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meriva
2 years ago
8

How many oxygens are used per glucose molecule in glycolysis?

Chemistry
1 answer:
dimulka [17.4K]2 years ago
7 0

Answer:

0

Explanation:

You might be interested in
The solubility of Cr(NO3)3⋅9H2O in water is 208 g per 100 g of water at 15 ∘C. A solution of Cr(NO3)3⋅9H2O in water at 35 ∘C is
Zinaida [17]

Answer:

102g of crystals

Explanation:

When the Cr(NO₃)₃⋅9H₂O is dissolved in water at 15°C, the maximum mass that water will dissolve in the equilibrium is 208 g per 100g of water. When you heat the water, this mass will increases.

In this problem, at 35°C the water dissolves 310g in 100g of water, as in the equilibrium at 15°C the maximum mass is 208g, the mass of crystals that will form is:

310g - 208g = <em>102g of crystals</em>

<em>-Crystals are the Cr(NO₃)₃⋅9H₂O that is not dissolved-.</em>

I hope it helps!

5 0
3 years ago
Where does the term carbohydrate come from? Explain by referring to the parts of this word
Troyanec [42]
A carbohydrate comes from a chain of carbon atoms with an H2O associated with each other 
5 0
4 years ago
How many milliliters of 0.0050 N KOH are required to neutralize 41 mL of 0.0050 M H2SO4?
Kipish [7]

Answer:

V KOH = 41 mL

Explanation:

for neutralization:

  • ( V×<em>C </em>)acid = ( V×<em>C </em>)base

∴ <em>C </em>H2SO4 = 0.0050 M = 0.0050 mol/L

∴ V H2SO4 = 41 mL = 0.041 L

∴ <em>C</em> KOH = 0.0050 N = 0.0050  eq-g/L

∴ E KOH = 1 eq-g/mol

⇒ <em>C</em> KOH = (0.0050 eq-g/L)×(mol KOH/1 eq-g) = 0.0050 mol/L

⇒ V KOH = ( V×<em>C </em>) acid / <em>C </em>KOH

⇒ V KOH = (0.041 L)(0.0050 mol/L) / (0.0050 mol/L)

⇒ V KOH = 0.041 L

4 0
3 years ago
The solubility of copper chloride at 20 °C is 73 g/100 g of water. Kiera adds 100 g of copper chloride to 100 g of water and sti
OLEGan [10]

Answer:

m_{undissolved}=27g

Explanation:

Hello there!

In this case, according to the given information of the solubility of copper chloride, as the maximum amount of this salt one can dissolve without having a precipitate, we infer that since just 73 grams are actually dissolved, the following amount will remain solid as a precipitate:

m_{undissolved}=100g-73g\\\\m_{undissolved}=27g

Best regards!

3 0
2 years ago
what is ( H+) and (OH-) in a healthy person's blood that has a PH of 7.50 assume that the temperature of the blood is 298 K.
AlekseyPX

pH = - log [H⁺]

pH = 7.5

\tt [H^+]=10^{-7.5}=3.16\times 10^{-8}

pH + pOH = 14

7.5 + pOH = 14

pOH = 6.5

\tt [OH^-]=10^{-6.5}=3.16\times 10^{-7}

5 0
2 years ago
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