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3 years ago

How does a catalyst increase the rate of a reaction?

Chemistry

1 answer:

Norma-Jean [14]3 years ago

6 0

Answer:

C) It provides a lower activation energy for the reaction is the correct answer.

Explanation:

A catalyst increases the rate of the chemical reaction by lowering the activation energy for a reaction.

Catalyst is used to increase the reaction rate, it remains unchanged in the chemical reaction and it does not change the equilibrium constant.

Activation energy is a minimum amount of energy required to initiate the reaction.

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Condensation point and freezing point of argon in KELVIN.

vesna_86 [32]

Answer:

Condensation: 423.3 K

Freezing: 83.96 K

(this is all i could figure out :) hope it helps)

4 0

3 years ago

How much ice at a temperature of -17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.

Tanya [424]

Answer:

An insulated beaker with negligible mass contains liquid water with a mass of 0.205kg and a temperature of 79.9 °C How much ice at a temperature of −17.5 °C must be dropped into the water so that the final temperature of the system will be 31.0 °C? Take the specific heat for liquid water to be 4190J/Kg.K, the specific heat for ice to be 2100J/Kg.K, and the heat of fusion for water to be 334000J/kg.

The answer to the above question is

Therefore 0.1133 kg ice at a temperature of -17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.0 °C

Explanation:

To solve this we proceed by finding the heat reaquired to raise the temperature of the water to 31.0 C from 79.9 C then we use tht to calculate for the mass of ice as follows

ΔH = m×c×ΔT

= 0.205×4190×(79.9 -31.0) = 42002.655 J

Therefore fore the ice, we have

Total heat = mi×L + mi×ci×ΔTi = mi×334000 + mi × 2100 × (0 -−17.5) = 42002.655 J

370750×mi = 42002.655 J

or mi = 0.1133 kg

Therefore 0.1133 kg ice at a temperature of -17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.0 °C

5 0

3 years ago

This is a reaction going on in your muscle cells right this very minute:The enzyme triose phosphate isomerase catalyzes this rea

Arisa [49]

Explanation:

The given reaction is as follows.

$E + S \rightleftharpoons ES \xrightarrow[]{k_{2}} E + P$

Here, [E] = triose phosphate isomerase = 0.1 $nm = 0.1 \times 10^{-9}m$

[S] = Dihydroxy acetone phosphate = 5 $\mu m = 5 \times 10^{-6}m$

[P] = Glyceraldehyde-3-phosphate = 2 $\mu m = 2 \times 10^{-6}m$

Therefore, velocity of the reaction will be as follows.

v = $\frac{d[P]}{dt}$ = $\frac{K_{2}[E][S]}{K_{M} + [S]}$

where, $K_{M}$ = Michaelic menten constant = $\frac{K_{1} + K_{2}}{K_{1}}$

v = $\frac{900 \times 0.1 \times 10^{-9}m \times 5 \times 10^{-6}m}{10^{-5} + 5 \times 10^{-6}}$

= $30 \times 10^{-9} m$

or, = 30 nm/s

Hence, we can conclude that the actual velocity of the forward reaction under physiologic conditions if KM = 10 μM is 30 nm/s.

5 0

3 years ago

Pls help me here, thanks in advance

Marat540 [252]

Answer:

What's the problem???

4 0

3 years ago

Read 2 more answers

Calculate the enthalpy of combustion, δh∘comb, for c6h14. you'll first need to determine the balanced chemical equation for the

Vedmedyk [2.9K]

<span>Answer: Enthalpy Change = (6 x -393.5) + (7 x -285.8) - (-204.6) + (19/2) 0.....??? like.. (6 x Enth CO2) + ( 7 x Enth H2O) - (Enth C6H14) + (19/2) Enth O2</span>

3 0

4 years ago