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Sagot :
Answer:
11.2g of the 45% pure silver and 2.8g of the 50% pure silver.
Explanation:
14g of 46% pure silver contains:
14g*46% = 6.44g of silver are required. Thus, we can write:
6.44 = X*50% + Y*45% (1)
Where X is the mass of 50% pure silver and Y the mass of 45% pure silver.
As the mass of the sample must be 14g:
14g = X + Y (2)
Replacing (2) in (1):
6.44 = (14-Y)*50% + Y*45%
6.44 = 7 - 0.5Y + 0.45Y
-0.56 = -0.05Y
Y = 11.2g of the 45% pure silver
And:
14g-11.2g = X
X =
2.8g of the 50% pure silver
The required amount to obtain 14g of 46% pure silver is 2.8g of 50% silver and 11.2g of 45% silver.
To solve this question, we can assume it's a mathematical problem more than a chemistry problem.
let x represent the amount of 45% of pure Ag
let y represent the amount of 50% of pure Ag
But from the question, we only need 14g from both x and y
We can write an equation to represent this
[tex]x+y=14...equation(i)[/tex]
likewise, we can say that
[tex]0.45x+0.50y=14*0.46\\0.45x+0.50y=6.44...equation(ii)[/tex]
Solving both equations simultaneously,
[tex]x+y=14...equation(i)\\0.45x+0.5y=6.44...equation(ii)[/tex]
from equation (i)
[tex]x+y=14\\x=14-y...equation(iii)[/tex]
substitute equation (iii) into equation (ii)
[tex]0.45x+0.5y=6.44\\x=14-y\\0.45(14-y)+0.5y=6.44\\6.3-0.45y+0.5y=6.44\\[/tex]
collect like terms
[tex]-0.45y+0.50y=6.44-6.3\\0.05y=0.14\\y=\frac{0.14}{0.05}\\y=2.8[/tex]
Now we know the value of the 50% Ag which is 2.8g. Let's substitute it's value into equation (i) and solve for x
[tex]x+y=14\\x+2.8=14\\x=14-2.8\\x=11.2g[/tex]
Learn more about definite proportions here
https://brainly.com/question/1496091
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