I think this is what you mean:
H H H H
H-C-C-C-C-H
H H H H
OR
<span>CH3CH2CH2CH3
</span>
If not, clarify and I will be happy to help.
Answer:
Haven't evaporated all of the water
Explanation:
One of the main sources of error that occur in a formula of a hydrate lab is that all of the water is not evaporated. We can see at the end of the video that half of the CoCl2 is a light blue colors and the other half is a dark blue color. This indicates that all of the water still has not been evaporated off, resulting in the actual mass of the salt to be greater than the predicted value.
Answer:
87.9%
Explanation:
Balanced Chemical Equation:
HCl + NaOH = NaCl + H2O
We are Given:
Mass of H2O = 9.17 g
Mass of HCl = 21.1 g
Mass of NaOH = 43.6 g
First, calculate the moles of both HCl and NaOH:
Moles of HCl: 21.1 g of HCl x 1 mole of HCl/36.46 g of HCl = 0.579 moles
Moles of NaOH: 43.6 g of NaOH x 1 mole of NaOH/40.00 g of NaOH = 1.09 moles
Here you calculate the mole of H2O from the moles of both HCl and NaOH using the balanced chemical equation:
Moles of H2O from the moles of HCl: 0.579 moles of HCl x 1 mole of H2O/1 mole of HCl = 0.579 moles
Moles of H2O from the moles of NaOH: 1.09 moles of HCl x 1 mole of H2O/1 mole of NaOH = 1.09 moles
From the calculations above, we can see that the limiting reagent is HCl because it produced the lower amount of moles of H2O. Therefore, we use 0.579 moles and NOT 1.09 moles to calculate the mass of H2O:
Mass of H2O: 0.579 moles of H2O x 18.02 g of H2O/1 mole of H2O = 10.43 g
% yield of H2O = actual yield/theoretical yield x 100= 9.17 g/10.43 g x 100 = 87.9%
Answer:
0.282 M
General Formulas and Concepts:
<u>Chemistry - Solutions</u>
- Reading a Periodic Table
- Using Dimensional Analysis
- Molarity = moles of solute / liters of solution
Explanation:
<u>Step 1: Define</u>
5.85 g KI
0.125 L
<u>Step 2: Identify Conversions</u>
Molar Mass of K - 39.10 g/mol
Molar Mass of I - 126.90 g/mol
Molar Mass of KI - 39.10 + 126.90 = 166 g/mol
<u>Step 3: Convert</u>
<u />
= 0.035241 mol KI
<u>Step 4: Find Molarity</u>
M = 0.035241 mol KI / 0.125 L
M = 0.281928
<u>Step 5: Check</u>
<em>We are given 3 sig figs. Follow sig fig rules and round.</em>
0.281928 M ≈ 0.282 M