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Effectus [21]
3 years ago
15

4. What is the substance called that dissolves the other substance in a solution?

Chemistry
1 answer:
blagie [28]3 years ago
3 0

Answer:

Solvent

Explanation:

When one substance dissolves into another, a solution is formed. A solution is a homogeneous mixture consisting of a solute dissolved into a solvent

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4. Cuanto electrones se necesitan para<br> pesarlo mismo que un proton?
sdas [7]

Answer:

aproximadamente dos mil

Explanation:

masa_proton/masa_electron=

1.673e-24 / 9.11e-28 = 1836.443468715697

7 0
3 years ago
How many grams are in 2.3 x 10^-4 moles of Ca3(PO4)2
OLEGan [10]
Molar mass 

Ca₃(PO₄)₂ = 310 g/mol

1 mole -------------------> 310 g
2.3x10⁻⁴ mole ---------> ?

m = 2.3x10⁻⁴ * 310 / 1

m = 0.0713 g 

hope this helps!
3 0
3 years ago
Calculate the moles of calcium chloride (CaCl2) needed to react in order to produce 85.00 grams of calcium carbonate (CaCO3). us
BartSMP [9]

Answer:

0.85 mole

Explanation:

Step 1:

The balanced equation for the reaction of CaCl2 to produce CaCO3. This is illustrated below:

When CaCl2 react with Na2CO3, CaCO3 is produced according to the balanced equation:

CaCl2 + Na2CO3 -> CaCO3 + 2NaCl

Step 2:

Conversion of 85g of CaCO3 to mole. This is illustrated below:

Molar Mass of CaCO3 = 40 + 12 + (16x3) = 40 + 12 + 48 = 100g/mol

Mass of CaCO3 = 85g

Moles of CaCO3 =?

Number of mole = Mass /Molar Mass

Mole of CaCO3 = 85/100

Mole of caco= 0.85 mole

Step 3:

Determination of the number of mole of CaCl2 needed to produce 85g (i.e 0. 85 mole) of CaCO3.

This is illustrated below :

From the balanced equation above,

1 mole of CaCl2 reacted to produced 1 mole of CaCO3.

Therefore, 0.85 mole of CaCl2 will also react to produce 0.85 mole of CaCO3.

From the calculations made above, 0.85 mole of CaCl2 is needed to produce 85g of CaCO3

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3 years ago
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3 years ago
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The process by which DNA is copied during the cell cycle
Eduardwww [97]
Cell replication in somatic cells is called mitosis.
Cell replication in gametes (sex cells) is called meiosis.
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