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2 years ago

The oxidation of glucose by enzymes gives

Chemistry

1 answer:

Lorico [155]2 years ago

4 0

Answer:

Carbondioxide, water and ATP

Explanation:

The reaction is called aerobic respiration because oxygen from the air is needed for it to work.

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A sample contains 2.2 g of the radioisotope niobium-91 and 15.4 g of its daughter isotope, zirconium-91. how many half-lives hav

dybincka [34]

Answer: 3

Explanation: This is a radioactive decay and all the radioactive process follows first order kinetics.

Equation for the reaction of decay of $_{19}^{40}\textrm{K}$ radioisotope follows:

$Moles of zirconium=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{15.4}{91}=0.17moles$

$Moles of niobium=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{2.2}{91}=0.024moles$

$_{41}^{91}\textrm{Nb}\rightarrow _{40}^{91}\textrm{Zr}+_{+1}^0e$

By the stoichiometry of above reaction,

1 mole of $_{40}^{91}\textrm{Zr}$ is produced by 1 mole $_{41}^{91}\textrm{Nb}$

So, 0.17 moles of $_{40}^{91}\textrm{Zr}$ will be produced by = $\frac{1}{1}\times 0.17=0.17\text{ moles of }_{40}^{91}\textrm{Nb}$

Amount of $_{82}^{212}\textrm{K}$

decomposed will be = 0.17 moles

Initial amount of $_{40}^{91}\textrm{Nb}$ will be = Amount decomposed + Amount left = (0.17 + 0.024)moles =0.194 moles

$a=\frac{a_o}{2^n}$

where,

a = amount of reactant left after n-half lives = 0.024

$a_o$ = Initial amount of the reactant = 0.194

n = number of half lives= ?

Putting values in above equation, we get:

$0.024=\frac{0.194}{2^n}$

$n=3$

Therefore, 3 half lives have passed.

3 0

2 years ago

Read 2 more answers

In an experiment, 4.14 g of phosphorus combined with chlorine to produce 27.8 g of a white solid compound. what is the empirical

Umnica [9.8K]

Grams of Phosphorus = 4.14 grams
Grams of white compound = 27.8 grams
Grams of Chlorine would be = 27.8 - 4.14 = 23.66 grams
Calculating moles which would be grams / molar mass
Molar mass of P = 30.97 grams / moles; Molar mass of Cl = 35.45 grams / moles
Moles of Phosphorus = 4.14 grams / 30.97 grams / moles = 0.1337 moles
Moles of Chlorine = 23.66 grams / 35.45 grams / moles = 0.6674 moles
Calculating the ratios by dividing with the small entity
P = 0.1337 moles / 0.1337 moles = 1
Cl = 0.6674 moles / 0.1337 moles = 5
So the empirical formula would be PCl5

3 0

2 years ago

Which of the following mixtures, with each component present at a concentration of 0.1 M, has a pH closest to 7? O A. HCIO(aq) a

eduard

for HClO, pKa = 7.54

for HNO_2, pKa = 3.15

for CH3 COOH, pKa = 4.74

Explanation:

The concentration of the solution given is 0.1 M has a pH closest to 7

The mixtures are weak acids and their salts except

HNO_3 and NaNO_3 pH = pH is near to '1'

for buffers( acidic) pH = pKa + log [salt] / [acid]

therefore [salt] = [acid] = 0.1 pH = pKa + log 0.1 / 0.1 = pKa

pH = pKa

for HClO, pKa = 7.54

for HNO_2, pKa = 3.15 therefore HClO and NaclO

mixture hs a pH closest to '7'

for CH3 COOH, pKa = 4.74

5 0

2 years ago

Read 2 more answers

How would you combine reactions A–C, shown below, to obtain the overall reaction: $$ Please select all that apply.

snow_lady [41]

Answer:

By eating a sandwich

Explanation:

your welcome nigel

4 0

2 years ago

A solution surrounding a cell is hypertonic solution if: a. It contains fewer nonpenetrating solute particles than the interior

arlik [135]

Answer:

The answer is B. It contains more non-penetrating solute particles than the interior of the cell.

Explanation:

This means that it has a greater concentration or number of solute particles outside a membrane than there are inside it.

A typical example is Saline solution.

8 0

2 years ago