During photosynthesis the chemical bonds between are broken

A vision for the public reading answers

nignag [31]

Can you say that again

3 0

3 years ago

Write a chemical equation depicting what happens to the sugar in water.

zysi [14]

Assuming that you mean table sugar (sucrose), then at room temperature and without any catalyst, there is no reaction.

However if you elevate and hold the temperature of the aqueous solution at 50 to 60 °C (especially in the presence of a suitable catalyst, like mineral acid) the sucrose dimer will split into glucose and fructose. This is called hydrolysis and the resulting solution is called an invert sugar solution.

The reaction could be written as:

C12H22O11 (sucrose) + H2O (water) → C6H12O6 (glucose) + C6H12O6 (fructose)

or

C12H22O11 (aq) + H2O (l) → C6H12O6 (aq) + C6H12O6 (aq)

Notice that both of the produced sugars have the same empirical formula. Check with your instructor or in your textbook to see if more exact formulas are needed.

5 0

3 years ago

Element in the compound are always _____ in fixed proportions.

AlekseyPX

(BRIANLIEST PLZZ NOT FORCING)

(〃▽〃)

5 0

3 years ago

PLEASE HELP! Will give brainliest answer for the right answer! ||| The specific types and sequence of amino acids determine the

vovangra [49]

Amino acids are Proteins,
Lipids are Fatty acids
Nucleic acid which have nucleotides and, build RNA and DNA
Starches are a cause of Carbohydrates, after lots of Polymers.

So, therefore the answer is a)

8 0

4 years ago

Be sure to answer all parts. For the titration of 10.0 mL of 0.250 M acetic acid with 0.200 M sodium hydroxide, determine the pH

DaniilM [7]

Explanation:

$Molarity=\frac{moles}{Volume(L)}$

Molarity of the acetic acid = 0.250 M

Volume of the acetic acid solution = 10.0 mL = 0.010 L( 1 mL =0.001L)

Moles of acetic acid ;

$n=0.250 M\times 0.010 L=0.0025 mol$

Molarity of the NaOH = 0.200 M

a) Volume of the NaOH solution = 10.0 mL = 0.010 L( 1 mL =0.001L)

Moles of NaOH : $0.200M\times 0.010 L=0.002 mol$

$CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O$

1 mole NaOH neutralizes 1 mole of acetic acid , then 0.002 moles of NaOH will neutralize 0.002 mol of acetic acid.

Moles of acetic acid left un-neutralized = 0.0025 mol - 0.002 = 0.0005 mol

1 mole of acetic acid gives 1 mole of hydrogen ion, then 0.0005 mole of acetic acid will give 0.0005 mole of hydrogen ions.

Moles of hydrogen ion= 0.0005 mol

Volume of the solution = 0.010 L+ 0.010 L = 0.020 L

$[H^+]=\frac{0.0005 mol}{0.020 L}=0.025 M$

The pH of the 10.0 mL of base added to acetic acid solution :

$pH=-\log[H^+]=-\log[0.025 M]=1.60$

b) Volume of the NaOH solution = 12.0 mL = 0.012 L( 1 mL =0.001L)

Moles of NaOH : $0.200M\times 0.012 L=0.0024 mol$

$CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O$

1 mole NaOH neutralizes 1 mole of acetic acid , then 0.0024 moles of NaOH will neutralize 0.0024 mol of acetic acid.

Moles of acetic acid left un-neutralized = 0.0025 mol - 0.0024 = 0.0001 mol

1 mole of acetic acid gives 1 mole of hydrogen ion, then 0.0001 mole of acetic acid will give 0.0001 mole of hydrogen ions.

Moles of hydrogen ion= 0.0001 mol

Volume of the solution = 0.010 L+ 0.012 L = 0.022 L

$[H^+]=\frac{0.0001 mol}{0.022 L}=0.0045 M$

The pH of the 12.0 mL of base added to acetic acid solution :

$pH=-\log[H^+]=-\log[0.0045 M]=2.34$

c) Volume of the NaOH solution = 15.0 mL = 0.015 L( 1 mL =0.001L)

Moles of NaOH : $0.200M\times 0.015 L=0.003 mol$

$CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O$

1 mole NaOH neutralizes 1 mole of acetic acid , then 0.003 moles of NaOH will neutralize 0.003 mol of acetic acid.

All the moles of acetic acid will get neutralized by NaOH and un-neutralized sodium hydroxide will left over.

Moles of NaOH left un-neutralized = 0.003 mol - 0.0025 = 0.0005 mol

1 mole of NaOH gives 1 mole of hydroxide ion, then 0.0005 mole of NaOH acid will give 0.0005 mole of hydroxide ions.

Moles of hydroxide ion= 0.0005 mol

Volume of the solution = 0.010 L+ 0.015 L = 0.025 L

$[OH^-]=\frac{0.0005 mol}{0.025 L}=0.02 M$

The pOH of the 15.0 mL of base added to acetic acid solution :

$pOH=-\log[OH^-]=-\log[0.02 M]=1.70$

The pH of the 15.0 mL of base added to acetic acid solution :

$pH=14-pOH=14-1.70=12.3$

7 0

3 years ago