Which one of the following represents a chemical change?

Identify 5 things that would happen if there is not electricity.

Brilliant_brown [7]

If the power is out long enough even the city folks will run out of water. Many homes are all electric, so as soon at the lights are out they have no heat, no hot water and they can't cook. ... If the power is out, gas stations can't pump gas. Once generators run out of gas, those people will be in the dark too.

7 0

3 years ago

Read 2 more answers

Please help I’m torn between A and B. I’ve asked many and people say A, but I asked my science teacher and she said B. And also

astraxan [27]

If your science teacher says B, it’s probably because water has a negative and positive end, heat is just a form of energy, as other atoms can’t leave (they’re attracted to the ends) they are being insulated; but notice that ice will melt into gas (where atoms have tons of space) for other atoms to escape. Hence ice and gas aren’t ideal. (Air is a gas here.)
It’s not a 100% but hopefully it helps with some kind of analogy.

4 0

3 years ago

At equilibrium, the concentrations in this system were found to be [N21 O20.200 M and [NO]0.500 M. N2(8) 02e) 2NO(g) If more NO

erastovalidia [21]

Answer : The concentration of $NO$ at equilibrium is 0.9332 M

Solution : Given,

Concentration of $N_2$ and $O_2$ at equilibrium = 0.200 M

Concentration of $N_2$ and $O_2$ at equilibrium = 0.500 M

First we have to calculate the value of equilibrium constant.

The given equilibrium reaction is,

$N_2(g)+O_2(g)\rightleftharpoons 2NO(g)$

The expression of $K_c$ will be,

$K_c=\frac{[NO]^2}{[N_2][O_2]}$

$K_c=\frac{(0.500)^2}{(0.200)\times (0.200)}$

$K_c=6.25$

Now we have to calculate the final concentration of NO.

The given equilibrium reaction is,

$N_2(g)+O_2(g)\rightleftharpoons 2NO(g)$

Initially 0.200 0.200 0

.800

At equilibrium (0.200-x) (0.200-x) (0.800+2x)

The expression of $K_c$ will be,

$K_c=\frac{[NO]^2}{[N_2][O_2]}$

$K_c=\frac{(0.800+2x)^2}{(0.200-x)\times (0.200-x)}$

By solving the term x, we get

$x=2.6\text{ and }-0.0666$

From the values of 'x' we conclude that, x = 2.6 can not more than initial concentration. So, the value of 'x' which is equal to 2.6 is not consider.

And the negative value of 'x' shows that the equilibrium shifts towards the left side (reactants side).

Thus, the concentration of $NO$ at equilibrium = (0.800+2x) = 0.800 + 2(0.0666) = 0.9332 M

Therefore, the concentration of $NO$ at equilibrium is 0.9332 M

5 0

3 years ago

Hydrogen and Methanol have both been proposed as alternatives to hydrocarbon fuels. Write balanced reactions for the complete co

Arlecino [84]

Answer:

The order of energy released per mass is

CH₃OH (-2.268 × 10⁴ kJ/kg) < C₈H₁₈ (-4.826 × 10⁴ kJ/kg) < H₂ (-2.835 × 10⁵ kJ/kg)

Explanation:

In order to calculate the enthalpy of a reaction (ΔH°r) we can use the following expression.

ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)

where

ΔH°f(i) are the enthalpies of formation of reactants and products

ni are the moles of reactants and products

Combustion of hydrogen

H₂(g) + 1/2 O₂(g) ⇒ H₂O(l)

ΔH°r = 2 mol × ΔH°f(H₂O) - 1 mol × ΔH°f(H₂) - 1/2 mol × ΔH°f(O₂)

ΔH°r = 2 mol × (-285.8 kJ/mol) - 1 mol × 0 - 1/2 mol × 0

ΔH°r = -571.6 kJ

571.6 kJ are released when 1 mole of H₂ is burned. The amount of heat released per kilogram is:

$\frac{-571.6kJ}{1molH_{2}} .\frac{1molH_{2}}{2.016gH_{2}} .\frac{10^{3}gH_{2} }{1kgH_{2}} =-2.835 \times 10^{5} kJ/kgH_{2}$

Combustion of methanol

CH₃OH(l) + 3/2 O₂(g) ⇒ CO₂(g) + 2 H₂O(l)

ΔH°r = 1 mol × ΔH°f(CO₂) + 2 mol × ΔH°f(H₂O) - 1 mol × ΔH°f(CH₃OH) - 3/2 mol × ΔH°f(O₂)

ΔH°r = 1 mol × (-393.5 kJ/mol) + 2 mol × (-285.8 kJ/mol) - 1 mol × (-238.4 kJ/mol) - 3/2 mol × 0

ΔH°r = -726.7 kJ

726.7 kJ are released when 1 mole of CH₃OH is burned. The amount of heat released per kilogram is:

$\frac{-726.7kJ}{1molCH_{3}OH} .\frac{1molCH_{3}OH}{32.04gCH_{3}OH} .\frac{10^{3}gCH_{3}OH }{1kgCH_{3}OH} =-2.268 \times 10^{4} kJ/kgCH_{3}OH$

Combustion of octane

C₈H₁₈(l) + 12.5 O₂(g) ⇒ 8 CO₂(g) + 9 H₂O(l)

ΔH°r = 8 mol × ΔH°f(CO₂) + 9 mol × ΔH°f(H₂O) - 1 mol × ΔH°f(C₈H₁₈) - 12.5 mol × ΔH°f(O₂)

ΔH°r = 8 mol × (-393.5 kJ/mol) + 9 mol × (-285.8 kJ/mol) - 1 mol × (-208.4 kJ/mol) - 12.5 mol × 0

ΔH°r = -5511.8 kJ

5511.8 kJ are released when 1 mole of C₈H₁₈ is burned. The amount of heat released per kilogram is:

$\frac{-5511.8kJ}{1molC_{8}H_{18}} .\frac{1molC_{8}H_{18}}{114.2gC_{8}H_{18}} .\frac{10^{3}gC_{8}H_{18} }{1kgC_{8}H_{18}} =-4.826 \times 10^{4} kJ/kgC_{8}H_{18}$

5 0

3 years ago

Why is water vapor (H2O) also considered a greenhouse gas?

zepelin [54]

Answer:

Water vapour is the most abundant greenhouse gas in the atmosphere, both by weight and by volume (1), (2). Water vapour is also an effective greenhouse gas, as it does absorb longwave radiation and radiates it back to the surface, thus contributing to warming

3 0

2 years ago