Use inversenormal on a calculator and type in .1 , 70 , 10 . That percent is for A and will determine at what mark an A will be. Do the same for the rest of the grades but change the first argument in the calculation
Answer:
.25
Explanation:
H+=10^-pH
- Hope that helps! Please let me know if you need further explanation.
Answer:
HCOOH(aq) + OH-(aq) —> HCOO-(aq) + H2O(l)
Explanation:
HCOOH is a weak acid and so will not ionised completely in solution.
KOH is a strong base and will ionised completely as shown below
KOH(aq) –> K+(aq) + OH-(aq)
The overall reaction can be written as follow:
HCOOH(aq) + K+(aq) + OH-(aq) —> HCOO-(aq) + K+(aq) + H2O(l)
Cancel out the K+ to obtain the net ionic equation as shown below
HCOOH(aq) + OH-(aq) —> HCOO-(aq) + H2O(l)
Answer:
milligram, centigram, decigram
<em>Hope</em><em> this</em><em> answer</em><em> correct</em><em> </em><em>:</em><em>)</em>
Answer:
i = 2.483
Explanation:
The vapour pressure lowering formula is:
Pₐ = Xₐ×P⁰ₐ <em>(1)</em>
For electrolytes:
Pₐ = nH₂O / (nH₂O + inMgCl₂)×P⁰ₐ
Where:
Pₐ is vapor pressure of solution (<em>0.3624atm</em>), nH₂O are moles of water, nMgCl₂ are moles of MgCl₂, i is Van't Hoff Factor, Xₐ is mole fraction of solvent and P⁰ₐ is pressure of pure solvent (<em>0.3804atm</em>)
4.5701g of MgCl₂ are:
4.5701g ₓ (1mol / 95.211g) = 0.048000 moles
43.238g of water are:
43.238g ₓ (1mol / 18.015g) = 2.400 moles
Replacing in (1):
0.3624atm = 2,4mol / (2.4mol + i*0.048mol)×0.3804atm
0.3624atm / 0.3804atm = 2,4mol / (2.4mol + i*0.048mol)
2.4mol + i*0.048mol = 2.4mol / 0.9527
2.4mol + i*0.048mol = 2.5192mol
i*0.048mol = 2.5192mol - 2.4mol
i = 0.1192mol / 0.048mol
<em>i = 2.483</em>
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I hope it helps!
