The volume of 0.160 M sulfuric acid necessary to react completely with 75.0 g sodium hydroxide is 0.15L
The reaction of sulphuric acid and sodium hydroxide is expressed as:
Determine the moles of NaOH present
Molar mass of NaOH = 23 + 16 + 1 = 40g/mole
Given that mass of NaOH is 75.0g
Mole of NaOH = 75/40 = 1.875moles
Determine the number of moles needed to react with 1.875 moles NaOH
Moles of =
Recall that volume = Molar mass * number of moles
Volume of = 0.160 * 0.9375
Volume of = 0.15L
Hence the volume of 0.160 M sulfuric acid necessary to react completely with 75.0 g sodium hydroxide is 0.15L
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Answer:
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Explanation:
Answer:
option C is correct (250 g)
Explanation:
Given data:
Half life of carbon-14 = 5700 years
Total amount of sample = 1000 g
Sample left after 11,400 years = ?
Solution:
First of all we will calculate the number of half lives passes during 11,400 years.
Number of half lives = time elapsed/ half life
Number of half lives = 11,400 years/5700 years
Number of half lives = 2
Now we will calculate the amount left.
At time zero = 1000 g
At first half life = 1000 g/2 = 500 g
At second half life = 500 g/2 = 250 g
Thus, option C is correct.