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2 years ago

13

If adequate o2 is present in the mitochondria, pyruvic acid will be converted to ______; if conditions are anaerobic, pyruvic ac

id will be converted to ______.

1 answer:

9966 [12]2 years ago

5 0

Presence of Oxygen in mitochondria is called Aerobic conditions and under this aerobic conditions Pyruvic acid is converted into Acetyl Co.A which enters Krebs' cycle
while in absence of oxygen which called Anaerobic conditions pyruvic acid is converted into Lactic acid to avoid accumulation of NADH⁺ which inhibit glycolysis and provide NAD⁺ required for glycolysis
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2 years ago

When 9.36 g sodium hydroxide is dissolved in enough water to make a total volume of 965 mL, what is the concentration of the res

SSSSS [86.1K]

Answer:

0.2425 M

Explanation:

Given mass = 9.36 g

Molar mass of NaOH = 39.997 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{9.36\ g}{39.997\ g/mol}$

$Moles\ of\ NaOH= 0.234\ mol$

Given that:- Volume = 965 mL = 0.965 L ( 1 mL = 0.001 L )

So,

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity=\frac{0.234\ mol}{0.965\ L}=0.2425\ M$

5 0

3 years ago

Read 2 more answers

Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)

Brums [2.3K]

Answer:1) Volume of $AgNO_3$ required is 55.98 mL.

2) 0.62577 grams of $Ag_3PO_4$ is produced.

Explanation:

$3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)$

1) Molarity of $AgNO_3,M_1=0.225 M$

Volume of $AgNO_3.V_1=?$

Molarity of $Na_3PO_4,M_2=0.135 M$

Volume of $Na_3PO_4,V_2=31.1 mL=0.0311 L$

$Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}$

$\text{number of moles }Na_3PO_4=M_2\times V_2=0.135 mol/L\times 0.0311 L=0.0041985 moles$

According to reaction, 1 mole of $Na_3PO_4$ reacts with 3 mole of $AgNO_3$ , then, 0.0041985 moles of $Na_3PO_4$ will react with:

$\frac{3}{1}\times 0.0041985$ moles of $AgNO_3$ that is 0.0125955 moles.

$M_1=0.225 M=\frac{\text{number of moles of }AgNO_3}{V_1}$

$V_1=\frac{0.0125955 moles}{0.225 M}=0.05598 L=55.98 mL$

Volume of $AgNO_3$ required is 55.98 mL.

2)

$Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}$

Number of moles of $AgNO_3=0.195\times 0.023 L=0.004485 moles$

According to reaction, 3 moles of $AgNO_3$ gives 1 mole of $Ag_3PO_4$ , then 0.004485 moles of $AgNO_3$ will give: $\frac{1}{3}\times 0.004485$ moles of $Ag_3PO_4$ that is 0.001495 moles.

Mass of $Ag_3PO_4$ =

Moles of $Ag_3PO_4$ × Molar Mass of $Ag_3PO_4$

= 0.001495 moles × 418.58 g/mol = 0.62577 g

0.62577 grams of $Ag_3PO_4$ is produced.

7 0

3 years ago