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3 years ago

Adding a(n) [Select] ' will slow down a chemical reaction.

Chemistry

1 answer:

leva [86]3 years ago

4 0

Inhibitor

Explanation:

Adding an inhibitor will slow down a chemical reaction. Inhibitors works like a negative catalyst.

The addition of a foreign body can influence the rate of a reaction.

When a foreign body slows down a reaction, it is called an inhibitor.
An inhibitor is a negative catalyst that will slow down a reaction.
Such substances do not allow a reaction to proceed freely and very fast.
They work in the opposite way a normal catalyst would.
It is good to know conditions and chemicals that can inhibit a chemical reaction.

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Catalysts brainly.com/question/8413755

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Light travels in _______.

Thepotemich [5.8K]

Should be C. I think

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3 years ago

What is the solubility of m(oh)2 in a 0.202 m solution of m(no3)2? ksp = 9.05×10−18?

saveliy_v [14]

M(NO3)2 ==> [M2+] + 2 [NO3-] 0.202 M ==> 0.202 M M(OH)2 ==> [M2+] + 2[OH-] 5.05*10^-18 ===> s + [2s]^2 5.05*10^-18 ===> 0.202 + [2s]^2 5.05*10^-18 = 0.202 * 4s^2 4s^2 = 25*10^-18 s^2 = 6.25*10^-18 s = 2.5*10^-9 So, the solubility is 2.5*10^-9

5 0

3 years ago

In order to determine if 2 atoms are cooper what must be the same for each

galina1969 [7]

-number of electrons
-atom size
-charge of the atom
-number of protons

4 0

3 years ago

Dissolution of KOH, ΔHsoln:

swat32

Using Hess's law we found:

1) By adding reaction 10.2 with the reverse of reaction 10.1 we get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) ΔH (10.3)

2) The ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy (ΔH).

The reactions of dissolution (10.1) and neutralization (10.2) are:

KOH(s) → KOH(aq) ΔHsoln (10.1)

KOH(s) + HCl(aq) → H₂O(l) + KCl(aq) ΔHneut (10.2)

1) According to Hess's law, the total change in enthalpy of a reaction resulting from differents changes in various reactions can be calculated as the sum of all the enthalpies of all those reactions.

Hence, to get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)

We need to add reaction 10.2 to the reverse of reaction 10.1

KOH(s) + HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq) + KOH(s)

Canceling the KOH(s) from both sides, we get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)

2) The change in enthalpy for reaction 10.3 can be calculated as the sum of the enthalpies ΔHsoln and ΔHneut:

$\Delta H = \Delta H_{soln} + \Delta H_{neut}$

The enthalpy of reaction 10.1 (ΔHsoln) changed its sign when we reversed reaction 10.1, so:

$\Delta H = \Delta H_{neut} - \Delta H_{soln}$

Therefore, the ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy ΔH.

Learn more here:

brainly.com/question/2082986?referrer=searchResults

brainly.com/question/1657608?referrer=searchResults

I hope it helps you!

6 0

3 years ago

Potassium hydroxide is classified as an arrhenius base because koh contains

mixas84 [53]

Answer: hydroxide ions

Explanation:

According to the Arrhenius concept, an acid is a substance that ionizes in the water to give hydronium ion or hydrogen ion and a bases is a substance that ionizes in the water to give hydroxide ion .

According to the Bronsted Lowry conjugate acid-base theory, an acid is defined as a substance which donates protons and a base is defined as a substance which accepts protons.

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.

As KOH can give hydroxide ions on dissociation , it is considered as arrhenius base.

$KOH\rightarrow K^++OH^-$

8 0

3 years ago