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wlad13 [49]
3 years ago
11

The energy of moving atoms is

Chemistry
2 answers:
Lana71 [14]3 years ago
5 0
Kinetic energy
Hope this helps :)

btw make me brainliest?
balandron [24]3 years ago
4 0

The answer is Thermal Energy :)

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The vaporization of 1 mole of liquid water (system) at 100.9 C, 1.00 atm, is endothermic.
yan [13]

Answer:

a) The work done is 10.0777 kJ

b) The water's change in internal energy is -122.1973 kJ

Explanation:

Given data:

1 mol of liquid water

T₁ = temperature = 100.9°C

P = pressure = 1 atm

Endothermic reaction

T₂ = temperature = 100°C

1 mol of water vapor

VL = volume of liquid water = 18.8 mL = 0.0188 L

VG = volume of water vapor = 30.62 L

3.25 moles of liquid water vaporizes

Q = heat added to the system = -40.7 kJ

Questions: a) Calculate the work done on or by the system, W = ?

b) Calculate the water's change in internal energy, ΔU = ?

Heat for 3.25 moles:

Q_{1} =3.25*(-40.7)=-132.275kJ

The work done:

W=-nP*delta(V)=-3.25*101.33*(0.0188-30.62)=10077.6637J=10.0777kJ

The change in internal energy:

delt(U)=W+Q_{1} =10.0777-132.275=-122.1973kJ

7 0
3 years ago
If 54 g of Al reacted with 160 g of O2, find out the weight of product​
Oxana [17]

Answer:

number of Al atom =54÷27=2atom of Al. number of o atom = 160÷16=10 atom of o . =214U MASS of ALO2.

4 0
3 years ago
What is the general trend for valence electrons as you go across the periodic table left to right
Margarita [4]
The atomic radius decreases across a period from left to right and increases down a given group. The atoms with the largest atomic radii are located in group l and at the bottom of groups. Moving from left to right across a period, electrons are added one at a time to the outer energy shell. Hope this helps!
6 0
3 years ago
What happens to dissolving rate when you increase KE?
TEA [102]
In general, solubility increases with temperature. When you increase the temperature of a solvent, you increase the kinetic energy (or energy of movement) of the molecules, and this greater energy helps dissolve more of the solute molecules.
4 0
4 years ago
Consider a sample of 3.5 mol of N2(g) at T1 = 350 K, that undergoes a reversible and adiabatic change in pressure from p1 = 1.50
devlian [24]

Answer:

Part A is just T2 = 58.3 K

Part B ∆U = 10967.6 x C_{V} You can work out C_{V}

Part C

Part D

Part E

Part F

Explanation:

P = n (RT/V)

V = (nR/P) T

P1V1 = P2V2

P1/T1 = P2/T2

V1/T1 = V2/T2

P = Pressure(atm)

n = Moles

T = Temperature(K)

V = Volume(L)

R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.

bar = 0.986923 atm

N = 14g/mol

N2 Molar Mass 28g

n = 3.5 mol N2

T1 = 350K

P1 = 1.5 bar = 1.4803845 atm

P2 = 0.25 bar = 0.24673075 atm

Heat Capacity at Constant Volume

Q = nCVΔT

Polyatomic gas: CV = 3R

P = n (RT/V)

0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))

V = (nR/P) T

V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K

V = (0.28721/1.4803845) x 350

V = 0.194 x 350

V = 67.9036 L

So V1 = 67.9036 L

P1V1 = P2V2

1.4803845 atm x 67.9036 L = 0.24673075 x V2

100.52343693 = 0.24673075 x V2

V2 = P1V1/P2

V2 = 100.52343693/0.24673075

V2 = 407.4216 L

P1/T1 = P2/T2

1.4803845 atm / 350 K = 0.24673075 atm / T2

0.00422967 = 0.24673075 /T2

T2 = 0.24673075/0.00422967

T2 = 58.3 K

∆U= nC_{V} ∆T

Polyatomic gas: C_{V} = 3R

∆U= nC_{V} ∆T

∆U= 28g x C_{V} x (350K - 58.3K)

∆U = 28C_{V} x 291.7

∆U = 10967.6 x C_{V}

5 0
3 years ago
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