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4 years ago

The radioactive isotope used in the calvin experiments was ____.

1 answer:

4 0

The radioactive isotope used in the calvin experiments was Carbon 14. It was used as a tracer. The experiments performed by Melvin Calvin explained the process that plants use to make food. Hope this answers the question. Have a nice day.

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The initial volume of HCl was 1.25 ml and LiOH was 2.65 ml. The final volume of HCL was 13.60 ml and LiOH was 11.20 ml. If the L

Dennis_Churaev [7]

Q1)
This is a strong acid- strong base base reaction, balanced equation for the reaction is as follows;
LiOH + HCl ---> LiCl + H₂O
stoichiometry of acid to base is 1:1
volume of HCl used up - 13.60 - 1.25 = 12.35 mL
volume of LiOH used up - 11.20 - 2.65 = 8.55 mL
molarity of LiOH - 0.140 M
The number of LiOH moles reacted - $\frac{0.140 mol/L*8.55mL}{1000mL}$ = 0.001197 mol
according to stoichiometry, number of LiOH moles = number of HCl moles
Therefore number of HCl moles reacted - 0.001197 mol
The number of HCl moles in 12.35 mL - 0.001197 mol
Then number of HCl moles in 1000 mL - $\frac{0.001197*1000mL}{12.35mL}$
Molarity of HCl - 0.0969 M

Q2)
Volume of HCl used - 12.35 mL
Volume of LiOH used - 8.55 mL
Molarity of HCl - 0.140 M
In 1 L solution of HCl there are 0.140 mol of HCl
Therefore number of HCl moles in 12.35 mL - $\frac{0.140mol*12.35 mL}{1000mL}$
Number of HCl moles reacted - 0.001729 mol
since molar ratio of acid to base is 1:1
the number of LiOH moles that reacted - 0.001729 mol
Therefore number of moles in 8.55 mL - 0.001729
Then number of LiOH moles in 1000 mL - $\frac{0.001729mol*1000mL}{8.55 mL}$
molarity of LiOH - 0.202 M

5 0

3 years ago

Determine the elements with these electron configurations

lakkis [162]

Use a periodic table. Count the exponent for your atomic number. So, 1s2 2s2 2p5= Fluorine
1s2 2s2 2p6 3s2 3p5= Chlorine
1s2 2s2 2p6 3s2 3p6 4s2 3d6= Iron

7 0

3 years ago

Why does grinding solid crystals increase the rate of dissolving for a solid solute in water? smaller crystals are more polar. s

Sedbober [7]

Answer:

The answer is...

Explanation:

Smaller crystals have more surface area.

4 0

3 years ago

The molecular formula and molar mass of two straight-chain hydrocarbons are listed in the table above. Based on the information

nikdorinn [45]

Answer:

~Shoto todorki here~

The following table illustrates some of the factors that influence the strength of intermolecular attractions. The formula of each entry is followed by its formula weight in parentheses and the boiling point in degrees Celsius. First there is molecular size. Large molecules have more electrons and nuclei that create van der Waals attractive forces, so their compounds usually have higher boiling points than similar compounds made up of smaller molecules. It is very important to apply this rule only to like compounds. (Hope this helps!)

5 0

3 years ago

A 36.165 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion analysis ap

anyanavicka [17]

Answer:

C6H12SO2

Explanation:

Step 1:

Data obtained from the question.

First experiment:

Mass of the compound = 36.165 mg

Mass of CO2 = 64.425 mg

Mass of H2O = 26.373 mg

Second experiment:

Mass of compound = 47.029 mg

Mass of SO2 = 20.32 mg

Step 2:

Determination of the mass of C, H and S. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 64.425 Mass of C = 17.57 mg

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 26.373

Mass of H = 2.93 mg

Molar Mass of SO2 = 32 + (16x2) = 64g/mol

Mass of S in SO2 = 32/64 x 20.32

Mass of S = 10.16 mg

Step 3:

Determination of the the percentage composition of C, H, S and O. This is illustrated below:

% of C = 17.57/36.165 x 100 = 48.58%

% of H = 2.93/36.165 x 100 = 8.10%

% of S = 10.16/47.029 x 100 = 21.60%

% of O = 100 - (48.58 + 8.1 + 21.6)

% of O = 21.72%

Step 4:

Divide by their molar mass

C = 48.58x10^-3/12 = 4.0483

H = 8.10/1 = 8.1

S = 21.60/32 = 0.675

O = 21.72/16 = 1.3575

Step 5:

Divide by the smallest:

C = 4.0483/0.675 = 6

H = 8.1/0.675 = 12

S = 0.675/0.675 = 1

O = 1.3575/0.675 = 2

Therefore, the empirical formula is C6H12SO2

5 0

3 years ago