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Zepler [3.9K]
3 years ago
5

Which equation is derived from the combined gas law?

Chemistry
2 answers:
Aloiza [94]3 years ago
5 0

Answer:

The correct answer is V1/T1=V2/T2.

Explanation:

Just took the test

nikklg [1K]3 years ago
3 0

Answer:

Its A on edge :)

Explanation:

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Choose the incorrect statement from the following:
Oksana_A [137]

Answer:

d

Explanation:

Generally, it is transported through pipes so I think statement d is incorrect.

4 0
2 years ago
A serving of Cheez-Its releases 130 kcal (1 kcal = 4.18 kJ) when digested by your body. If this same amount of energy were trans
Brut [27]

The final temperature : 78.925°C

<h3>Further explanation  </h3>

Heat can be calculated using the formula:  

Q = mc∆T  

Q = heat, J  

m = mass, g  

c = specific heat, joules / g ° C  

∆T = temperature difference, ° C / K  

Energy releases = 130 kcal = 130 x 4.18 kJ=543.4 kJ

The final temperature :

\tt \Delta T=\dfrac{Q}{m.c}\\\\\Delta T=\dfrac{543.4}{2.5\times 4.186~kJ/kg^oC}\\\\\Delta T=51.925^oC

Final temperature :

ΔT=final-initial

51.925°c=final-27°c

final = 51.925+27=78.925°C

7 0
2 years ago
The enzyme, phosphoglucomutase, catalyzes the interconversion
Fittoniya [83]

Answer:

K_{eq = 19

ΔG° of the reaction forming glucose 6-phosphate =  -7295.06 J

ΔG° of the reaction  under cellular conditions = 10817.46 J

Explanation:

Glucose 1-phosphate     ⇄     Glucose 6-phosphate

Given that: at equilibrium, 95% glucose 6-phospate is  present, that implies that we 5% for glucose 1-phosphate

So, the equilibrium constant K_{eq can be calculated as:

K_{eq = \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}

K_{eq= \frac{0.95}{0.05}

K_{eq = 19

The formula for calculating ΔG° is shown below as:

ΔG° = - RTinK

ΔG° = - (8.314 Jmol⁻¹ k⁻¹ × 298 k ×  1n(19))

ΔG° = 7295.05957 J

ΔG°≅ - 7295.06 J

b)

Given that; the concentration  for  glucose 1-phosphate = 1.090 x 10⁻² M

the concentration of glucose 6-phosphate is 1.395 x 10⁻⁴ M

Equilibrium constant  K_{eq can be calculated as:

K_{eq = \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}

K_{eq}= \frac{1.395*10^{-4}}{1.090*10^{-2}}

K_{eq} = 0.01279816514  M

K_{eq} = 0.0127 M

ΔG° = - RTinK

ΔG° = -(8.314*298*In(0.0127)

ΔG° = 10817.45913 J

ΔG° = 10817.46 J

5 0
3 years ago
1. It is the place near the coast where sea water and fresh water mixes and is called “nurseries
lawyer [7]

Answer:

Estuaries are often called the “nurseries of the sea”

Explanation:

Because so many marine animals reproduce and spend the early part of their lives there.

4 0
2 years ago
A weather balloon was initially at a pressure of 0.950 atm, and its volume was 35.0 L. The pressure decreased to 0.750 atm, with
puteri [66]

Answer: The change in volume will be 9.30 L

Explanation:

To calculate the new pressure, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

P_1V_1=P_2V_2

where,

P_1\text{ and }V_1 are initial pressure and volume.

P_2\text{ and }V_2 are final pressure and volume.

We are given:

P_1=0.950atm\\V_1=35.0L\\P_2=0.750atm\\V_2=?L

Putting values in above equation, we get:

0.950\times 35.0L=0.750\times V_2\\\\V_2=44.3L

The change in volume will be (44.3-35.0)L = 9.30 L

7 0
3 years ago
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