Answer:
I think it will be (B) only because you have to do it repeatedly and analyze it and not noted, in order for the results are accepted. I May be wrong. Thanks for letting me help you.
decalescent energy-absorbing endothermal heat-absorbing endoergic.
Hope this helps!
<span>Answer:
Mass % KCL:
Add the grams of both compounds (31.0 g KCL + 225 g water) to find total mass and then divide the grams of KCL over the total mass, then multiply by 100: ( 31.0 g KCL / 31.0 g + 225 G) * 100%
Mole fraction KCL
Calculate the moles of KCL and water and add them to find the total moles (Moles of KCL + moles of water). Then, divide the number of KCL moles over the total moles.
moles of KCL/ moles kcl + moles water= mole fraction of KCL</span>
Answer:
30 moles of ethanol are needed to prepare a 25 m solution, using 1200 g of water
Explanation:
This a short problem of solutions:
Our solute is ethanol: C₂H₅OH
Our solvent is water-
25 m, is our solution's concentration. m means molality (moles of solute contained in 1kg of solvent).
Our solvent mass is 1200g. We convert them to kg
1200 g . 1kg / 1000g = 1.2 kg
m = mol/kg → mol = m . kg
mol = 25 mol/kg . 1.2kg →30 moles
Find the number of moles of sodium you have:
<span>n = m/M where m is your 20g of sodium and M is 22.99 g/mol. </span>
<span>Look at the stoichiometry of the equation - it's 2:2 when you are producing NaOH. So if you took 1 mole of Na, it'd produce 1 mole of NaOH (as the ratio is equal). </span>
<span>That means that your moles of sodium is equal to the moles of NaOH produced. Use the molar mass of NaOH - which is 39.998 g/mol along with your calculated number of moles to get the mass (the formula rearranges to m = nM). </span>
<span>This figure is the theoretical yield - what you would get if every last mole of sodium was converted into NaOH. </span>
<span>What you get in practice is the experminetal yield, and the percentage yield is the experimental yield divided by the theoretical yield - and then multiplied by 100%.</span>
