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podryga [215]
2 years ago
8

How does a catalyst increase reaction rate?

Chemistry
1 answer:
Effectus [21]2 years ago
7 0
By increasing the action energy
You might be interested in
A box with a mass of 3.5 kg is pushed along a table. If the net force on the box is 35 Newtons, what is the acceleration of the
Vesnalui [34]

Answer:

<h2>10 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

m is the mass

f is the force

From the question

mass = 3.5 kg

force = 35 N

We have

a =  \frac{35}{3.5}  \\  = 10

We have the final answer as

<h3>10 m/s²</h3>

Hope this helps you

8 0
2 years ago
What assumptions do we make in order to use the Henderson-Hasselbalch equation? a. Both the weak acid and its conjugate base are
zepelin [54]

Answer:

The final and initial concentration of the acid and it's conjugate base are approximately equal, that is we use the weak acid approximation.

Explanation:

The Henderson-Hasselbalch is used to calculate the pH of a buffer solution. It depends on the weak acid approximation.

Since the weak acid ionizes only to a small extent, then we can say that [HA] ≈ [HA]i

Where [HA] = final concentration of the acid and [HA]i = initial concentration of the acid.

It also follows that [A^-] ≈ [A^-]i where [A^-] and[A^-]i refer to final and initial concentrations of the conjugate base hence the answer above.

7 0
2 years ago
Complete this equation for the dissociation of NH4NO3(aq).
Delvig [45]
Dissociation
NH₄NO₃(aq) = NH₄⁺(aq) + NO₃⁻(aq)


the hydrolysis of the cation
NH₄⁺(aq) + H₂O(l) = NH₃(aq) + H₃O⁺(aq)
pH<span><7</span>

5 0
3 years ago
Read 2 more answers
7. NH2CO2NH4(s) when heated to 450 K undergoes the following reaction to produce a system which reaches equilibrium: NH2CO2NH4(s
Taya2010 [7]

Answer:

Value of equilibrium constant is 0.0888

Explanation:

Both NH_{3} and CO_{2} are gaseous. Hence equilibrium constant depends upon partial pressures of NH_{3} and CO_{2}.

Initially no NH_{3} and CO_{2} were present.

Hence mole fraction of NH_{3} and CO_{2} at equilibrium can be calculated from coefficient of NH_{3} and CO_{2} in balanced equation.

Mole fraction of NH_{3} = (number of moles of NH_{3})/(total number of moles of NH_{3} and CO_{2}) = \frac{2moles}{(2+1)moles}=\frac{2}{3}

Mole fraction of CO_{2} = (number of moles of CO_{2})/(total number of moles of NH_{3} and CO_{2}) = \frac{1moles}{(2+1)moles}=\frac{1}{3}

Let's assume both CO_{2} and NH_{3} behaves ideally.

Therefore partial pressure of NH_{3}, P_{NH_{3}}= x_{NH_{3}}.P_{total} and P_{CO_{2}}= x_{CO_{2}}.P_{total}

Where x represents mole fraction

So, P_{NH_{3}}=\frac{2}{3}\times 0.843atm=0.562atm

P_{CO_{2}}=\frac{1}{3}\times 0.843atm=0.281atm

So, K_{p}=P_{NH_{3}}^{2}.P_{CO_{2}}=(0.562)^{2}\times 0.281=0.0888

4 0
3 years ago
.Which of the following is true about elements?
Ludmilka [50]

Answer:

C

Explanation:

6 0
2 years ago
Read 2 more answers
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