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STALIN [3.7K]
3 years ago
12

What is the purpose of a catalyst?

Chemistry
1 answer:
marusya05 [52]3 years ago
8 0

Answer:

option c

Explanation:

as catalyst increases the reaction while activation energy is inversely proportional to rate of reaction, so option c is correct

You might be interested in
True or False: Bald eagles have white heads, which is an inherited trait.
vredina [299]

Answer:

A true

Explanation:

Bald eagles have white heads distinguishing them from the rest of their body(which is dark brown

8 0
3 years ago
If the conjugate base of a molecule has a pkb of 1.4, what would you expect the molecule to be?
maw [93]

If the conjugate base of a molecule has a pKb of 1.4, the molecule should be a Weak Acid.

Notice this question gives us the pKb of the molecule, not the pKa. Because of this, the pH scale basically gets reversed, so lower numbers in pKb correlate with stronger bases, and higher numbers in pKb correlate with stronger acids - the exact opposite of the pH scale.

It's important to make sure you completely understand the terms of conjugate base, conjugate acid, pKb, pKa, and how they all relate. It's easy to mix up the meanings of these definitions.

Here are the two other pieces of information you need to know to correctly answer this question:

  • Strong acids have a weak conjugate base.
  • Strong bases have a weak conjugate acid.

So if the problem says you have a strong conjugate base, then the molecule must be a weak acid. To illustrate this, think of ammonium, NH4+. Ammonium is a weak acid, but the conjugate base of ammonium is ammonia, NH3, which is a reasonably good base.

Learn more about conjugate base here : brainly.com/question/22514615

#SPJ4

3 0
2 years ago
What kind (ionic or covalent) bond is FeCl2 + Na2CO3? (iron (II) chloride + sodium carbonate).
kipiarov [429]

Answer:

corbonate

Explanation:

i think so. not sure

6 0
3 years ago
chegg write a net ionic equation describing the oxidation of no2 2 to no3 2 by o2 in a basic solution.
Marina86 [1]

When the same species undergoes both oxidation and reduction in a single redox reaction, this is referred to as a disproportionation. Therefore, divide it into two equal reactions.

NO2→NO^−3

NO2→NO

and do the usual changes

First, balance the two half reactions:

3. NO2 +H2O →NO^−3 + 2 H^+ + e−

4. NO2 +2 H^+ + 2e− → NO + H2O

Now multiply one or both half-reactions to ensure that each has the same number of electrons. Here, Eqn (3) x 2 results in each half-reaction having two electrons:

5. 2 NO2 + 2 H2O → 2 NO^−3 + 4H^+ + 2e−

Now add Eqn 4 and 5 (the electrons now cancel each other):

3NO2 + 2H^+ + 2H2O → NO + 2 NO−3 + H2O + 4H+

and cancel terms that’s common to both sides:

3NO2 + H2O → NO + 2NO^−3 + 2H+

This is the net ionic equation describing the oxidation of NO2 to NO3 in basic solution.

Learn more about balancing equation here:

brainly.com/question/26227625

#SPJ4

7 0
2 years ago
Suppose a 2.95 g of potassium iodide is dissolved in 350. mL of a 62.0 m M aqueous solution of silver nitrate. Calculate the fin
STALIN [3.7K]

Answer : The final molarity of iodide anion in the solution is 0.0508 M.

Explanation :

First we have to calculate the moles of KI and AgNO_3.

\text{Moles of }KI=\frac{\text{Mass of }KI}{\text{Molar mass of }KI}

Molar mass of KI = 166 g/mole

\text{Moles of }KI=\frac{2.95g}{166g/mole}=0.0178mole

and,

\text{Moles of }AgNO_3=\text{Concentration of }AgNO_3\times \text{Volume of solution}=0.0620M\times 0.350L=0.0217mole

Now we have to calculate the limiting and excess reagent.

The given chemical reaction is:

KI+AgNO_3\rightarrow KNO_3+AgI

From the balanced reaction we conclude that

As, 1 mole of KI react with 1 mole of AgNO_3

So, 0.0178 mole of KI react with 0.0178 mole of AgNO_3

From this we conclude that, AgNO_3 is an excess reagent because the given moles are greater than the required moles and KI is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of AgI

From the reaction, we conclude that

As, 1 mole of KI react to give 1 mole of AgI

So, 0.0178 moles of KI react to give 0.0178 moles of AgI

Thus,

Moles of AgI = Moles of I^- anion = Moles of Ag^+ cation = 0.0178 moles

Now we have to calculate the molarity of iodide anion in the solution.

\text{Concentration of }AgNO_3=\frac{\text{Moles of }AgNO_3}{\text{Volume of solution}}

\text{Concentration of }AgNO_3=\frac{0.0178mol}{0.350L}=0.0508M

Therefore, the final molarity of iodide anion in the solution is 0.0508 M.

3 0
3 years ago
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