Answer:
C2H2O4
Explanation:
To get the molecular formula, we first get the empirical formula. This can be done by dividing the percentage compositions by the atomic masses. The percentage compositions are shown as follows :
C = 26.86%
H = 2.239%
O = 100 - ( 26.86 + 2.239) = 70.901%
We then proceed to divide by their atomic masses. Atomic mass of carbon is 12 a.m.u , H = 1 a.m.u , O = 16 a.m.u
The division is as follows:
C = 26.86/12 = 2.2383
H = 2.239/1 = 2.239
O = 70.901/16 = 4.4313
We now divide each by the smallest number I.e 2.2383
C = 2.2383/2.2383 = 1
H = 2.239/2.2383 = 1
O = 4.4313/2.2383 = 1.98 = 2
Thus, the empirical formula is CHO2.
To get the molecular formula, we use the molar mass .
(CHO2)n = 90
We add the atomic masses multiplied by n.
(12 + 1 + 2(16))n = 90
45n = 90
n = 90/45 = 2.
Thus , the molecular formula is C2H2O4
M = 22.1 g
V = 52.3 mL
D = ?
D = m/V
= 22.1/52.3
= 22.1*10/52.3*10
= 221/523
= 0.4
There. I’m sorry i forgot what exactly was the S.I. unit of density :(
Answer:
81.26% is the percent yield
Explanation:
Based on the reaction:
CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃
<em>Where 1 mole of CaCl₂ in excess of sodium carbonate produces 1 mole of calcium carbonate.</em>
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To solve this question we must find the moles of CaCl2 added = Moles CaCO₃ produced (Theoretical yield). The percent yield is:
Actual yield (0.366g) / Theoretical yield * 100
<em>Moles CaCl₂ = Moles CaCO₃:</em>
0.0500L * (0.0900moles / L) = 0.00450 moles of CaCO₃
<em>Theoretical mass -Molar mass CaCO₃ = 100.09g/mol-:</em>
0.00450 moles of CaCO₃ * (100.09g / mol) = 0.450g of CaCO₃
Percent yield = 0.366g / 0.450g * 100
81.26% is the percent yield
If the glasses and glove were wrong then I would chose the fire extinguisher and the power source should be the correct answer.
That's just what I would do though.
