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maksim [4K]
3 years ago
10

A jeweler has five rings, each weighing g, made of an alloy of silver and gold. She decides to melt down the rings and add enoug

h silver to reduce the gold content to . How much silver should she add
Chemistry
1 answer:
Kryger [21]3 years ago
5 0

Complete question :

A jeweler has five rings, each weighing 18 g, made of an alloy of 10% silver and 90% gold. She decides to melt down the rings and add enough silver to reduce the gold content to 75%. How much silver should she add?

Answer:

Kindly check explanation

Explanation:

Given the following :

Weight of each alloy = 18g

Initial % of Silver = 10% = 0.1

Initial % of Gold = 90% = 0.9

New % of gold = 75% = 0.75

Hence, new % of silver will be (1 - 0.75) = 0.25

Let the amount of silver to be added = s

Hence,

Initial gold content = final gold content + added silver

18 * 0.9 = 0.75(18 + s)

16.2 = 13.5 + 0.75s

16. 2 = 13.5 + 0.75s

16.2 - 13.5 = 0.75s

2.7 = 0.75s

s = 2.7 / 0.75

s = 3.6

S = 3.6

Hence the amount of silver needed per ring in other to reduce the gold content = 3.6g

Total for the 5 rings = (3.6 * 5) = 18g

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Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.76
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The given question is incomplete, the complete question is:

Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.763g of oxygen. For a given mass of oxygen, what is the lowest whole-number mass ratio of lead in the two compounds that combines with a given mass of oxygen?

Answer:

The lowest whole-number mass ratio in the two compounds is 1:2.

Explanation:

There is a need to find the mole ratio between lead and oxygen atoms in order to find the whole-number mass ratio of lead in the two compounds. In the first compound, the given mass of lead is 2.98 grams, the molar mass of lead is 207.2 gram per mole.  

The no. of moles can be determined by using the formula,  

moles = mass/molecular mass

moles = 2.98 g/207.2 g/mol

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The mass of oxygen in the compound I is 0.461 grams, the molecular mass of oxygen is 16 gram per mol.  

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The ratio between the lead and oxygen in the compound I is 0.0144/0.0288 = 1:2

On the other hand, in the compound II, the mass of lead given is 9.89 grams, therefore, the moles of lead in compound II is,  

moles = 9.89 g / 207.2 g/mol

= 0.0477 moles

The mass of oxygen given in compound II is 0.763 grams, the moles of oxygen present in the compound II is,  

moles = 0.763 g / 16 g

= 0.0477 moles

The ratio between the lead and oxygen in the compound II is, 0.0477 moles lead /0.0477 moles oxygen = 1:1

Hence, of the two compounds, the lowest ratio is found in the compound I, that is, 1:2.  

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