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3 years ago

Help it’s a quiz!! And it’s timed?

Chemistry

2 answers:

yawa3891 [41]3 years ago

8 0

Sorry man i sad to see i’m late

patriot [66]3 years ago

8 0

Izkskdmdod

Good luck oh say

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Show how the following are formed

bogdanovich [222]

I think that beryllium is formed in alcohol I think

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3 years ago

A hydrate is a compound that has water molecules within its crystal structure. Magnesium sulfate heptahydrate, MgSO4·7H2O, is a

Alenkasestr [34]

Answer:

We are heating the sample repeatedly to become a pure compound of only MgSO4 (withot H2O) and a constant mass.

Explanation:

Step 1: Data given

Mass of MgSO4·7H2O = 5.06 grams

The remaining MgSO4 had a constant mass of 2.47 grams.

Step 2: Explain why the sample in the crucible was heated repeatedly until the sample had a constant mass.

Before heating the compound has magnesium sulfate and water.

The total mass of this compound is 5.06 grams

By heating we try to eliminate the water.

After heating there remain mgSO4 with a mass of 2.47 grams

This means 5.06 - 2.47 = 2.59 grams is water. All of this is eliminated.

The heating process happens repeatedly to make sure the final compound is pure. So the 2.47 grams os only MgSO4. If the mass would not be constant. It means the compound is not pure, the not all the water is eliminated yet.

So we are heating the sample repeatedly to become a pure compound of only MgSO4 (withot H2O) and a constant mass.

4 0

3 years ago

wariber [46]

The relationship between independent and dependent variables is constant. The independent variables are not affected and do not change while dependent variables rely on something or a particular change.

Hope this helps and have a good day/night!

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3 years ago

In order to become stable, Potassium, K, will _______ and have a resulting charge of _____.

svlad2 [7]

Answer:

Explanation:

it doesn't matter how many protons k has all the answers have electron with the positive charge when is negative the question is trying to distract you D is the only one that would use the electrons sign correctly

4 0

2 years ago

A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati

Sedaia [141]

Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.

$\text{calculating the initial HI}= \frac{mol}{V}$

$=\frac{9.3 \times 10 ^ -3}{2}$

$=0.00465 \ Mol$

$\text{Similarly}\ \ I_2 \ \ \text{follows} \ \ H_2 = 0 }$

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

$HI = 0.00465 - 2x\\\\ I_{2} \ eq = H_2 \ eq = 0 + x \\\\$

It is defined that:

$I_2 = 6.29 \times 10^{-4} \ M \\\\x = I_2 \\\\$

$HI \ eq= 0.00465 - 2x \\$

$=0.00465 -2 \times 6.29 \times 10^{-4} \\\\ = 0.00465 -\frac{25.16 }{10^4} \\\\ = 0.003392\ M$

Now, we calculate the position:

For the reaction $H 2(g) + I 2(g)\rightleftharpoons 2HI(g)$ , you can calculate the value of Kc at 1000 K.

data expression for Kc

$2HI \rightleftharpoons H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}$

$= \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386$

calculating the reverse reaction

$H_2(g) + I_2(g)\rightleftharpoons 2HI(g)$

$Kc = \frac{1}{Kc} \\\\$

$= \frac{1}{0.034386}\\ \\= 29.081\\$

7 0

3 years ago