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vitfil [10]
3 years ago
13

100x100 wsbntxrdfdmxnjhdtnvj

Chemistry
2 answers:
CaHeK987 [17]3 years ago
6 0

Answer:

10000

Explanation:

just add the zeros

hope dis helps ^-^

almond37 [142]3 years ago
4 0
It’s just 10,000... :)
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the standard change in Gibbs free energy is Δ????°′=7.53 kJ/molΔG°′=7.53 kJ/mol . Calculate Δ????ΔG for this reaction at 298 K29
Vera_Pavlovna [14]

Answer:

ΔG = 16.218 KJ/mol

Explanation:

  • dihydroxyacetone phosphate ↔ glyceraldehyde-3-phosphate
  • ΔG = ΔG° - RT Ln Q

∴ ΔG° = 7.53 KJ/mol * ( 1000 J / KJ ) = 7530 J/mol

∴  R = 8.314 J/K.mol

∴ T = 298 K

∴ Q = [glyceraldehyde-3-phosphate] / [dihydroxyacetone phosphate]

⇒ Q = 0.00300 / 0.100 = 0.03

⇒ ΔG = 7530J/mol - (( 8.314 J/K.mol) * ( 298 K ) * Ln ( 0.03 ))

⇒ ΔG = 16217.7496 J/mol ( 16.218 KJ/mol )

5 0
3 years ago
Temperature of the water to the nearest degree:___ °C
forsale [732]

Answer:

From the image the answer is 24.

8 0
3 years ago
Read 2 more answers
If 3 moles of a compound use 12 J of energy in a reaction, what is the Hreaction in kJ/mol
igomit [66]

Answer:

\Delta _RH=4x10^{-3}\frac{kJ}{mol}

Explanation:

Hello,

In this case, the molar enthalpy of reaction is obtained by dividing the involved energy by the reacting moles:

\Delta _RH=\frac{12J}{3mol} =4\frac{J}{mol}

Thus, it is important to notice that the compound "uses" the energy, it means that it absorbs the energy, for that reason the sign is positive. Moreover, computing the result in kJ/mol we finally obtain:

\Delta _RH=4\frac{J}{mol}*\frac{1kJ}{1000J} =4x10^{-3}\frac{kJ}{mol}

Best regards.

5 0
2 years ago
10) What are the four main spheres of planet?
Tomtit [17]

Answer:

the four main spheres of the earth are geosphere, hydrosphere, atmosphere and biosphere

Explanation:

geosphere consists of all rocks on Earth

atmosphere which are the gases that surrounds the earth

hydrosphere which is all the water on the earth

biosphere which are the living things on the earth

5 0
3 years ago
The relative atomic mass of Chlorine is 35.45. Calculate the percentage abundance of the two isotopes of Chlorine, 35Cl and 37Cl
nata0808 [166]

Answer:

35Cl = 75.9 %

37Cl = 24.1 %

Explanation:

Step 1: Data given

The relative atomic mass of Chlorine = 35.45 amu

Mass of the isotopes:

35Cl = 34.96885269 amu

37Cl = 36.96590258 amu

Step 2: Calculate percentage abundance

35.45 = x*34.96885269 + y*36.96590258

x+y = 1  x = 1-y

35.45 = (1-y)*34.96885269 + y*36.96590258

35.45 = 34.96885269 - 34.96885269y +36.96590258y

0.48114731 = 1,99704989‬y

y = 0.241 = 24.1 %

35Cl = 34.96885269 amu = 75.9 %

37Cl = 36.96590258 amu = 24.1 %

3 0
3 years ago
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