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Rama09 [41]
2 years ago
10

How many atoms of Ca are present in 80.156 grams of Ca? (4 points) 1.2044 × 1024 2.4088 × 1024 4.8270 × 1025 1.9346 × 1027

Chemistry
2 answers:
Alika [10]2 years ago
6 0

Answer: 1.2044 × 10^24

Explanation:

1 mole of calcium is 40 gram

Based on Avogadro's law:

1 mole of any substance has 6.02 x 10^23 atoms, also 1 mole of calcium is 40gram

So, 1 mole of calcium = 6.02 x 10^23 atoms

Also, 1 mole of calcium is 40gram

40 grams of calcium = 6.02 x 10^23 atoms

80.156 grams of calcium = Z atoms

To get the value of Z, cross multiply:

Z atoms x 40 grams = 6.02 x 10^23 atoms x 80.156 grams

40Z = 482.54 x 10^23

Z = (482.54 x 10^23/40)

Z = 12.06 x 10^23

Put result in standard form

Z = 1.206 x 10^24 atoms

Thus, 1.206 x 10^24 atoms of Ca are present in 80.156 grams of Ca

Anika [276]2 years ago
3 0

Answer:

1.204x10^24atoms

Explanation:

From Avogadro's hypothesis, 1mole of any substance contains 6.02x10^23 atoms.

This implies that 1mole of Ca also contains 6.02x10^23 atoms.

1 mole of Ca = 40.078g

Now, if 40.078g of Ca contains 6.02x10^23 atoms,

then 80.156g of Ca will contain = (80.156 x 6.02x10^23)/40.078 = 1.204x10^24atoms

Therefore, 80.156g of Ca contains 1.204x10^24 atoms

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mihalych1998 [28]

Answer:

0.08097 grams of nitrate ions are there in the final solution.

Explanation:

Moles of cobalt(II) nitrate ,n= \frac{4.00 g}{245 g/mol}=0.01633 mol

Volume of the cobalt(II) nitrate solution, V = 100.0 mL = 0.1 L

Molarity=\frac{n}{V(L)}

Let the molarity of the solution be M_1

M_1=\frac{0.01633 mol}{0.1 L}=0.1633 M

A students then takes 4 .00 mL of M_1 solution and dilute it to 275 ml.

M_1=0.1633 M

V_1=4.00 mL

M_2=? (molarity after dilution)

V_2=275 mL (after dilution)

M_1V1=M-2V_2

M_2=\frac{M_1V_1}{V_2}=\frac{0.1633 M\times 4.00 mL}{275 mL}=0.002375 M

Molarity of the of solution after dilution is 0.002375 M.

Co(NO_3)_2(aq)\rightarrow Co^{2+}(aq)+2NO_3^{-}(aq)

1 mol of cobalt(II) nitrate gives 2 moles of nitrate ions. Then 0.002375 M solution of cobalt (II) nitrate will give:

[NO_3^{-}]=\frac{2}{1}\times 0.002375 M=0.004750 M

Moles of nitrate ions = n

Volume of the solution = 275 mL = 0.275 L

Molarity of the nitrate ions = [NO_3^{-}]=0.004750 M

[NO_3^{-}]=\frac{n}{0.275 L}

n = 0.001306 mol

Mass of 0.001306 moles of nitrate ions:

0.001306 mol × 62 g/mol= 0.08097 g

0.08097  grams of nitrate ions are there in the final solution.

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You wish to extract an organic compound from an aqueous phase into an organic layer (three to six extractions on a marco scale).
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Answer:

Explanation:

It will be better to use solvents that are lighter than water, because their density has an influence on the miscibility . This will give you a better separation during extraction.

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On a trip to the natural history museum you find two minerals that are similar in color. You can see from their chemical formula
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Answer:Yes they are in the same mineral group

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2 years ago
A graduated cylinder is filled to 10.0 mL with water and a piece of granite is placed in the cylinder displacing the level to 23
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8 0
2 years ago
Calculate the pka of hypochlorous acid. The ph of a 0.015 m solution of hypochlorous acid has a ph of 4.64.
o-na [289]

Answer:

  • pKa = 7.46

Explanation:

<u>1) Data:</u>

a) Hypochlorous acid = HClO

b) [HClO} = 0.015

c) pH = 4.64

d) pKa = ?

<u>2) Strategy:</u>

With the pH calculate [H₃O⁺], then use the equilibrium equation to calculate the equilibrium constant, Ka, and finally calculate pKa from the definition.

<u>3) Solution:</u>

a) pH

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  • 4.64 = - log [H₃O⁺]

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b) Equilibrium equation: HClO (aq) ⇄ ClO⁻ (aq) + H₃O⁺ (aq)

c) Equilibrium constant: Ka =  [ClO⁻] [H₃O⁺] / [HClO]

d) From the stoichiometry: [CLO⁻] = [H₃O⁺] = 2.29 × 10 ⁻⁵ M

e) By substitution: Ka = (2.29 × 10 ⁻⁵ M)² / 0.015M = 3.50 × 10⁻⁸ M

f) By definition: pKa = - log Ka = - log (3.50 × 10 ⁻⁸) = 7.46

5 0
2 years ago
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