Chemistry, 26.12.2019 04:31 girlwholikesanime

What volume of 0.600m h2so4 would be required to reach the equivalence point in a titration with a solution containing 2.50g of nahco3

What volume of 0.600m h2so4 would be required to reach the equivalence point in a titration with a s

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Answer from: 20emmanuelg1030

Hey there!:

Reaction stoichiometry :

Number of moles NaHCO3 :

Molar mass NaHCO3 = 84.007 g/mol

n = m / mm

n = 2.50 / 84.007

n = 0.0297moles of NaHCO3

2 NaHCO3 + H2SO4 = Na2SO4 + 2 CO2 + 2 H2O

2 moles NaHCO3 1 mole H2SO4
0.0297 moles NaHCO3 moles H2SO4

moles H2SO4 = 0.0297 * 1 / 2

moles H2SO4 = 0.0297 / 2

= 0.01485 moles of H2SO4

Therefore:

Molarity ( H2SO4 ) = moles H2SO4 / volume

0.600 M = 0.01485 / V

V = 0.01485 / 0.600

V = 0.02475 L of H2SO4

hope this helps!

Answer from: Quest

odds are decent that your handwriting right now looks a little bit different than it did five years ago, and will look different again in five more. when i look back through my notebooks from middle school to now, i see a long, slow progression from meticulous and teeny letters to bigger, sloppier ones, plus a few side detours: sometime in high school, for example, i decided those loopy cursive lowercase f’s were just the coolest, adopting them as the lone rep for cursive in my otherwise all-print handwriting.

explanation:

Answer from: Quest

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form( $n_2$ ) .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

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