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g What amounts of 45% pure silver and 50% pure silver should be mixed to obtain 14 grams of 46% pure silver

Sagot :

Jufsanabriasaidn

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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