All categories

3 years ago

Prove how the Law Conservation of Mass and Enery apply to Cellular Respiration

Chemistry

1 answer:

snow_lady [41]3 years ago

7 0

Answer:

mass and energy is reused when those things happen

You might be interested in

A sample of CO2 gas has a volume of 2.7 L at 78.5 kPa. At what pressure would this sample of gas have a volume of 4.0L? Temperat

lisabon 2012 [21]

Answer:

52.99 kPa

Explanation:

Initial volume V1 = 2.7 L

Initial Pressure P1 = 78.5 kPa

Final Volume V2 = 4.0L

Final Pressure P2 = ?

Temperature is constant

The relationship between these quantities is given by the mathematical expression of Boyles law. This is given as;

V1P1 = V2P2

P2 = V1P1 / V2

P2 = 2.7 * 78.5 / 4.0

P2 = 52.99 kPa

7 0

3 years ago

Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti

gayaneshka [121]

Answer : The value of $K$ for this reaction is, $2.6\times 10^{15}$

Explanation :

The given chemical reaction is:

$CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)$

Now we have to calculate value of $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]$

where,

$\Delta G^o$ = Gibbs free energy of reaction = ?

n = number of moles

$\Delta G^0_{(HCH_3CO_2(g))}$ = -389.8 kJ/mol

$\Delta G^0_{(CH_3OH(g))}$ = -161.96 kJ/mol

$\Delta G^0_{(CO(g))}$ = -137.2 kJ/mol

Now put all the given values in this expression, we get:

$\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]$

$\Delta G^o=-89.4kJ/mol$

The relation between the equilibrium constant and standard Gibbs, free energy is:

$\Delta G^o=-RT\times \ln K$

where,

$\Delta G^o$ = standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol

R = gas constant = 8.314 J/L.atm

T = temperature = $30.0^oC=273+30.0=303K$

$K$ = equilibrium constant = ?

Now put all the given values in this expression, we get:

$-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K$

$K=2.6\times 10^{15}$

Thus, the value of $K$ for this reaction is, $2.6\times 10^{15}$

4 0

3 years ago

As the sun heats the surface of the ocean, two layers of water result. What separates the layers.

pychu [463]

Question: As the sun heats the surface of the ocean, two layers of water result. What separates the layers.?

Answer: The Thermocline

Hope this helps!.

~~~~~~~~~~~~~~~~~~~

~A.W~ZoomZoom44

4 0

3 years ago

وزن الملي مكافئ لحامض الخليك

Natalka [10]

Answer:

hshssytdtctdyeheb

Explanation:

yye6d66d6d6dududyydydydyehwj2

3 0

3 years ago

Why do you think large differences in temperature between the wet bulb and dry bulb thermometers are possible only at higher tem

olganol [36]

Answer:

This is the temperature indicated by a moistened thermometer bulb exposed to the air flow. The evaporation is reduced when the air contains more water vapor. The wet bulb temperature is always lower than the dry bulb temperature but will be identical with 100% relative humidity.

Explanation:

7 0

3 years ago

Prove how the Law Conservation of Mass and Enery apply to Cellular Respiration​

Prove how the Law Conservation of Mass and Enery apply to Cellular Respiration