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Lunna [17]
3 years ago
7

Which number is NOT a perfect square? 25 8 4 1

Chemistry
2 answers:
Artemon [7]3 years ago
5 0
All of them are perfect squares.
notka56 [123]3 years ago
4 0

Answer:

8

Explanation:

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Which statement is true about shape?
slega [8]
B: Liquid take the shape of their container.
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Where is most of the fresh water on earth found?
borishaifa [10]

Answer:

<h3> GLACIERS AND ICE CAPS </h3>

Explanation:

According to the U.S. Geological Survey, most of that three percent is inaccessible. Over 68 percent of the fresh water on Earth is found in icecaps and glaciers, and just over 30 percent is found in ground water. Only about 0.3 percent of our fresh water is found in the surface water of lakes, rivers, and swamps.

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3 years ago
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What happens if atoms lose energy during a change of state?
kvasek [131]

Answer:

The answer is D :)

Explanation:

7 0
3 years ago
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When an aqueous solution of magnesium nitrate is mixed with an aqueous solution of potassium carbonate, ____________.?
spin [16.1K]
  <span>Ca(NO3)2 + Na2CO3 = CaCO3 + 2NaNO3 
Yes a precipitate of Calcium Carbonate is formed since it is insoluble in water. 
Mol Wt of Calcium Nitrate is 164. And that of Calcium Carbonate is 100. 
One mole of Calcium Nitrate produces one mole of Calcium Carbonate. 
i.e. 164 gms will produce 100gms of precipitate 
So, 1.74gms of Calcium Carbonate will be obtained from 2.85gms Calcium Nitrate present in the original solution.</span>
3 0
3 years ago
A 8.249 gram sample of copper is heated in the presence of excess fluorine. A metal fluoride is formed with a mass of 13.18 g. D
levacccp [35]

Answer:

CuF_2 the empirical formula of the metal fluoride.

Explanation:

Mass of copper heated = 8.249 g

Mass of copper fluoride formed = 13.18 g

Mass of fluorine gas in copper fluoride = x

13.18 g = 8.249 g + x\\x= 13.18 - 8.249 g = 4.931 g

Moles of copper :

= \frac{8.249 g}{63.546 g/mol}=0.1298 mol

Moles of fluorine:

= \frac{4.931 g}{18.998 g/mol}=0.2596 mol

For the empirical formula divide the smallest mole of an element with all the moles of elements present in the compound.

Copper= \frac{0.1298 mol}{0.1298 mol}=1\\Fluorine = \frac{0.2596 mol}{0.1298 mol}=2

The empirical formula of the copper fluoride = CuF_2

CuF_2 the empirical formula of the metal fluoride.

6 0
3 years ago
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