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MAVERICK [17]
3 years ago
11

Materials expand when heated. Consider a metal rod of length L0 at temperature T0. If the temperature is changed by an amount ΔT

, then the rod’s length changes by ΔL=αL0ΔT, where α is the thermal expansion coefficient.
For steel, α=1.24×10−5∘C−1.(a) A steel rod has length L0=40cm at T0=40∘C. What is its length at T=180∘C?
(Use decimal notation. Give your answer to three decimal places.)
L≈cm
(b) Find its length at T=90∘C if its length at T0=120∘C is 65 in.
(Use decimal notation. Give your answer to three decimal places.)
L≈ in.
(c) Express length L as a function of T and α if L0=60in. at T0=150∘C.
(Use symbolic notation and fractions where needed.) L(T,α)=
Chemistry
1 answer:
marysya [2.9K]3 years ago
7 0

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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kodGreya [7K]

Answer:

18 minutes is the half life

8 0
3 years ago
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CO(g)+2H2(g)⇌CH3OH(g) A reaction mixture in a 5.23-L flask at a certain temperature initially contains 27.2 g CO and 2.36 g H2.
irga5000 [103]

Answer: 5.70M

Explanation:

Molar mass of CO = 28.01 g/mol

Molar mass of H2 = 2.02 g/mol

Molar mass of CH3OH = 32.05 g/mol.

To determine the amount of each compound in the reaction mixture we use the formula.

Amount in mol = reacting mass/molar mass.

Inputing the given values we have,

26.6 g CO x (1 mol / 28.01 g ) = 0.9496608354 mol of CO.

To calculate the concentration of CO we use C=n/v, where n=amount and v= volume of CO.

Inputing the values in the formula

[CO] = 0.9496608354 mol CO / 5.23 L = 0.18158 M CO

Repeating thesame procedure for H

Amount of H=2.36 g H2 x ( 1 mol / 2.02 g ) = 1.168316832 mol of H2

Concentration of H2 in the mixture

[H2] = 1.168316832 mol H2 / 5.23 L = 0.223388 M of H2

Amount of CH3OH is determine similarly using rmass/molar mass

8.66 CH3OH x (1 mol / 32.05 g ) = 0.2702028081 mol of CH3OH

Concentration of

[CH3OH] = 0.2702028081 mol CH3OH/ 5.23 L = 0.051664 M CH3OH

Now equilibrium constant is determined by

Kc = [CH3OH] / [CO] [H2]^2

=0.051664/0.18158×0.223388×0.223388.

=5.70

7 0
3 years ago
If 6.00 g of CaCl2 • 2 H2O and 5.50 g of Na2CO3 are allowed to react in aqueous solution, what mass of CaCO3 will be produced? P
Andre45 [30]

Answer:

6.00 g CaCl₂ .2H₂O /1 × 1 mol CaCl₂ .2H₂O / 147 g CaCl₂ .2H₂O × 0.04 mol CaCO₃/ 0.04 mol of  CaCl₂ .2H₂O  × 100 g CaCO₃ /  1 mole CaCO₃ = 4 g

5.50 g Na₂CO₃   /1 × 1 Na₂CO₃  / 106 g Na₂CO₃ × 0.05 mol CaCO₃/ 0.05 mol of Na₂CO₃  × 100 g CaCO₃ /  1 mole CaCO₃ = 5 g

Explanation:

Given data:

Mass of CaCl₂.2H₂O = 6.00 g

Mass of Na₂CO₃ = 5.50 g

Mass of CaCO₃ produced = ?

Solution:

Number of moles of CaCl₂.2H₂O.

Number of moles = mass/ molar mass

Number of moles = 6.00 g/ 147 g/ mol

Number of moles = 0.04 mol

Number of moles of Na₂CO₃:

Number of moles = mass/ molar mass

Number of moles = 5.50 g/ 106 g/ mol

Number of moles = 0.05 mol

Chemical equation:

CaCl₂  +  Na₂CO₃   →   CaCO₃ + 2NaCl

Now we will compare the moles of CaCO₃  with Na₂CO₃  and CaCl₂ through balanced chemical equation .

                      CaCl₂              :               CaCO₃

                             1                :                1

                       0.04               :            0.04

Mass of CaCO₃:

Mass = number of moles × molar mass

Mass = 0.04 mol× 100 g/mol

Mass = 4 g

6.00 g CaCl₂ .2H₂O /1 × 1 mol CaCl₂ .2H₂O / 147 g CaCl₂ .2H₂O × 0.04 mol CaCO₃/ 0.04 mol of  CaCl₂ .2H₂O  × 100 g CaCO₃ /  1 mole CaCO₃ = 4 g

                     Na₂CO₃            :            CaCO₃

                          1                   :                1

                       0.05               :            0.05

Mass of CaCO₃:

Mass = number of moles × molar mass

Mass = 0.05 mol× 100 g/mol

Mass = 5 g

5.50 g Na₂CO₃   /1 × 1 Na₂CO₃  / 106 g Na₂CO₃ × 0.05 mol CaCO₃/ 0.05 mol of Na₂CO₃  × 100 g CaCO₃ /  1 mole CaCO₃ = 5 g

                     

5 0
3 years ago
Carbon dioxide, when it is at -145 degreesC has a density of 2.54 g/L. What is the pressure in torr??
Black_prince [1.1K]

Answer: The pressure in torr is 461 torr

Explanation:

To calculate the relation of density and molar mass of a compound, we use the ideal gas equation:

PV=nRT

P = pressure

V = Volume

n = number of moles

R = gas constant = 0.0821 Latm/Kmol

T = temperature =-145^0C=(273-145)K=128K

Number of moles (n) can be written as:

n=\frac{m}{M}

where, m = given mass

M = molar mass  = 44 g/mol

PV=\frac{m}{M}RT\\\\PM=\frac{m}{V}RT

where,

\frac{m}{V}=d

where d = density = 2.54 g/L

The relation becomes:

PM=dRT  

P=\frac{dRT}{M}

P=\frac{2.54\times 0.0821\times 128}{44}=0.607atm

P=461torr    (760torr=1atm)

Thus the pressure in torr is 461 torr

5 0
3 years ago
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