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lyudmila [28]
3 years ago
13

How many grams of AgNO3 are needed to prepare a 75.00 mL solution that is 6.5% (m/v)?

Chemistry
1 answer:
FrozenT [24]3 years ago
5 0

Answer:

Volume of AgNO₃ = 4.9 ml (Approx)

Explanation:

Given:

Total solution = 75 ml

Volume of AgNO₃ = 6.5%

Find:

Volume of AgNO₃

Computation:

Volume of AgNO₃ = Total solution x Volume of AgNO₃

Volume of AgNO₃ = 75 x 6.5%

Volume of AgNO₃ = 4.875

Volume of AgNO₃ = 4.9 ml (Approx)

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Calculate each of the following quantities:<br> (a) Total number of ions in 38.1 g of SrF₂
adelina 88 [10]

The total number of ions in 38.1 g of SrF₂ is 5.479 x 10²³.

<h3>What are ions?</h3>

Ions are the elements with a charge on them. It happens when they share electrons with other atoms to form a compound.

We have to calculate the total number of ions in 38.1 g of .

The molar mass of SrF₂ = 125.62 g/mol

The number of moles = 38.1 g of  1.0 mol / 125.62  = 0.30329 moles

Given that, total moles of SrF₂ ions in  = 1.0 mol of + 2.0 moles of  = 3.0 moles

Total moles of ions in 0.30329 moles of

= (0.30329 moles of SrF₂) x 3.0 / 1.0  = 0.90988 mol ions

We know that,

1.0 mole of ions = 6.023 x 10²³ ions

Thus, the number of total ions = ( 0.90988 mol ions) x  6.023 x 10²³ / 1.0 mol = 5.479 x 10²³ ions

Thus, the number of ions is in 38.1 g of 5.479 x 10²³ ions

To learn more about ions, refer to the link:

brainly.com/question/14295820

#SPJ4

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1 year ago
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