • λ = 40.6kJ/mol
  • n = 10.1g / 18.015g/mol = 0.5606mol

      L=40.6kJ/mol\times 0.5604mol=22.76214kJ=22,762.14J

<u>3. Total energy</u>

       1,899.81J+22,762.14J=24,661.95J\approx24,662J\approx24.6kJ

7 0
3 years ago
In the following net ionic equation, identify each reactant as either a Bronsted-Lowry acid or a Bronsted-Lowry base. HCN(aq) H2
vfiekz [6]

Answer:

Explanation:

The definition of acids and bases by Arrhenius Theory  was modified and extended by  Bronsted-Lowry.

Bronsted-Lowry defined acid as a molecule or ion which donates a proton while a base is a molecule or ions that accepts the proton. This definition can be extended to include acid -base titrations in non-aqueous solutions.

In this theory, the reaction of an acid with a base constitutes a transfer of a proton from the acid to the base.

From the given information:

\mathsf{HCN _{(aq)} + H_2O_{(l)} \to CN^{-}_{(aq)} + H_3O_{(aq)}}

From above:

We will see that HCN releases an H⁺ ion, thus it is a Bronsted-Lowry acid

H_2O accepts the H⁺ ion ,thus it is a Bronsted-Lowry base.

The formula of the reactant that acts as a proton donor is <u>HCN</u>

The formula of the reactant that acts as a proton acceptor is <u>H2O</u>

8 0
2 years ago
Which of the following is NOT a derived unit?<br><br>a) <img src="https://tex.z-dn.net/?f=cm%5E%7B3%7D" id="TexFormula1" title="
seraphim [82]

Answer:

c) kg

Explanation:

Kilograms stands alone. It has to be hooked up to another unit for it to be a derived unit.

I am joyous to assist you anytime.

4 0
2 years ago
Read 2 more answers
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