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faltersainse [42]
3 years ago
5

How many moles of sodium cyanide are needed to react completely with 2.6 moles of sulfuric acid

Chemistry
1 answer:
Tanya [424]3 years ago
4 0

7.2 mol NacN



Is the answer



Hope i helped <3

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Sodium hydrogen carbonate NaHCO3, also known as sodium bicarbonate or "baking soda", can be used to relieve acid indigestion. Ac
olga2289 [7]

Answer:

1.4952 grams of sodium bicarbonate she  would need to ingest to neutralize this much HCl.

Explanation:

Moles (n)=Molarity(M)\times Volume (L)

Moles of hydrochloric acid = n

Volume of hydrochloric acid solution = 200.0 mL = 0.200 L

Molarity of the hydrochloric acid = 0.089 M

n=0.089 M\times 0.200 L=0.0178 mol of HCL

HCl(aq)+NaHCO_3(aq)\rightarrow NaCl(aq)+H_2O(l)+CO_2(g)

According to reaction, 1 mole of HCl is neutralized by 1 mole of sodium bicarbonate.

Then 0.0178 moles of HCl wil be neutralized by :

\frac{1}{1}\times 0.0178 mol=0.0178 mol of sodium bicarbonate

Mass of 0.0178 moles of sodium bicarbonate:

0.0178 mol × 72 g/mol = 1.4952 g

1.4952 grams of sodium bicarbonate she  would need to ingest to neutralize this much HCl.

8 0
3 years ago
According to our textbook, write the formula for the anhydrous compound that was part of the mixture called natron that was used
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Answer:

The formula for the anhydrous compound that was part of the mixture called natron that was used by the Egyptians is Na2(CO3)10(H2O).

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Compound of sodium carbonate and sodium bicarbonate was the name of the resulting hydrate that formed.

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What tests can you do to identify gases, based on their chemical properties?
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Tests for gases
Hydrogen, oxygen, carbon dioxide, ammonia and chlorine can be identified using different tests.
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Oxygen. A glowing wooden splint relights in a test tube of oxygen.
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Answer:

Ok; just use PV=nRT solve for n.

Explanation:

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A certain protein was found to contain 0.292% manganese by mass. Determine the minimum molecular mass of the protein.
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55u \ \ \ \ \ \ \Rightarrow \ \ \ \ \ 0,292\%\\&#10;x \ \ \ \ \ \ \ \ \ \Rightarrow \ \ \ \ \ 100\%\\\\&#10;x = \frac{55u*100\%}{0,292\%}\approx18835,62u
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3 years ago
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