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MaRussiya [10]
2 years ago
10

How many grams are of sodium chlorate are present in a 2.45 mole sample?

Chemistry
2 answers:
natima [27]2 years ago
8 0

Answer:

260 grams sodium chlorate

Explanation:

moles to grams multiply by formula weight.

grams to moles divide by formula weight.

sodium chlorate => NaClO₃

=> formula wt = Na + Cl + 3O = [1(23) + 1(35) + 3(16)]g/mole = 106g/mole

∴ grams NaClO₃ = 2.45 mole x 106 g/mole = 259.7 grams ≈ 260 grams  

DedPeter [7]2 years ago
4 0

Answer:

The SI base unit for amount of substance is the mole. 1 mole is equal to 1 moles NaCl, or 58.44277 grams.

Explanation:

I hope this helps!!

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3 years ago
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posledela

Answer:

A.  Boyle's Law

B.  Charles' Law

C. Gay-Lussac's Law

Explanation:

An air bag inflates due to the decomposition of sodium azide or NaN₃ to completely fill the bag with nitrogen gas which is an example of Boyle's law, which states that the pressure of a given mass of gas is inversely proportional to its volume, hence due to the estricted volume of the airbag, the pressure of the nitrogen gas in the bag increses protecting the occupants of a cr from injuries in a crash

Helium balloon decrease in sice in a freezer is an example of Charlles law which states that the volume of a given mass of gas is nverslely proportionl to its temperature at constant pressure

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3 years ago
How many moles of Au2S3 is required to form 56 grams of H2S at STP?
Law Incorporation [45]

Answer:

0.55 mol Au₂S₃

Explanation:

Normally, we would need a balanced equation with masses, moles, and molar masses, but we can get by with a partial equation, if the S atoms are balanced.

1. Gather all the information in one place:

M_r:                          34.08

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m/g:                             56  

2. Calculate the moles of H₂S

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                      = 1.64 mol H₂S

3. Calculate the moles of Au₂S₃

The molar ratio is 1 mol Au₂S₃/3 mol H₂S.

Moles of Au₂S₃ = 1.64 mol H₂S × (1 mol Au₂S₃/3 mol H₂S)

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7 0
3 years ago
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mariarad [96]
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Now, concentration of the solution of sulfuric acid is 0.0980 g/L.
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