Answer:
N2 is lost from the craft 6.9% faster than O2 is lost.
Explanation:
<u>Step 1:</u> Data given
0.500 atm of N2
0.500 atm of 02
Molar mass of 02 = 2*16 g/mole = 32 g/mole
Molar mass of N2 = 2* 14 g/mole = 28g/mole
<u>Step 2:</u> Calculate the rate of loss
r1N2 / r2O2 = √(M2(02)/M1(N2)) = √(32/28) = 1.069
1.069-1)*100= 6.9%
This means N2 is lost from the craft 6.9% faster than O2 is lost.
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Answer the correct answer for the question is option a Increase in temperature of water.
Explanation- When the temperature of water is increased the energy of water is increased at molecular level, this increased energy causes vibration between the hot water molecules bit faster than in normal water which also allows the sugar molecules to vibrate and associate with water at molecular level as it gets more space between the molecules.
Hence this action would make sugar more soluble in water.
Answer:
1) 0.0625 g.
2) 0.0125 g.
Explanation:
<em>1) A solution of NaOH has a concentration of 25.00% by mass. What mass of NaOH is present in 0.250 g of this solution?</em>
mass% of NaOH = [(mass of NaOH)/(mass of solution)] x 100.
mass% of NaOH = 25.0%, mass of NaOH = ??? g, mass of solution = 0.250 g.
∴ mass of NaOH = (mass% of NaOH)(mass of solution)/100 = (25.0%)(0.250 g)/100 = 0.0625 g.
<em>2) What mass of NaOH must be added to the solution to increase the concentration to 30.00% by mass?</em>
We can use the relation:
mass% of NaOH = [(mass of NaOH)/(mass of solution)] x 100.
mass% of NaOH = 30.0%, mass of NaOH = ??? g, mass of solution = 0.250 g.
∴ mass of NaOH = (mass% of NaOH)(mass of solution)/100 = (30.0%)(0.250 g)/100 = 0.075 g.
∴ The mass of NaOH should be added = 0.075 - 0.0625 = 0.0125 g.
Answer:
Too high a value
Explanation:
HA + NaOH ⟶ NaA +H₂O
If the student has gone slightly past the equivalence point, they have added too much base.
The moles of HA are directly proportional to the moles of NaOH, so the moles of acid that the student calculates will be too high.
The calculated concentration of acid will also be too high.
