D. propane and hydrogen gass
Answer:
Molecular formula = C20H30
Explanation:
NB 440mg = 0.44g, 135mg= 0.135g
From the question, moles of CO2= 0.44/44= 0.01mol
Since 1 mol of CO2 contains 1mol of C, it implies mol of C = 0.01
Also from the question, moles of H2O = 0.135/18= 0.0075mole
Since 1 mol of H2O contains 2mol of H, it implies mol of H = 0.0075×2= 0.015 mol of H
To get the empirical formula, divide by smallest number of mole
Mol of C = 0.01/0.01=1
Mol of H = 0.015/0.01= 1.5
Multiply both by 2 to obtain a whole number
Mol of C =1×2 = 2
Mol of H= 1.5×2 = 3
Empirical formula= C2H3
[C2H3] not = 270
[ (2×12) + 3]n = 270
27n = 270
n=10
Molecular formula= [C2H3]10= C20H30
The initial equation is <span>F = (9/5)C + 32. We are asked to get the formula to solve for C. So the idea here is to isolate C on the other side, where F is currently located.
1. Subtract 32 on both sides.
2. Divide 9/5 on both sides.
You should get </span><span>C = 5 over 9(F – 32) as your answer. </span>
Answer:
<em>The accurate label for the solution is</em><em> </em><u><em>2.00 mol/L NaCl.</em></u>
Explanation:
Molrity = mass ÷ molar mass
Molar mass of NaCL = 58.44 g/mol
Weighed mass = 116.9g
⇒ Molarity of the solution = 
= 2.0M
<em />
<em>Therefore, the accurate label for the solution is </em><u><em>2.00 mol/L NaCl.</em></u>
