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vaieri [72.5K]
4 years ago
14

What is a molecule containing only carbon and hydrogen called?

Chemistry
2 answers:
pogonyaev4 years ago
4 0

Answer:

d

Explanation:

it's called the hydrocarbon

aalyn [17]4 years ago
4 0

Answer:

A hydrocarbon

Explanation:

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Ray Of Light [21]
Total cost: $603.42

Tickets (total): 134.23*4= $536.92

603.42-536.92= $66.50

Hot dog (total)= $9

66.5-9= $57.50

Popcorn (total)= $17.50

57.50-17.50= $40

They bought 2 parking passes

40/2=20

Answer: Each pass costs $20
4 0
4 years ago
Calculate the change in entropy when 10.0 g of CO2 isothermally expands from a volume of 6.15 L to 11.5 L. Assume that the gas b
USPshnik [31]

Answer:

The change in entropy of the carbon dioxide is 1.183\times 10^{-3} kilojoules per Kelvin.

Explanation:

By assuming that carbon dioxide behaves ideally, the change in entropy (\Delta S), measured in kilojoules per Kelvin, is defined by the following expression:

\Delta S = m\cdot \bar c_{v}\cdot \ln \frac{T_{f}}{T_{o}}+m\cdot \frac{R_{u}}{M}\cdot \ln \frac{V_{f}}{V_{o}} (1)

Where:

m - Mass of the gas, measured in kilograms.

\bar c_{v} - Isochoric specific heat of the gas, measured in kilojoules per kilogram-Kelvin.

T_{o}, T_{f} - Initial and final temperatures of the gas, measured in Kelvin.

V_{o}, V_{f} - Initial and final volumes of the gas, measured in liters.

R_{u} - Ideal gas constant, measured in kilopascal-cubic meter per kilomole-Kelvin.

M - Molar mass, measured in kilograms per kilomole.

If we know that T_{o} = T_{f}, m = 0.010\,kg, R_{u} = 8.315\,\frac{kPa\cdot m^{3}}{kmol\cdot K}, M = 44.010\,\frac{kg}{kmol}, V_{o} = 6.15\,L and V_{f} = 11.5\,L, then the change in entropy of the carbon dioxide is:

\Delta S = \left[\frac{ (0.010\,kg)\cdot \left(8.315\,\frac{kPa\cdot m^{3}}{kmol\cdot K} \right)}{44.010\,\frac{kg}{kmol} } \right]\cdot \ln \left(\frac{11.5\,L}{6.15\,L}\right)

\Delta S = 1.183\times 10^{-3}\,\frac{kJ}{K}

The change in entropy of the carbon dioxide is 1.183\times 10^{-3} kilojoules per Kelvin.

4 0
3 years ago
For the reaction A+B↽−−⇀C+D, assume that the standard change in free energy has a positive value. Changing the conditions of the
77julia77 [94]

Explanation:

a. Adding a catalyst

no effect .( Catalyst can only change the activation energy but not the free energy).

b. increasing [C] and [D]

Increase the free energy .

c. Coupling with ATP hydrolysis

decrease the free energy value .

d.Increasing [A] and [B]

decrease the free energy.

7 0
3 years ago
Choose the correct statement regarding the behavior of water.
AfilCa [17]

Answer:

d. The energy required to evaporate 1 kg of liquid water equals the energy released when 1 kg of water vapor condenses into liquid.

Explanation:

Hello,

Since we're considering the same amount of water, the vapor phase has a higher energy content than the liquid phase, thus, for the specified amount of water particles (those contained in the given 1 kg) the energy MUST be same when taking them either to a gaseous phase or to a liquid phase, the only difference is the sign which is negative from gaseous to liquid (heat withdrawal) and positive from liquid to gaseous (heat adding).

Best regards.

5 0
3 years ago
Question - Complete and balance the following chemical equations:
Mekhanik [1.2K]
Answer is because
Please give feedback
6 0
3 years ago
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