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Llana [10]
3 years ago
6

I need help with this pls

Chemistry
1 answer:
coldgirl [10]3 years ago
8 0

Answer:

C Is The Correct Answer Because The Deeper The Rock Is The Older It Is So The Top Is The Youngest And The Bottem Is The Oldest

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PLEASE HELP!!!!!!
jeyben [28]

<u>Answer:</u> The correct option is 1 mole of acetic acid was required for 2 moles of sodium bicarbonate

<u>Explanation:</u>

We are given:

Average number of drops of sodium bicarbonate = 142

The chemical equation for the reaction of sodium bicarbonate and acetic acid in vinegar follows:

NaHCO_3+CH_3COOH\rightarrow CH_3COONa+H_2CO_3

From the stoichiometry of the reaction:

1 mole of sodium carbonate reacts with 1 mole of acetic acid in vinegar.

Hence, the correct option is 1 mole of acetic acid was required for 2 moles of sodium bicarbonate

4 0
3 years ago
A compound is a substance composed of two or more elements that are
Anastaziya [24]
Chemically bonded together
3 0
4 years ago
How many moles of water are produced from 373 mol Al??
stepladder [879]
I cannot come up with a reaction which you can convert directly from Al to H2O.

But you can convert from Al2O3 to H2O by adding HCL solvent.
5 0
3 years ago
Read 2 more answers
A certain crunch cereal contains 11.0 grams of sugar(sucrose, C12H22O11) per serving size of 60.0 gram. How many
Paladinen [302]

Answer:

51.859 grams

Explanation:

Given:

Mass of sugar in the crunch cereal per serving= 11.0 grams

Mass of cereal served per serving = 60.0 grams

Molar mass of the sugar (C₁₂H₂₂O₁₁) = ( 12 × 12 + 22 × 1 + 16 × 11 ) = 342 grams

Number of moles of sugar per serving = Mass / Molar mass

= 11.0 / 342

= 0.03216 moles

Now,

0.03216 moles sugar requires 60 grams cereal

for 0.0278 moles of sugar, cereal required = (60 / 0.03216) × 0.0278

or

for 0.0278 moles of sugar, cereal required = 51.859 grams

5 0
4 years ago
Elements in group via (group 16) have similar properties because they all have __________.
Ksivusya [100]
They all have 2 elections in the outer S orbital, and 4 electrons in the P orbitals.
6 0
3 years ago
Read 2 more answers
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