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densk [106]
3 years ago
8

A material has an ASTM grain size number of 7. Determine the magnification, if the number of grains per square inch observed is:

i. 64 grains/in ii. 500 grains/in? iii. 20 grains/in-
Chemistry
1 answer:
Luden [163]3 years ago
8 0

Answer:

A) M = 100X

B) M = 36X

C) M = 178.88X

Explanation:

Given data:

ASTM grain size number 7

a) total grain per inch^2 - 64 grain/inch^2

we know that number of grain per square inch is given as

Nm = 2^{n-1} (\frac{100}{M})^2

where M is magnification, n is grain size

therefore we have

64 = 2^{7-1}(\frac{100}{M})^2

solving for M we get

M = 100 X

B)  total grain per inch^2 = 500 grain/inch^2

we know that number of grain per square inch is given as

Nm = 2^{n-1} (\frac{100}{M})^2

where M is magnification, n is grain size

therefore we have500 = 2^{7-1}(\frac{100}{M})^2

solving for M we get

M = 36 X

C) Total grain per inch^2 = 20 grain/inch^2

we know that number of grain per square inch is given as

Nm = 2^{n-1} (\frac{100}{M})^2

where M is magnification, n is grain size

therefore we have20 = 2^{7-1}(\frac{100}{M})^2

solving for M we get

M = 178.88 X

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A 0.3146 g sample of a mixture of NaCl ( s ) and KBr ( s ) was dissolved in water. The resulting solution required 45.70 mL of 0
Masteriza [31]

Answer:

The answer to the question is

The mass percentage of NaCl(s) in the mixture is 49.7%

Explanation:

The given variables are

mass of sample of mixture = 0.3146 g

Volume of AgNO₃ required to react comletely with the chloride ions = 45.70 mL

Concentration of the AgNO₃ added = 0.08765 M

The equations for the reactions oare

NaCl(aq) + AgNO₃ (aq) = AgCl(s) + NaNO₃(aq)

AgNO₃ (aq) + KBr (aq) → AgBr (s) + KNO₃

The equation for the reaction shows one mole of NaCl reacts with one mole of AgNO₃ to form one mole of AgCl

Thus 45.70 mL of 0.08765 M solution of AgNO₃ contains\frac{45.7}{1000} (0.08765) = 0.004 moles

Therefore the sum of the number of moles of Br⁻ and Cl⁻

precipitated out of the solution =  0.004 moles

Thus if the mass of NaCl in the sample = z then the mass of KBr = y

However the mass of the sample is given as 0.3146 g which  means the molarity of the solution is 0.004 moles

given by

\frac{z}{58.44} + \frac{y}{119} = 0.004 moles  and z + y = 0.3146

Therefore z = 0.3146 - y which gives

\frac{(0.3146-y)}{58.44} + \frac{y}{119} = 0.004 moles

-8.7×10⁻³y +0.54×10⁻³ = 0.004

or 8.7×10⁻³y = 1.37769× 10⁻³

y = 0.158 g and z = 0.156 Thus the mass of NaCl = 0.156 g and the mass percentage = 0.156/0.3146×100 = 49.7% NaCl

The mass percentage of NaCl(s) in the mixture is 49.7%

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