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melisa1 [442]
3 years ago
9

1. 5.0 g of copper was heated from 20°C to 80°C. How much energy was used to heat Cu? (Specific heat capacity of Cu is 0.092 cal

/g °C) show work
Chemistry
1 answer:
iren2701 [21]3 years ago
5 0

 The amount  of energy  that  was  used to heat Cu  is 27.6 cal

 

<u><em>calculation</em></u>

Heat (Q) = M ( mass)  x c(specific heat capacity)  x ΔT( change in temperature)

where;

 Q=?

 M = 5.0 g

C=0.092 cal/g°C

ΔT = 80°c-20°c=60°C


Q is therefore = 5.0 g  x 0.092 cal / g°c x 60°c =27.6 cal

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Answer:

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b) 99.4 nm

Explanation:

a) The energy of radiation is given by:

E = h*f

Where h is the Planck constant (6.626x10⁻³⁴ J.s), and f is the frequency. To have the highest frequency, the energy must be the highest too, because they're directly proportional. So we must use E = -E1 = 20x10⁻¹⁹ J

20x10⁻¹⁹ = 6.626x10⁻³⁴xf

f = 3.02x10¹⁵ s⁻¹

The wavelenght is the velocity of light (3.00x10⁸ m/s) divided by the frequency:

λ = 3.00x10⁸/3.02x10¹⁵

λ = 9.94x10⁻⁸ m = 99.4 nm

b) To have the shortest wavelength, it must be the highest energy and frequency, so it would be the same as the letter a) 99.4 nm.

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3 years ago
Help me please| (• ◡•)|​
nika2105 [10]

Answer:

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6 0
3 years ago
Consider the Gibbs energies at 25 ∘C.
zysi [14]

Answer:

A)Δ​G​r​x​n​∘​=​55.7​k​J​/​m​o​l

B)Ksp=1.75×10^−10

C)Δ​G​r​x​n​∘​=​70.0​k​J​/​m​o​l

D)Ksp=5.45×10^−13

Explanation:

ΔGrxn∘=​Δ​G​f​,​p​r​o​d​u​c​t​s​∘​−​Δ​G​f​,​r​e​a​c​t​a​n​t​s​∘

To calculate for the

Ksp

of the dissolution reaction can be claculated

ΔGrxn∘=−RTlnKsp

where R is the proportionality constant equal to 8.3145 J/molK.

A)

Δ​G​r​x​n​∘​=​[​Δ​G​f​,​A​g​(​a​q​)​+​∘​+​Δ​G​f​,​C​l​(​a​q​)​−​∘​]​−​Δ​G​f​,

A​g​C​l​(​s​)​⇌​A​g​(​a​q​)​+​+​C​l​(​a​q​)​−

ΔG∘rxn=[77.1kJ/mol+(−131.2kJ/mol)]−(−109.8kJ/mol)

Δ​G​r​x​n​∘​=​55.7​k​J​/​m​o​l

b) Calculate the solubility-product constant of AgCI.

ΔGrxn∘=−RTlnKsp

55.7​k​J​/​m​o​l​=​−​(​8.3145​×​10​−​3​J​/​m​o​l​K​)​(​298.15​K)InKsp

Ksp=1.75×10^−10

c) Calculate

To calculate ΔG°rxn

for the dissolution of AgBr(s).

Δ​G​r​x​n​∘​=​[​Δ​G​f​,​A​g​(​a​q​)​+​∘​+​Δ​G​f​,​B​r​(​a​q​)​−​∘​]​−​Δ​G​f​,

Δ​G​r​x​n​∘​=​[​77.1​k​J​/​m​o​l​+​(​−​104.0​k​J​/​m​o​l​)​]​−​(​−96.90kj/mol

Δ​G​r​x​n​∘​=​70.0​k​J​/​m​o​l

d)To Calculate the solubility-product constant of AgBr.

ΔGrxn∘=−RTlnKsp

70.0kJ/mol=−(8.3145×10−3J/molK)(298.15K)lnKsp

70.0​k​J​/​m​o​l​=​−​(​8.3145​×​10​−​3​J​/​m​o​l​K​)​(​298.15​K

Ksp=5.45×10^−13

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raketka [301]

Answer:

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