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murzikaleks [220]
2 years ago
7

How many inches is in 10 cm? Use dimensional analysis to show your work. Include the work in your answer.

Chemistry
1 answer:
Serhud [2]2 years ago
3 0

Answer:

5 inches

Explanation:

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If 60 grams of a liquid takes up 120ml how dense is the liquid
Black_prince [1.1K]

Answer:

<h2>Density = 0.5 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

<h3>Density =  \frac{mass}{volume}</h3>

From the question

mass = 60 g

volume = 120 mL

Substitute the values into the above formula and solve

That's

<h3>Density =  \frac{60}{120}  \\  =  \frac{1}{2}</h3>

We have the final answer as

<h3>Density = 0.5 g/mL</h3>

Hope this helps you

4 0
3 years ago
What is the percent by mass of water in Na2SO4 • 10H2O?
bulgar [2K]
We determine the percent by mass of water in the compound by dividing the mass of  water by the total mass. The total mass of Na2SO4.10H2O is equal to 322 g. The mass of water is 180 g. 
       
                    percent by mass of water = (180 / 322)*(100 %) = 55.9%
5 0
3 years ago
Read 2 more answers
Identify two ways that temperature plays a role in chemical changes.
sergejj [24]

Answer:

Increasing the temperature will cause chemical changes to occur faster. Decreasing the temperature, causes the particles to lose energy which causes them to move around less and slower. The less they move, the less collisions occur, and the less reactions occur between the chemicals = slower reaction rate.

Explanation:

3 0
3 years ago
Does anyone know how to do this? the solubility of barium carbonate, BaCO3, is 0.0100 g/L. Its molar mass is 197.3 g/mol. What i
g100num [7]
It's simple, just follow my steps.

1º - in 1 L we have 0.0100~g of BaCO_3

2º - let's find the number of moles.

\eta=\frac{m}{MM}

\eta=\frac{0.0100}{197.3}

\boxed{\boxed{\eta=5.07\times10^{-5}~mol}}

3º - The concentration will be

C=5.07\times10^{-5}~mol/L

But we have this reaction

BaCO_3\rightleftharpoons Ba^{2+}+CO_3^{2-}

This concentration will be the concentration of Ba^{2+}~~and~~CO_3^{2-}

K_{sp}=\frac{[Ba^{2+}][CO_3^{2-}]}{[BaCO_3]}

considering [BaCO_3]=1~mol/L

K_{sp}=[Ba^{2+}][CO_3^{2-}]

and

[Ba^{2+}]=[CO_3^{2-}]=5.07\times10^{-5}~mol/L

We can replace it

K_{sp}=(5.07\times10^{-5})*(5.07\times10^{-5})

K_{sp}\approx25.70\times10^{-10}

Therefore the K_{sp} is:

\boxed{\boxed{\boxed{K_{sp}\approx2.57\times10^{-11}}}}
4 0
3 years ago
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What group of substances does this substance belong to?
HACTEHA [7]
In comparison see it is very easy in goolge
4 0
3 years ago
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