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3 years ago

PLEASE ANSWER...........WILL MARK BRAINLIEST!!!!!!!

Chemistry

1 answer:

ANTONII [103]3 years ago

3 0

Answer:

<h2>The Alkali metal halide may precipitate or there may be no change at all</h2>

Explanation:

Alkali metal cations are positively charged. Halogen anions are negatively charged. When a solution of Alkali metal cations is added to a solution of Halogen anions, there are two possibilities :

The alkali metal halide( salt formed from reation of the two ions) may precipitate if the Ionic product is higher than the Solubility product.
However, if it can remain in the solution, it will remain so. No chemical changes happen with respect to these both ions. Nothing willl happen.

There is no reaction happening in either of the cases because both species are already in ionic form before addition, hence they continue to be in this form.

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The activation energy of a certain uncatalyzed reaction is 64 kJ/mol. In the presence of a catalyst, the Ea is 55 kJ/mol. How ma

Ksivusya [100]

Answer:

About 5 times faster.

Explanation:

Hello,

In this case, since the Arrhenius equation is considered for both the catalyzed reaction (1) and the uncatalized reaction (2), one determines the relationship between them as follows:

$\frac{k_1}{k_2}=\frac{Aexp(-\frac{Ea_1}{RT} )}{Aexp(-\frac{Ea_2}{RT})} \\\frac{k_1}{k_2}=\frac{exp(-\frac{Ea_1}{RT} )}{exp(-\frac{Ea_2}{RT})}$

By replacing the corresponding values we obtain:

$\frac{k_1}{k_2}=\frac{exp(-\frac{55000J/mol}{8.314J/molK*673.15K} )}{exp(-\frac{64000J/mol}{8.314J/molK*673.15K} )} =4.8$

Such result means that the catalyzed reaction is about five times faster than the uncatalyzed reaction.

Best regards.

4 0

3 years ago

Which of the following is the correct formula for barium (II) sulfate?

nasty-shy [4]

BaSO4 is the correct formula for barium (ll) sulfate

5 0

3 years ago

Read 2 more answers

What volume of 6.58M HCI is needed to make 500. mL of 3.00M HCI?

zmey [24]

Answer:

228 mL

Explanation:

M1*V1 = M2*V2

M1 = 6.58 M

V1 = ?

M2 = 3.00 M

V2 = 500 mL

V1 = M2*V2/M1 = 3.00M*500.mL/6.58 M = 228 mL

3 0

3 years ago

How much heat do you need to raise the temperature of 150 g of ice from -30°C to -15°c?

OLga [1]

To solve this question, you must use the formula: q=mc(change in temperature), where q is heat, m is mass, C is specific heat and temperature change is temperature change. The specific heat for ice is 2.1kJ/Kg x K (given). The change in temperature is 15 degrees Celsius (which you should change to kelvins so you can cancel out units), or 273 + 15 = 288K. The mass is 150 grams, which is 0.15 kg. Now, we can solve for q, heat. We will do this by substituting variables into the formula. After simplifying and cancelling out units, the answer we get is: 90.72kJ.

4 0

3 years ago

How many moles of water would form the reaction of exactly 58.3 grams of magnesium hydroxide

Marat540 [252]

Answer:

$\boxed{\text{2.00 mol}}$

Explanation:

We know we will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

You don't tell us what the reaction is, but we can solve the problem so long as we balance the OH.

M_r: 58.32

Mg(OH)₂ + … ⟶ … + 2HOH

m/g: 58.3

(a) Moles of Mg(OH)₂

$\text{Moles of Mg(OH)$_{2}$} =\text{58.3 g Mg(OH)$_{2}$} \times \dfrac{\text{1 mol Mg(OH)$_{2}$}}{\text{58.32 g Mg(OH)$_{2}$}}\\\\=\text{0.9997 mol Mg(OH)$_{2}$}$

(b) Moles of H₂O

The molar ratio is 2 mol H₂O = 1 mol Mg(OH)₂.

$\text{Moles of H$_{2}$O}= \text{0.9995 mol Mg(OH)$_{2}$} \times \dfrac{\text{2 mol {H$_{2}$O}}}{ \text{1 mol Mg(OH)$_{2}$}}\\\\= \textbf{2.00 mol H$_{2}$O}$

The reaction will form $\boxed{\textbf{2.00 mol}}$ of water.

6 0

3 years ago