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Nostrana [21]
2 years ago
11

How many atoms of hydrogen are in 4Ca(OH)2​

Chemistry
1 answer:
inn [45]2 years ago
5 0

Explanation:

there are 8 atoms of hydrogen

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Which of the following values are not equal to 1 mole?
Gekata [30.6K]

Answer:

none of them are equal to one mole

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Write the complete atomic structure for the Carbon atom. The atomic and mass numbers for carbon are 6 and 12 reapectively.​
sergiy2304 [10]

Answer:

C-12 or C with a 12 superscripted on Upper left and 6 Subscripted on bottom left

Explanation:

Isotopic notation

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Which layer of the atmosphere contains a substance that was created from a product of living things and that protects living thi
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<h2>Answer : Option B) Ozone</h2><h3>Explanation :</h3>

The layer of atmosphere which contains a substance(Ozone) that was created from a product of living thins (oxygen) and protects living things (from harmful UV rays).

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8 0
3 years ago
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Classify the following compounds as ionic or covalent:. a.H2O. b.C6H12O6. c.CuBr2. d.KMnO4. e.C2H6. f.Fe(C2H3O2)3. . How do you
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Ionic compounds are compounds composed of a metal as the cation (positive charge) and non-metals as anion (negative charge). Covalent compounds are composed of both non-metals. In this respect, a) water is covalent b) covalent c) ionic d) ionic e) covalent f) ionic. 
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3 years ago
how many moles of iodine should be added to 750 grams of carbon tetrachloride to prepare a 0.24 m solution
aleksley [76]

Answer:

0.18 mol

Explanation:

Given data

  • Mass of carbon tetrachloride (solvent): 750 g
  • Molality of the solution: 0.24 m
  • Moles of iodine (solute): ?

Step 1: Convert the mass of the solvent to kilograms

We will use the relationship 1 kg = 1,000 g.

750g \times \frac{1kg}{1,000g} =0.750kg

Step 2: Calculate the moles of the solute

The molality is equal to the moles of solute divided by the kilograms of solvent. Then,

m = \frac{moles\ of\ solute }{kilograms\ of\ solvent} \\moles\ of\ solute = m \times kilograms\ of\ solvent = \frac{0.24mol}{kg}  \times 0.750kg = 0.18 mol

6 0
3 years ago
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