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tekilochka [14]
2 years ago
11

Please help ill give brainly !

Chemistry
2 answers:
nlexa [21]2 years ago
6 0
Amount of wind in each area of the earth
i hope it’s correct, if it’s not sorry !
soldi70 [24.7K]2 years ago
3 0

Answer:

C

Explanation:

The Earth is a sphere and around the equator is where the closest parts of the Earth are to the Sun, while places like Antarctica are the furthest away from the Sun.

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The percent composition by mass of nitrogen in NH OH (gram-formula mass = 35 grams/mole) is equal to
Temka [501]

Answer:

the percentage composition of mass of nitrogen in NH OH is 42.86 %

7 0
3 years ago
Heat transfer putting a heating pad on leg
beks73 [17]

ANSWER IS CONDUCTION. HOPE THIS HELPED!

8 0
3 years ago
Which substance is a heterogeneous mixture
Anton [14]
Mixtures of sand and water or sand and iron filings, a conglomerate rock, water and oil, a portion salad, trail mix, and concrete (not cement).
6 0
3 years ago
Read 2 more answers
A block of metal has a volume of 14.0in3 and weighs 5.26lb .
natulia [17]
Mass of block, m = 5.26 lb  
1 lb = 453.592 g
 5.26 lb = 2385.896 g
V = 14 in3
 1 in = 2.5 cm
 1 in3 = 15.625 cm3
 14 in3 = 218.75 cm3


Density = mass/volume
 = 2358.896 / 218.75
 = 10.783 g/cm3
6 0
3 years ago
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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
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