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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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jasenka [17]

Answer:

B

Explanation:

Cuz they are supported once discovered and then widely spread then accepted

4 0
3 years ago
Read 2 more answers
500 mL of a solution contains 1000 mg of CaCl2. Molecular weight of CaCl2 is 110 g/mol. Specific gravity of the solution is 0. C
dangina [55]

Answer:

a) 0,2% w/v

b) r=500

c) 0,0182 M

d) 0,0145 m

e) 0,0137 equivalents

Explanation:

a) % w/v means mass of solute in grams per 100 mililiter of solution. Thus:

%w/v= \frac{1,000 g CaCl2}{500mL}×100 = 0,2%w/v

b) Ratio strength is a way to express concentration.  For w/v is in 1g of solute <em>r</em> mililiters of solution have. Thus, r = 500 because we have in the first 1 g of CaCl₂ in 500 mL of solution.

c) Molarity is moles of solute per liter of solution, thus:

1,000 g of CaCl₂ × \frac{1mol}{110g} = 9,09×10⁻³ moles of CaCl₂

500 mL of solution  × \frac{1L}{1000mL} = 0,500 L of solution

M = \frac{9,09x10^{-3} moles }{0,500 L} = 0,0182 M

d) Molality is moles of solute per kg of solution.

Specific gravity is the ratio between density of the solution and density of a reference substance (Usually water). With a specific gravity of 0,8:

kg of solution = 0,500 L of solution × \frac{0,8 kg}{1L} =<em> </em><em>0,625 kg of solution</em>

m = \frac{9,09x10^{-3}moles }{0,625 kg} = 0,0145 m

e)  In a salt, equivalents are the number of moles ables to replace one mole of charge. In CaCl₂ is ¹/₂ because with  ¹/₂ moles of CaCl₂ it is possible to replace 1 mole of charges. Thus, in 1,5 L there are:

1,5 L ×\frac{0,0182 CaCl2 moles}{1L} × \frac{1equivalent}{2 moles} = 0,0137 equivalents

I hope it helps!

7 0
3 years ago
Explain why the lattice energy of CaSe is approximately 4 times as large as that of KBr. Check all that apply. Check all that ap
Elza [17]

Answer:

According to Coulomb’s law, the Ca and Se ions have 4 times the attractive force (2+ × 2-) than that of the K and Br ions (1+ × 1-).

Explanation:

From Coulomb's law, the attractive force between calcium and selenium ions is four times the attractive force between potassium and bromide ions.

This has something to do with size and magnitude of charge. Calcium ions and selenide ions are smaller and both carry greater charge magnitude than potassium and bromide ions. This paves way for greater electrostatic attraction between them when the distance of the charges apart is minimal. Hence a greater lattice energy.

4 0
3 years ago
A fixed mass of a gas occupies 1000cm
Serhud [2]

Answer:

the pressure of gas is 100.0 the volume is 500.0

Explanation:

4 0
3 years ago
Part 1. A chemist reacted 18.0 liters of F2 gas with NaCl in the laboratory to form Cl2 gas and NaF. Use the ideal gas law equat
Alika [10]

Answer:

Part 1

The mass of the NaCl that reacted with F₂ at 290.K and 1.5 atm is approximately 132.6 gams

Part 2

The mass of NaCl that can react with the same volume of gas at STP is approximately 93.77 grams

Explanation:

Part 1

The volume of F₂ gas in the reaction, V = 18.0 liters

The ideal gas equation is P·V = n·R·T

∴ n = P·V/(R·T)

The pressure, P = 1.5 atm

The temperature, T = 290 K

The universal gas constant, R = 0.0820573 L·atm/(mol·K)

∴ n = 1.5×18/(0.0820573 × 290) ≈ 1.134615

The number of moles of F₂ in the reaction n ≈ 1.134615 moles

The chemical reaction is given as follows;

F₂ + 2NaCl → Cl₂ + 2NaF

1 mole of F₂ reacts with 2 moles of NaCl

Therefore;

1.134615 moles of F₂ reacted with 2 × 1.134615 moles ≈ 2.26923 moles of NaCl

1 mole of NaCl = The molar mass of NaCl, MM = 58.44 g/mol

The mass, of 2.26923 moles of NaCl, m = Number of moles × MM

∴ m ≈ 2.26923 moles × 58.44 g/mol ≈ 132.6 grams

The mass of the NaCl ≈ 132.6 gams

Part 2

The volume occupied by 1 mole of all gases at STP = 22.4 l/mole

Therefore, the number of moles of F₂ in 18.0 L of F₂ = 18.0 L/(22.4 L/mole) ≈ 0.804 moles

Therefore;

The number of moles of NaCl, in the reaction n = 2 × The number of moles of F₂ ≈ 2×0.804 moles = 1.608 moles

The number of moles of NaCl, in the reaction n ≈ 1.608 moles

The mass of NaCl in the reaction, m = n × MM

∴ m ≈ 1.608 moles × 58.44 g/mol ≈ 93.97 grams

The mass of NaCl that can react with the same volume of gas at STP ≈ 93.77 grams

8 0
3 years ago
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