2.22x10^-3 would be the answer to the question
please mark as brainliest answer
Its total charge is zero but for the elements:
Sn===> Sn4+ positive
Cl===> Cl- negative
<span>The half-life of a first-order reaction is determined as follows:
</span>t½<span>=ln2/k
From the equation, we can calculate the </span><span>first-order rate constant:
</span>k = (ln(2)) / t½ = 0.693 / 90 = 7.7 × 10⁻³
When we know the value of k we can then calculate concentration with the equation:
A₀ = 2 g/100 mL
t = 2.5 h = 150min
A = A₀ × e^(-kt) =2 × e^(-7.7 × 10⁻³ × 150) = 0.63 g / 100ml
= 6.3 × 10⁻⁴ mg / 100ml
Answer:
-) 3-bromoprop-1-ene
-) 2-bromoprop-1-ene
-) 1-bromoprop-1-ene
-) bromocyclopropane
Explanation:
In this question, we can start with the <u>I.D.H</u> (<em>hydrogen deficiency index</em>):
In the formula we have 3 carbons, 5 hydrogens, and 1 Br, so:
We have an I.D.H value of one. This indicates that we can have a cyclic structure or a double bond.
We can start with a linear structure with 3 carbon with a double bond in the first carbon and the Br atom also in the first carbon (<u>1-bromoprop-1-ene</u>). In the second structure, we can move the Br atom to the second carbon (<u>2-bromoprop-1-ene</u>), in the third structure we can move the Br to carbon 3 (<u>3-bromoprop-1-ene</u>). Finally, we can have a cyclic structure with a Br atom (<u>bromocyclopropane</u>).
See figure 1
I hope it helps!
Answer:
1/3
Explanation:
Pyruvate is produced by the glycolysis in cytoplasm. The oxidation of pyruvate takes place in mitochondrial matrix.
Pyruvate is converted to acetyl-CoA in the reaction given below:
Pyruvate + NAD⁺ + CoA-SH ⇒ acetyl-CoA + NADH + CO₂
1 molecule of carbon dioxide is eliminated from 1 molecule of pyruvate.
Also,
2 molecules of carbon dioxide is eliminated from 2 molecules of pyruvate (as glucose on glycolysis yields 2 molecules of pyruvate).
Also, acetyl-CoA further goes into the citric acid cycle and produces 2 molecules of carbon dioxide.
Thus pyruvate produces total 3 molecules of CO₂ and hence glucose produces 6 molecules of CO₂ (as glucose on glycolysis yields 2 molecules of pyruvate)
Thus,
<u>Fraction = 2/6 = 1/3</u>
