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3 years ago

2. What happens to the reaction rate if a catalyst is added?

Chemistry

2 answers:

Ghella [55]3 years ago

6 0

Answer:

A catalyst speeds up a chemical reaction

Explanation:

It increases the reaction rate by lowering the activation energy for a reaction.

mrs_skeptik [129]3 years ago

6 0

It will speed up the chemical reaction.

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Which expression is equivalent to (q Superscript 6 Baseline) squared?

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Answer:

The answer is c

Explanation:

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2 years ago

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Kilograms represented by the mass defect for oxygen-16: 2.20 × 10 -28 kg What is the nuclear binding energy for oxygen-16?

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1.98 × 10^-11 J i just took it this is the right awnser

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3 years ago

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2.What is the chemical formula for ammonium nitrite

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Answer:

NH4NO3

Explanation:

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2 years ago

At 25∘C, the decomposition of dinitrogen pentoxide, N2O5(g), into NO2(g) and O2(g) follows first-order kinetics with k=3.4×10−5

Annette [7]

Answer:

4600s

Explanation:

$2N_{2}O_{5}(g) - - -> 4NO_{2}(g) +O_{2}$

For a first order reaction the rate of reaction just depends on the concentration of one specie [B] and it’s expressed as:

$-\frac{d[B]}{dt}=k[B] - - - -\frac{d[B]}{[B]}=k*dt$

If we have an ideal gas in an isothermal (T=constant) and isocoric (v=constant) process.

PV=nRT we can say that P = n so we can express the reaction order as a function of the Partial pressure of one component.

$-\frac{d[P(B)]}{P(B)}=k*dt$

$-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=k*dt$

Integrating we get:

$\int\limits^p \,-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=\int\limits^ t k*dt$

$-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])=k(t_{2}-t_{1})$

Clearing for t2:

$\frac{-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])}{k}+t_{1}=t_{2}$

$ln[P(N_{2}O_{5})]=ln(650)=6.4769$

$ln[P(N_{2}O_{5})_{o}]=ln(760)=6.6333$

$t_{2}=\frac{-(6.4769-6.6333)}{3.4*10^{-5}}+0= 4598.414s$

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2 years ago

A solution of potassium cyanide, KCN, is made by dissolving 3.2 g of KCN in water. When this solution reacts with excess hydroch

Rama09 [41]

Answer:

22.7%

Explanation:

We must first put down the equation of reaction to guide our work while solving the problem.

KCN(aq) + HCl (aq)--> KCl(aq) + HCN(aq)

The questions specifically state that HCl is in excess so KCN is the limiting reactant.

Number of moles of KCN reacted= mass of KCN reacted / molar mass of KCN

Mass of KCN reacted= 3.2 g

Molar mass of KCN= 65.12 g/mol

Number of moles of KCN= 3.2/65.12 g/mol= 0.049 moles

Theoretical yield of HCN is obtained thus;

From the reaction equation;

1 mol of KCN produced 1 mole of HCN thus 0.049 moles of KCN will produce 0.049 moles of HCN.

Mass of HCN = number of moles ×molar mass

Molar mass of HCN= 27.0253 g/mol

Hence mass of HCN produced= 0.049mol × 27.0253 g/mol= 1.32g of HCN

Actual yield of HCN= 0.30g

% yield= actual yield/ theoretical yield ×100

% yield= 0.30/1.32 ×100

%yield= 22.7%

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3 years ago

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