Asolution of hno3 is standardized by reaction with pure sodium carbonate. 2h++na2co3⟶2na++h2o+co2 a volume of 23.45±0.05 ml of hno3 solution was required for complete reaction with 0.9616±0.0009 g of na2co3 , (fm 105.988±0.001 g/mol ). find the molarity of the hno3 solution and its absolute uncertainty. note: significant figures are graded for this problem. to avoid rounding errors, do not round your answers until the very end of your calculations.

(0,7738±0,0009) M of HNO₃

Explanation:

The reaction of standarization of HNO₃ with Na₂CO₃ is:

2 HNO₃ + Na₂CO₃ ⇒ 2 Na⁺ + H₂O + CO₂ + 2NO₃⁻

To obtain molarity of HNO₃ we need to know both moles and volume of this acid. The volume is (23.45±0.05) mL and to calculate the moles it is necessary to obtain the Na₂CO₃ moles and then convert these to HNO₃ moles, thus:

0,9616 g of Na₂CO₃ × ( 1 mole / 105,988 g) =

9,073×10⁻³ mol Na₂CO₃ × ( 2 moles of HNO₃ / 1 mole of Na₂CO₃) = 1,815×10⁻² moles of HNO₃

Molarity is moles divide liters, thus, molarity of HNO₃ is:

1,815×10⁻² moles / 0,02345 L = 0,7738 M of HNO₃

The absolute uncertainty of multiplication is the sum of relative uncertainty, thus:

ΔM = M× (0,0009/0,9616 + 0,001/105,988 + 0,05/23,45) =

0,7738 M× 1,16×10⁻³ = 8,9×10⁻⁴ M

Thus, molarity of HNO₃ solution and its absolute uncertainty is:

(0,7738±0,0009) M of HNO₃

I hope it helps!

the answer is na (sodium)

water at room temperature is flowing through a pipe with pressure taps located 10 ft apart along one section of the pipe, which are connected to a mercury (s=13.6) manometer.