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madreJ [45]
3 years ago
13

How much work (in JJ) is required to expand the volume of a pump from 0.0 LL to 2.5 LL against an external pressure of 1.1 atmat

m
Chemistry
1 answer:
stira [4]3 years ago
8 0

Answer:

- 278.85 J  

Explanation:

Given that:

Pressure = 1.1 atm

The initial volume V₁ = 0.0 L

The final volume V₂ = 2.5 L

The work that takes place in a reaction at constant pressure can be expressed by using the equation:

W = P(V₂ - V₁ )

Since the volume of the gas is expanded from 0 to 2.5 L when 1.1 atm pressure is applied. Then, the work can be given by the expression:

W = - P(V₂ - V₁ )

W = -1.1 atm ( 2.5 - 0.0) L

W = -1.1 atm (2.5 L)

W = -2.75 atm L

Recall that:

1 atm L = 101.4 J

Therefore;

-2.75 atm L = ( -2.75 × 101.4 )J

= -278.85 J  

Thus, the work required at the chemical reaction when the pressure applied is 1.1 atm  = - 278.85 J  

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Mass/volume = density
mass = (440 mg)*(1g)/(1000mg) = 0.440g
volume = (1000cm)(1000cm)(t)
where t = thickness
density = 2.70 g/cm^3 = (0.440g)/((1000cm)(1000cm)(t))

multiply both sides by 't' and divide both sides by (2.70g/cm^3)

t = (0.440) / ((1000cm)(1000cm)(2.70)) = 1.629x10^-7 cm

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3 years ago
A potted plant is placed under a grow lamp, which provides 6,400 J of energy to the plant and the soil over the course of an hou
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Answer:

823.7g

Explanation:

Using the formula as follows:

Q = m × c × ∆T

Where;

Q = amount of heat (J)

m = mass of substance (g)

c = specific heat capacity (J/g°C)

∆T = change in temperature (°C)

Using the information given in this question as follows:

Q = 6,400 J

m = ?

c of soil = 0.840 J/g°C

∆T = 9.25°C

Using Q = mc∆T

m = Q ÷ c∆T

m = 6,400 ÷ (0.840 × 9.25)

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3 years ago
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Answer: Igneous rocks may be simply classified according to their chemical/mineral composition as felsic, intermediate, mafic, and ultramafic, and by texture or grain size: intrusive rocks are course grained (all crystals are visible to the naked eye) while extrusive rocks may be fine-grained (microscopic crystals) or glass.

Explanation: Hope this helped! :)

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Answer: The rate constant is 0.334s^{-1}

Explanation ;

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = ?

t = age of sample  = 4.26 min

a =  initial amount of the reactant  = 2.56 mg

a - x = amount left after decay process  = 2.50 mg

Now put all the given values in above equation to calculate the rate constant ,we get

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k=\frac{2.303}{4.26}\log\frac{2.56}{2.50}

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Thus rate constant is [tex]0.334s^{-1}

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