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Ivan
3 years ago
10

Choose ALL TRUE statements about calorimetry from the choices below:

Chemistry
1 answer:
Ostrovityanka [42]3 years ago
6 0

Answer:

True statment

2) Styrofoam would make a good calorimeter

3) Insulating material would make a good calorimeter

Explanation:

The calorimeter is one which is insulated that is which will not absorb or let the heat to escape from it. the calorimeter is used to measure the heat change during a process so if it will allow to exchange heat with surrounding it will deviate the readings or observence.

Copper is a good conductor of heat so we cannot use it make a calorimeter.

Hence

1) Copper would make a good calorimeter : False

2) Styrofoam would make a good calorimeter: True

Styrofoam is a bad conductor or insulator so it can be and it is used for calorimeter.

3) Insulating material would make a good calorimeter : True

4) A good calorimeter should easily absorb heat : false

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Answer:

A water molecule consists of three atoms; an oxygen atom and two hydrogen atoms, which are bond together like little magnets. The atoms consist of matter that has a nucleus in the centre. one mole of water contains 6.02 x 1023 MOLECULES of water But each molecule of water contains 2 H and 1 O atom = 3 atoms, so there are approximately 1.8 x 1024 atoms in a mole of water.Feb 12, 2003

Explanation:

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Compared to CH4,NH3 has:
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a) NH₃ molecules have stronger intermolecular attractions than CH₄ molecules.

Explanation:

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3 years ago
Materials expand when heated. Consider a metal rod of length L0 at temperature T0. If the temperature is changed by an amount ΔT
marysya [2.9K]

Answer:

(a) The length at temperature 180°C is 40.070 cm

(b) The length at temperature 90°C is 64.976 inches

(c) L(T, α) = 60·α·T - 9000·α + 60

Explanation:

(a) The given parameters are

The thermal expansion coefficient, α for steel = 1.24 × 10⁻⁵/°C

The initial length of the steel L₀ = 40 cm

The initial temperature, t₀ = 40°C

The length at temperature 180°C = L

Therefore, from the given relation, for change in length, ΔL, we have;

ΔL = α × L₀ × ΔT

The amount the temperature changed ΔT = 180°C - 40°C = 140°C

Therefore, the change in length, ΔL, is found as follows;

ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 40 × 140°C = 0.07 cm

Therefore, L =  L₀ + ΔL = 40 + 0.07 = 40.07 cm

The length at temperature 180°C = 40.07 cm

(b) Given that the length at T = 120°C is 65 in., we have;

The temperature at which the new length is sought = 90°C

The amount the temperature changed ΔT = 90°C - 120°C = -30°C

ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 65 × -30°C = -0.024375 inches

The length, L at 90°C is therefore, L = L₀ + ΔL = 65 - 0.024375 = 64.976 in.

The length at temperature 90°C = 64.976 inches

(c) L = L₀ + ΔL  = L₀ +  α × L₀ × ΔT = L₀ +  α × L₀ × (T - T₀)

Therefore;

L = 60 +  α × 60 × (T - 150°C)

L = 60 + α × 60 × T - 9000 × α

L(T, α) = 60·α·T - 9000·α + 60

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Answer:

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Explanation:

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