Cellular respiration<span> is a set of metabolic reactions and processes that take place in the cells of organisms to convert biochemical energy from nutrients into adenosine triphosphate (ATP), and then release waste products. </span><span>do you get the c6h12o6 and o2 that is required for this process from the air we breath</span>
Carbon-Oxygen
We breathe in oxygen, we breathe out carbon dioxide.
Answer:
The statements that correctly describes pyruvate dehydrogenase includes:
- Several copies each of E 1 and E 3 surround E 2.
-A regulatory kinase and phosphatase are part of the mammalian PDH complex.
-E 2 contains three domains.
Explanation:
Pyruvate dehydrogenase is a hydrolase key enzyme in glucose metabolism which converts pyruvate to acetyl- ChoA. It also forms a complex that catalyzes an irreversible reaction that is the entry point of pyruvate into the TCA cycle. Pyruvate dehydrogenase complex contains E1, E2 and E3 enzymes that transform pyruvate, NAD+, coenzyme A into acetyl-CoA, CO2, and NADH. Also, A regulatory kinase and phosphatase are part of the mammalian PDH complex and E 2 contains three domains.
<h3><u>Answer</u>;</h3>
Actual yield = 46.44 g
<h3><u>Explanation;</u></h3>
1 mole of water = 18 g/mol
Therefore;
The experimental yield = 2.58 moles
equivalent to ; 2.58 × 18 = 46.44 g
The theoretical value is 47 g
Percentage yield = 46.44/47 × 100%
= 98.8%
The questions asks for actual yield = 46.44 g
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
