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stiv31 [10]
4 years ago
12

_Ca + _H20 --> _Ca(OH)2 + _H2

Chemistry
1 answer:
Liono4ka [1.6K]4 years ago
5 0
2Ca + 2H2O —-> Ca(OH)2 + H2
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2.7g of aluminum metal is placed into solution hydrochloric acid to form aluminum chloride and hydrogen gas.
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3 years ago
How do isotopes of a givin element similar
BlackZzzverrR [31]

Answer:

All isotopes of an element have the same number of protons and electrons, which means they exhibit the same chemistry. The isotopes of an element differ only in their atomic mass, which is given by the mass number (A), the sum of the numbers of protons and neutrons.

Explanation:

Hope this <em><u>Helped!</u></em> :D

5 0
3 years ago
Read 2 more answers
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

5 0
3 years ago
Read 2 more answers
What compound is formed when 2,2-dimethyloxirane (2-methyl-1,2-epoxypropane) is treated with ethanol containing sulfuric acid
vladimir1956 [14]

Answer:

2-ethoxy-2-methylpropan-1-ol

Explanation:

On this reaction, we have an "<u>epoxide"</u> (2-methyl-1,2-epoxypropane). Additionally, we have <u>acid medium</u> (due to the sulfuric acid H_2SO_4). The acid medium will produce the <u>hydronium ion</u> (H^+). This ion would be attacked by the oxygen of the epoxide. Then a <u>carbocation</u> would be produced, in this case, the most stable carbocation is the <u>tertiary one</u>. Then an <u>ethanol</u> molecule acts as a nucleophile and will attack the carbocation. Finally, a <u>deprotonation </u>step takes place to produce <u>2-ethoxy-2-methylpropan-1-ol</u>.

See figure 1

I hope it helps!

3 0
3 years ago
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