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3 years ago

Why do the isotopes of a particular element behave differently in nuclear reactions but the same in

Chemistry

2 answers:

Sphinxa [80]3 years ago

7 0

Answer:

A and C

Explanation:

AfilCa [17]3 years ago

4 0

The isotopes of a particular element behave differently in nuclear reactions but the same in chemical reactions because the nuclear reactions involve changes to protons and neutrons in an atom and isotopes differ in neutrons and so they react differently to nuclear changes. While for chemical reactions, the electrons in an atom are involved but the isotopes do not differ in electrons and so they react the same way during chemical changes.

Answer: Option A & C

<u>Explanation:</u>

The reactants of chemical and nuclear reactions are different. The reactants of chemical reactions are the electrons present in the outermost shell of the reactants, while the reactants of nuclear reactions are either the nuclei of reactants or a nuclei and any subatomic particles of other reactants.

Thus it can be understood that chemical reactions consider the electrons while nuclear reactions consider the nuclei or the number of protons and neutrons of the reactants.

The isotopes of elements contain different mass number or we can say different number of neutrons but the number of electrons are same, so they behave differently in nuclear reactions and similar in chemical reactions.

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The following data were obtained in a kinetics study of the hypothetical reaction A + B + C → products. [A]0 (M) [B]0 (M) [C]0 (

Vladimir [108]

Answer:

B. First order, Order with respect to C = 1

Explanation:

The given kinetic data is as follows:

A + B + C → Products

[A]₀ [B]₀ [C]₀ Initial Rate (10⁻³ M/s)

1. 0.4 0.4 0.2 160

2. 0.2 0.4 0.4 80

3. 0.6 0.1 0.2 15

4. 0.2 0.1 0.2 5

5. 0.2 0.2 0.4 20

The rate of the above reaction is given as:

$Rate = k[A]^{x}[B]^{y}[C]^{z}$

where x, y and z are the order with respect to A, B and C respectively.

k = rate constant

[A], [B], [C] are the concentrations

In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.

Order w.r.t A : Use trials 3 and 4

$\frac{Rate3}{Rate4}= [\frac{[A(3)]}{[A(4)]}]^{x}$

$\frac{15}{5}= [\frac{[0.6]}{[0.2]}]^{x}$

$3 = 3^{x} \\\\x =1$

Order w.r.t B : Use trials 2 and 5

$\frac{Rate2}{Rate5}= [\frac{[B(2)]}{[B(5)]}]^{y}$

$\frac{80}{20}= [\frac{[0.4]}{[0.2]}]^{y}$

$4 = 2^{y} \\\\y =2$

Order w.r.t C : Use trials 1 and 2

$\frac{Rate1}{Rate2}= [\frac{[A(1)]}{[A(2)]}]^{x}[\frac{[B(1)]}{[B(2)]}]^{y}[\frac{[C(1)]}{[C(2)]}]^{z}$

we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:

$\frac{160}{80}= [\frac{[0.4]}{[0.2]}]^{1}[\frac{[0.4]}{[0.4]}]^{2}[\frac{[0.2]}{[0.4]}]^{z}$

$1 = (0.5)^{z}$

z = 1

Therefore, order w.r.t C = 1

8 0

3 years ago

A family of four lives in a three-bedroom house and uses an average of 900 kWh of electricity per month. The family cools their

agasfer [191]

<h2>Answer:</h2><h2>The percentage of the family’s total annual electricity that is used to run the two air conditioners for the three summer months = 19.4 %</h2>

Explanation:

Average electricity consumed per month = 900 kWh

The family cools their house for three months during the summer with two window-unit air conditioners

The power consumed by one window-unit air conditioners = 350 kWh

The power consumed by two window-unit air conditioners = 350(2) = 700 kWh

Power consumed for two air conditioners for the three summer months = 700 (3) = 2100 kWh

Total power consumed for 1 year = 900 (12) = 10800kWh

The percentage of the family’s total annual electricity that is used to run the two air conditioners for the three summer months = $\frac{2100}{10800}100$ = 19.4 %

5 0

2 years ago

An example of a colloid which is composed of a liquid dispersed in a gas would be

Aleksandr-060686 [28]

Any colloid consisting of a solid dispersed in a gas is called a smoke. A liquid dispersed in a gas is referred to as a fog. So the answer would be smoke

6 0

2 years ago

Can someone please help me? :(

Dmitriy789 [7]

Answer:

Trade winds

Explanation:

I am not sure

7 0

1 year ago

Calculate the values of ΔU, ΔH, and ΔS for the following process:

ladessa [460]

Answer:

ΔU = 45.814 KJ

ΔH = 46.4375 KJ

ΔS = 18.76 J/K

Explanation:

H2O(l) → H2O(l) → H2O(steam)

298.15K, 1atm ΔHp 373.15K,1atm ΔHv 373.15K,1 atm

∴ ΔHp = Qp = nCpΔT

∴ n H2O = 1 mol

∴ Cp,n = 75.3 J/mol.K

∴ ΔT = 373.25 - 298.15 = 75 K

⇒ Qp = (1 mol)*(75.3 J/mol.K)*(75K) = 5647.5 J

⇒ ΔHp = 5647.5 J = 5.6475 KJ

⇒ ΔH = ΔHp + ΔHv

∴ ΔHv = 40.79 KJ/mol * 1 mol = 40.79 KJ

⇒ ΔH = 5.6475 KJ + 40.79 KJ = 46.4375 KJ

ideal gas:

∴ ΔH = ΔU + PΔV

∴ V1 = nRT1/P1 = ((1)*(0.082)*(298.15))/1 = 24.45 L

∴ V2 = nRT2/P2 = ((1)*(0.082)*(373.15))/ 1 = 30.59 L

⇒ ΔV = V2 - V1 = 6.15 L * (m³/1000L) = 6.15 E-3 m³

∴ P = 1 atm * (Pa/ 9.86923 E-6 atm) = 101325.027 Pa

⇒ ΔU = ΔH - PΔV = 46.4375 KJ - ((101325.027 Pa*6.15 E-3m³)*(KJ/1000J))

⇒ ΔU = 46.4375 KJ - 0.623 KJ

⇒ ΔU = 45.814 KJ

∴ ΔS = Cv,n Ln (T2/T1) + nR Ln (V2/V1)

⇒ ΔS = (75.3) Ln(373.15/298.15) + (1)*(8.314) Ln (30.59/24.45)

⇒ ΔS = 16.896 J/K + 1.863 J/K

⇒ ΔS = 18.76 J/K

3 0

3 years ago