Answer:
41.9(w/w) %
Explanation:
Based on the reaction:
Na₂C₂O₄(s) + 2HCl(aq) → H₂C₂O₄(aq) + 2NaCl(aq)
<em>Where 1 mole of sodium oxalate reacts with 2 moles of HCl</em>
Moles of HCl solution to reach end point are:
44.15mL = 0.04415L ₓ (0.250mol / L) = 0.01104 moles of HCl
As 2 moles of HCl reacts per mole of Na₂C₂O₄:
0.01104mol HCl ₓ (1 mol Na₂C₂O₄ / 2 mol HCl) = <em>5.519x10⁻³ moles Na₂C₂O₄</em> are in the sample.
Molar mass of Na₂C₂O₄ is 134g/mol; thus, mass of 5.519x10⁻³ moles Na₂C₂O₄ is:
5.519x10⁻³ moles Na₂C₂O₄ ₓ (134g / mol) = <em>0.740g of Na₂C₂O₄</em> in the sample.
Thus, percent by mass of sodium oxalate in the sample is:
0.740g of Na₂C₂O₄ / 1.766g ₓ 100 =
<h3>41.9(w/w) %</h3>
Answer:
A meso compound is a non-optically active member of a set of stereoisomers, at least two of which are optically active.
Answer:
by using ideal gas law
Explanation:
ideal gas law:
PV=nRT
where:
P is pressure measured in Pascal (pa)
V is volume measured in letters (L)
n is number of moles
R is ideal gas constant
T is temperature measured in Kelvin (K)
by applying the given:
P(initial) V(initial)=nRT(initial)
P(final) V(final)=nRT(final)
nR is constant in both equations since same gas
then,
P(initial) V(initial) / T(initial) = P(final) V(final) / T(final)
then by crossing multiply both equations
V (final)= { (P(initial) V(initial) / T(initial)) T(final) } /P (final)
P(initial)=P(final)= 1 atm = 101325 pa
V(initial)= 6 L
T(initial) = 28°c = 28+273 kelvin
T(final) = 39°c = 39+273 kelvin
by substitution
V(final) = 6.21926 L
Answer:
100 °C
Explanation:
Well it actually It depends on temperature and altitude
. The boiling point of water is 100 °C or 212 °F at 1 atmosphere of pressure (sea level). However, the value is not a consistent. The boiling point of water depends on the atmospheric pressure, which changes according to elevation.
It is called the Empirical Formula.