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LiRa [457]
3 years ago
7

ipt A wall that was once white is painted black. Which of the following is definitely true of the painted wall? A. Its chemical

properties have changed. B. Its physical properties have changed. C. Both its physical properties and chemical properties have changed. D. None of its characteristic properties have changed.​
Chemistry
2 answers:
Delvig [45]3 years ago
4 0
The answer would be C
Hope this helps
Julli [10]3 years ago
3 0

Answer:

B It's physical properties changed

Explanation:

Because they painted the outside

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How is Carbon returned to the atmosphere<br><br>Please help image below​
Harrizon [31]

Answer:

Carbon is released back into the atmosphere when organisms die, volcanoes erupt, fires blaze, fossil fuels are burned, and through a variety of other mechanisms.Humans play a major role in the carbon cycle through activities such as the burning of fossil fuels or land development.

8 0
2 years ago
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What kind of charge does a neutron have?
Nat2105 [25]
Neutrons actually don't carry an electrical charge, which is why they are called neutrons because they are "Neutral".


6 0
3 years ago
Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH
sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole..[1]

H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole..[2]

CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2  then adding all the equations, Using Hess's law:

We get :

C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole..[1]

2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol..[2]

CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole [3]

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3

\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)

\Delta H=-238.7kJ/mole

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

4 0
3 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
Molar mass of c16h19n305s
Masja [62]

Answer:

365.212

Explanation:

According to the given situation, the calculation of molar mass is shown below:-

Data provided

c_{16}\ h_{19}\ n_3\ o_5\ S

Molar mass = 16 × 12 + 19 × 1.008 + 3 × 14 + 5 × 16 + 32.06

= 192 + 19.152 + 42 + 80 + 32.06

Molar mass = 365.212

Therefore for determining the molar mass we simply solve the above equation.

So, the correct answer is 365.212

5 0
3 years ago
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