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yanalaym [24]
3 years ago
6

Which element has 2 valence electrons

Chemistry
1 answer:
dalvyx [7]3 years ago
8 0
Ca
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The mass of an object is described in what??
inessss [21]
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6 0
3 years ago
How can I balance this equation? ____ KClO3 ---&gt; ____ KCl + ____ O2
tensa zangetsu [6.8K]
__ KClO₃ → __ KCl + __ O₂

Left Side:
1 K
1 Cl
3 O

Right Side:
1 K
1 Cl
2 O

Since the least common multiple of 3 and 2 is 6, we need to multiply the compound with 2 oxygen by 3 and the compound with 3 oxygen by 2.

This gives us 2KClO₃ → __ KCl + 3O₂.

However, this equation is still not balanced.

Left Side:
2 K
2 Cl
6 O

Right Side:
1 K
1 Cl
6 O

In order to balance the K and Cl, we need to multiply the KCl compound on the right side by 2.

2KClO₃ → 2KCl + 3O₂
8 0
2 years ago
If you are asked to compare matter in solid, liquid, and gaseous states, which statement best defines a gas?
Elza [17]
I am pretty sure that <span>If I were asked to compare matter in solid, liquid, and gaseous states, the statement which would best defined a gas is </span>highest energy, highest molecular motion, and least dense packaging of molecules. I choose this one because it's not sensible to <span>heat CO2 (in case of safety) and in the last option the amount of energy is not satisfying.

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8 0
3 years ago
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A 6.0 M HCl solution is available. What volume is needed to make 6.00 L of 1.00 M<br> solution?
Ket [755]

Answer:

The volume of the stock solution needed is 1L

Explanation:

Step 1:

Data obtained from the question. This include the following:

Concentration of stock solution (C1) = 6M

Volume of stock solution needed (V1) =?

Concentration of diluted solution (C2) = 1M

Volume of diluted solution (V2) = 6L

Step 2:

Determination of the volume of the stock solution needed.

With the dilution formula C1V1 = C2V2, the volume of the stock solution needed can be obtained as follow:

C1V1 = C2V2

6 x V1 = 1 x 6

Divide both side by 6

V1 = 6/6

V1 = 1L

Therefore, the volume of the stock solution needed is 1L

5 0
2 years ago
What is typically displayed on the y-axis of a solubility curve
tekilochka [14]
<span>Grams of solute per 100 grams of water</span>
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