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astra-53 [7]
2 years ago
12

Chlorophyll a is one of the green pigments found in plants. Chlorophyll a has the molecular formula C55H72MgN4O5. How many atoms

are in this molecule?
Chemistry
1 answer:
algol132 years ago
8 0
The number of atoms in a molecule can be calculated by adding all of the number of elements from the chemical formula. For chlorophyll, the chemical formula would be <span>C55H72MgN4O5 adding all the elements we have 137 atoms. Hope this answers the question.</span>
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Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find
FromTheMoon [43]

<u>Answer:</u> The concentration of SO_4^{2-} at equilibrium is 0.00608 M

<u>Explanation:</u>

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

       H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)

            0.025          0.025       0.025

Equation for the second dissociation of sulfuric acid:

                    HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)

<u>Initial:</u>            0.025            0.025      

<u>At eqllm:</u>      0.025-x          0.025+x        x

The expression of second equilibrium constant equation follows:

Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}

We know that:

Ka_2\text{ for }H_2SO_4=0.01

Putting values in above equation, we get:

0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of SO_4^{2-} at equilibrium is 0.00608 M

4 0
3 years ago
Stoichiometry: Calculate the moles of H2 produced by .042 g Mg in the equation Mg + 2HCl &gt; MgCl2 + H2
Evgesh-ka [11]

Answer:

<em>Mg </em>(<em>s</em>) + 2<em>HCI2 </em>(<em>aq</em>) → <em>MgCI2 </em>(<em>aq</em>) + <em>H2 </em>(<em>g</em>)

I think this is the correct answer I not a 100% sure if it is correct.

Explanation:

Guessing

6 0
3 years ago
Which type of wave cannot move without a medium in which to travel?
Zanzabum

sound wave cannot travel without a medium.

5 0
2 years ago
How can I balance this equation? TELL ME HOW you did it and all the steps and you will be the brainliest.
lianna [129]

Answer:

3 CH3CH2OH + 4 H2CrO4 + 6 H2SO4 --> 3 CH3COOH + 2 Cr2(SO4)3 + 13 H2O

Explanation:

To balance, start of with the groups that are common on both sides of the reaction equation;

In this case, these are the SO4 groups treating them as single units;

There are 3 on the right side and 1 on the left side, so we put 3 in front of the H2SO4 to balance this first;

Next, deal with the Cr, there are 2 Cr on the right side and 1 on the other side, so we put a 2 in front of the H2CrO4 to balance that;

Thirdly, we notice the C are already balanced as there are 2 on each side so this is fine;

Lastly, we can deal with the O and H;

Bearing in mind the numbers that are in front of the molecules now from prior balancing, there are 9 O and 16 H on the left side and 3 O and 5 H on the right;

7 H2O on the right side would balance the O, but gives us 18 H, which is 2 too many H;

If we were to put 2 in front of the two organic molecules (the ones with C) on either side, we would balance the O by having 6 H2O, but this gives 2 fewer H than necessary;

In order for the H to balance, we need to have 13/2 (or 6.5) H2O, which means we need 3/2 (or 1.5) in front of each organic molecule;

Since, it is not sensible to have 13/2 water molecules or 3/2 organic molecules, we can just multiply everything by 2;

Thus we end up with:

3 CH3CH2OH + 4 H2CrO4 + 6 H2SO4 --> 3 CH3COOH + 2 Cr2(SO4)3 + 13 H2O

Rules of thumb:

- When there are common chemical groups (e.g. SO4) on both sides of a reaction equation, treat them as single units

- Start of with balancing these common groups

- Thereafter, balance the atoms that appear in only one reactant and one product

- Proceed to balance the atoms that appear in more than one reactant or product

- Typically, you should deal with the O and H last

4 0
3 years ago
MULTIPLE CHOICE
Flauer [41]

Answer:

its very simple ans we have 2 just multiply256

6 0
3 years ago
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