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NNADVOKAT [17]
2 years ago
13

Which of the following are the three basic types of static stresses to which a material can be subjected (three correct answers)

: a. yield b. tensile c. hardness d. compression e. shear f. reduction in area g. true stress
Chemistry
1 answer:
Tasya [4]2 years ago
4 0

Answer:

b, d, e

Explanation:

Three type basic static stresses to which a material can be subjected are

Tensile: Stress generated due to elongation force on the body.

Compressive: The stress generated when the body is compressed.

Shear:  The stress generated when a force parallel to cross-section are of the body.

Hardness, reduction in area and true stress are material properties.

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What type of chemical reaction is this? Cl2(g) + 2KBr(aq) - 2KCl(aq) + Br2(l)
Sindrei [870]
C. Single-replacement

Chlorine replaces Bromine in KBr.
7 0
2 years ago
A 1.0 g sample of a cashew was burned in a calorimeter containing 1000. g of water, and the temperature of the water changed fro
Savatey [412]

Answer:

The correct answer is option C.

Explanation:

1.0 g sample of a cashew :

Heat released on  combustion of 1.0 gram of cashew = -Q

We have mass of water = m = 1000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water :  Q

Q=1000 g\times 4.184 J/g^oC\times 5^oC=20,920 J

Heat released on  combustion of 1.0 gram of cashew is -20,920 J.

3.0 g sample of a marshmallows  :

Heat released on  combustion of 3.0 g sample of a marshmallows = -Q'

We have mass of water = m = 2000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water :  Q'

Q'=2000 g\times 4.184 J/g^oC\times 5^oC=41,840 J

Heat released on 3.0 g sample of a marshmallows= -Q' = -41,840 J

Heat released on 1.0 g sample of a marshmallows : q

q =\frac{-Q'}{3} = \frac{-41,840 J}{3}=-13,946.67 J

Heat released on  combustion of 1.0 gram of marshmallows -13,946.67 J.

-20,920 J. > -13,946.67 J

The combustion of 1.0 g of cashew releases more energy than the combustion of 1.0 g of marshmallow.

5 0
2 years ago
A penny has a mass of
Ahat [919]
You did not include the questions.

I did some research and found the questions:

<span> What is the mass of 1 mole of pennies? How many moles of pennies have a mass equal to the mass of the moon?

Solutions:

1) mass of 1 mole of pennies

Data: mass of 1 penny = 2.50 g

1 mole = 6.022 * 10^ 23 units

Proportion:

  1 penny      6.022 * 10^23 penny
-------------- = ----------------------------
   2.50 g                    x

Solve: x = 6.022 * 10^23 penny * 2.50g / 1 penny = 15.055* 10^23

Since 2.50 has 3 significant figures, the answer must use 3 significant figures => x = 15.1 * 10^ 23 g = 1.51 * 10^24 g

Answer: 1 mol of pennies have a mass of 1.51 * 10^24 g

2) How many moles of pennies have a mass equal to the same mass of the Moon

Convert the mass of the Moon grams: 7.35 * 10^22 kg = 7.35 * 10^ 25 g

       1 mol                            x
---------------------- =  ----------------------
1.51 * 10^ 24g          7.35 * 10^ 25 g

=> x = 7.35 * 10^ 25 g * 1 mol / (1.51 * 10^24 g)= 48.7 mol

Answer: 48.7 mol
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5 0
2 years ago
Read 2 more answers
Which aqueous solution would exhibit a lower freezing point, 0.35 m k2so4 or 0.50 m kcl? explain and show necessary calculations
Umnica [9.8K]
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b -  molality, moles of solute per kilogram of solvent.
i - </span>Van't Hoff factor.<span>
b(K</span>₂SO₄<span>) = 0.35 m.
</span>b(KCl) = 0.5 m.
i(K₂SO₄) = 3.
i(KCl) = 2.
ΔT(K₂SO₄) = 3 · 0.35 m · 1.86°C/m.
ΔT(K₂SO₄) = 1.953°C.
ΔT(KCl) = 2 · 0.5 m · 1.86°C/m.
ΔT(KCl) = 1.86°C.


8 0
2 years ago
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