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NISA [10]
3 years ago
14

How many calcium atoms would be in a 100 g sample of calcium

Chemistry
1 answer:
KengaRu [80]3 years ago
4 0

 The number of calcium  atoms  that would  be in a  100 g sample of calcium  is 1.505 x 10²⁴  atoms

     <u><em>calculation</em></u>

Step 1 ; find the  moles of calcium

moles = mass÷  molar mass

from periodic table the  molar mass of Ca =40 g/mol

moles is therefore = 100 g÷ 40 g/mol = 2.5 moles

Step 2: use the Avogadro's  law constant to determine the number of atoms  of Calcium

That is According  Avogadro's law   1 mole = 6.02 × 10²³  atoms

                                                           2.5 moles=? atoms

 {(2.5  moles  ×  6.02×10²³  atoms)  / 1 mole}  = 1.505  ×  10²⁴  atoms


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What is the h oh ph and poh of a 0.005m solution of calcium hydroxide?
fiasKO [112]

Answer: [OH^{-}]= 0.01M or 1.0\times 10^{-2}M

[H^{+}]= 1.0\times 10^{-12}M

pH = 12

pOH = 2

Explanation: Calcium hydroxide (Ca(OH)_{2}) is a strong base that dissociates completely.

Dissociation equation of Calcium hydroxide is :

Ca(OH)_{2} \rightarrow Ca^{+2} + 2OH^{-}

1. Concentration of [OH-]

1 mol Ca(OH)_{2} produces 2 mol OH- ions.

The given solution is 0.005M Ca(OH)_{2} , then  concentration of OH- would be twice the concentration of Ca(OH)_{2}

[OH^{-}] = 0.005\times 2 = 0.01M or 1.0\times 10^{-2}M

2.Concentration of [H+]

Concentration of [H+] can be calculated by the formula: [H^{+}] = \frac{Kw}{[OH^{-}]}

kw = ionic product of water and its values is (1\times 10^{-14})

[OH-] = 0.01 M or 1.0\times 10^{-2}M

[H^{+}] = \frac{(1\times 10^{-14})}{[OH^{-}]}

[H^{+}] = \frac{(1\times 10^{-14})}{[0.01]}

[H^{+}] = 1.0\times 10^{-12}M

3. pH value

pH is calculated by the formula : pH = -log[H^{+}]

pH = -log[1.0\times 10^{-12}]

pH = 12

4. pOH value

pOH is calculated by the formula : pOH = 14 - pH

pOH = 14 - 12

pOH = 2

pOH can also be calculated by using a different formula which is :

pOH = -log(OH^{-})

pOH = -log(0.01)

pOH = 2.


3 0
3 years ago
You react 2.33 g of iron (III) chloride with 50.0 mL of 0.500 M solution of sodium phosphate to
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Answer:

2.33g of iron (iii) chloride

50.0 mL of 5.00 M of sodium phosphate

FeCl3 + Na3PO4 > Fe(PO4) + 3NaCl

mol = conc × vol = 0.5 × 50/1000 = 0.025 mol Na3PO4

from the equation:

1 mol of Na3PO4 reacts with 1 mol FeCl3 = 3 mol of NaCl

0.025 mol = x

x = 0.0025 × 3 = 0.075 mol NaCl

mass = 0.075 g × 59 g/mol = 4.425 g NaCl

i guessed all of this so i dont know i it is correct

3 0
3 years ago
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8 0
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andreyandreev [35.5K]
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rounded to 2.09 meters.
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