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3 years ago

Write the rate law for the following elementary reaction:

Chemistry

1 answer:

Harrizon [31]3 years ago

6 0

Answer:

Rate law = k1 [N2O5]^2

Explanation:

Rate law only cares about reactants and rate law can only be determined experimentally or by using the coefficients of reactants in elementary reactions

If 2A---->B is an elementary reaction... rate law for this is rate=k[A]^2

So if we look at your question...2N2O5---->2N2O4 + 02 (I think you are missing O2 in question because otherwise equation is unbalanced)

Rate law = k1 [N2O5]^2

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g Write a molecular equation for the precipitation reaction that occurs (if any) when the following solutions are mixed. If no r

Mrac [35]

Answer:

CuBr₂(aq) + Pb(CH₃COO)₂(aq) → Cu(CH₃COO)₂(aq) + PbBr₂ (s)↓

Explanation:

We identify the reactants:

CuBr₂ and Pb(CH₃COO)₂

The products will be: Cu(CH₃COO)₂ and PbBr₂

You may know these information:

Salts from acetate are soluble.

Bromide can make solid salts with these cations: Ag⁺, Pb²⁺, Hg₂²⁺, Cu⁺

PbBr₂ is formed, so this will be our precipitate

The equation is:

CuBr₂(aq) + Pb(CH₃COO)₂(aq) → Cu(CH₃COO)₂(aq) + PbBr₂ (s)↓

8 0

2 years ago

If nitrogen (N) has 2 naturally occurring isotopes, nitrogen-14 (78.3%) and nitrogen-16 (21.7%), what is its average r.a.m.?

leva [86]

Answer:

14.434 r.a.m.

Explanation:

The atomic mass of an element is a weighted average of its isotopes in which the sum of the abundance of each isotope is equal to 1 or 100%.

∵ The atomic mass of N = ∑(atomic mass of each isotope)(its abundance)

∴ The atomic mass of N = (atomic mass of N-14)(abundance of N-14) + (atomic mass of N-16)(abundance of N-16)

atomic mass of N-14 = 14.0 r.a.m, abundance of N-14 = percent of N-14/100 = 78.3/100 = 0.783.

atomic mass of N-16 = 16.0 r.a.m, abundance of N-16 = percent of N-16/100 = 21.7/100 = 0.217.

∴ The atomic mass of N = (atomic mass of N-14)(abundance of N-14) + (atomic mass of N-16)(abundance of N-16) = (14.0 r.a.m)(0.783) + (16.0 r.a.m)(0.217) = 14.434 r.a.m.

5 0

2 years ago

Without constants you would not know which variable affected the

bonufazy [111]

Two independent variables could change at the same time, and you would not know which variable affected the dependent variable

4 0

2 years ago

Express 7.25 in scientific notation

sergeinik [125]

Considering that scientific notation has to move the decimal over so there is only one digit in the Tens place, that is the scientific notation.

8 0

2 years ago

Read 2 more answers

Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reactionP4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°r

weqwewe [10]

Answer:

Therefore $\bigtriangledown H^\circ_{rxn}$ = -1835 KJ

Explanation:

Enthalpy is denoted by H.

Enthalpy: Total heat change in a chemical reaction is called enthalpy.

The change of entalpy of a reaction is denoted by $\bigtriangledown H^\circ_{rxn}$

Hass's Law:The change in enthalpy of any process can be determined by calculating the sum of change in enthalpy of each of the steps involved in the process.

g= gas

S= solid

P₄(g)+10Cl₂(g)→ 4Cl₅(s) $\bigtriangledown H^\circ_{rxn}$ =?

PCl₅(s)→ PCl₃(g)+Cl₂(g) .......(1) $\bigtriangledown H^\circ_{rxn}$ = +157KJ

P₄(g)+6Cl₂(g)→ 4PCl₃(g).............(2) $\bigtriangledown H^\circ_{rxn}$ = -1207 KJ

If we flip a reaction the value of enthalpy will be change positive to negative or nagative to positive but the numerical value will be remain same.

We need rearrange the equation (1) because in the required equation Cl₂ is on the left side. So we flip the first equation.

PCl₃(g)+Cl₂(g)→PCl₅(s)......(3) $\bigtriangledown H^\circ_{rxn}$ = -157KJ

Multiplying 4 with equation (3)

4 PCl₃(g)+4Cl₂(g)→4PCl₅(s)......(4) $\bigtriangledown H^\circ_{rxn}$ =4×( -157)KJ= -628 KJ

Adding equation (2) and (4) we get

P₄(g)+6Cl₂(g)+4 PCl₃(g)+4Cl₂(g)→4PCl₃(g)+4PCl₅(s) $\bigtriangledown H^\circ_{rxn}$ =( -1207-628)KJ

⇒P₄(g)+10Cl₂(g)→4PCl₃(g)-4PCl₃(g)+4PCl₅(s) $\bigtriangledown H^\circ_{rxn}$ = - 1835KJ

⇒P₄(g)+10Cl₂(g)→ 4Cl₅(s) $\bigtriangledown H^\circ_{rxn}$ = -1835 KJ

Therefore $\bigtriangledown H^\circ_{rxn}$ = -1835 KJ

5 0

2 years ago