What is a property of a base?
A. bitter taste
B. watery feel
C. unreactive
D. nonelectrolyte
1) A, Bases definitely have a bitter taste (not that I have tasted them before lol) while acids have a sour taste (like Coca-cola). I'm not sure why they taste the way they do but for sure this answer is not B,C or D because bases are definitely electrolytes (they dissolve in water) and they are not unreactive. Watery feel is quite a funny answer because I highly recommend you to not ever touch a base (they can burn you just as bad as an acid can if it has a really high pH or low pOH).
Friday, July 13, 2012
Thursday, July 12, 2012
How do you dilute a solution that contains 50 g of sucrose dissolved in 1L of water?
If you have a solution that contains 50g of sucrose in 1L of water, you can simply dilute it by adding more water to it. This is because if you calculate the current molarity of this solution (convert 50g of sucrose to moles and then divide it over 1L to find the molarity) it will be about 0.146moles/liters. If you add more liters of water, you will notice that the denominator will become larger and hence, the total moles per liter will begin to decrease or dilute.
Wednesday, July 11, 2012
Calculation of pH of solution using pKa?
20cm³ of 1mol per dm³ nitrous acid (pKa = 3.34) is added to 40 cm³ of 0.5 mol per dm³ sodium nitrate solution. What is the pH of the solution? Help please because I am unsure as to whether im supposed to calculate the pH using the number of moles or mol per dm³. And can you explain why mole or concentration is used.
It appears you a dealing with a buffer solution. However, the mixture is not that of a acid and its conjugate base. Perhaps you meant sodium nitrite (not nitrate). In that case
pH = pKa + log[NO2-]/[HNO2]
[NO2-] = 0.040*0.5/0.060 M (there are 60 cc of solution after mixing)
[HNO2] = 0.020*1/0.060 M
Since the volumes after mixing are the same, you can use moles or concentration, but the formula is based on molar concentration.
[NO2-]/[HNO2] = 0.040*0.5 / 0.020*1 = 1.0; log1.0 = 0, so pH = pKa
It appears you a dealing with a buffer solution. However, the mixture is not that of a acid and its conjugate base. Perhaps you meant sodium nitrite (not nitrate). In that case
pH = pKa + log[NO2-]/[HNO2]
[NO2-] = 0.040*0.5/0.060 M (there are 60 cc of solution after mixing)
[HNO2] = 0.020*1/0.060 M
Since the volumes after mixing are the same, you can use moles or concentration, but the formula is based on molar concentration.
[NO2-]/[HNO2] = 0.040*0.5 / 0.020*1 = 1.0; log1.0 = 0, so pH = pKa
Tuesday, July 10, 2012
Which is the least inert gas among them? He Ar Xe Ne Kr
The reactivity of the noble gases increases with increasing atomic number. Xenon has the largest atomic number in the given group.
Monday, July 9, 2012
Calculate the weight of lime(CaO) obtained by heating 200kg of 95% pure limestone(CaCO3)?
95% of 200 kg = 190 kg.
The molar mass of CaCO₃ is 100 g/mol, so 190 kg = (190,000 g)/(100 g/mol)
= 1900 mol.
CaCO₃ → CaO + CO₂, so 1900 mol of CO₂ are removed by heating. The molar mass of CO₂ is 44 g/mol, so 1900 mol = (1900 mol)(44 g/mol) = 83 600 g = 83.6 kg.
190 kg - 83.6 kg = 106.4 kg CaO.
The molar mass of CaCO₃ is 100 g/mol, so 190 kg = (190,000 g)/(100 g/mol)
= 1900 mol.
CaCO₃ → CaO + CO₂, so 1900 mol of CO₂ are removed by heating. The molar mass of CO₂ is 44 g/mol, so 1900 mol = (1900 mol)(44 g/mol) = 83 600 g = 83.6 kg.
190 kg - 83.6 kg = 106.4 kg CaO.
Sunday, July 8, 2012
If 0.30 mol of Zn are added to HCl containing 0.52 mol of HCl , how many moles of hydrogen are produced?
Zinc and HCl react according to the following reaction:
Zn(s) + 2 HCl(aq)-------->ZnCl2(aq) + H2(g)
You need 2 moles of HCl to neutralize 1 mole of Zn
So 0.52 moles of HCl produce 0.52/2= 0.26 moles of gas H2
Zn(s) + 2 HCl(aq)-------->ZnCl2(aq) + H2(g)
You need 2 moles of HCl to neutralize 1 mole of Zn
So 0.52 moles of HCl produce 0.52/2= 0.26 moles of gas H2
Balance in acidic soln:Tin(II) is oxidized to tin(IV) with hydrogen peroxide, Sn2+(aq)+H2O2(aq)=H2O(l)+Sn4+?c
Tin (II) is oxidized to tin (IV) with hydrogen peroxide: Sn2+ (aq) + H2O2 (aq) -> H2O (l) + Sn 4+ (aq). Balance the equation in an acidic solution.
Sn2+ -----------> Sn4+ + 2e- oxidation
H2O2 + 2H+ + 2e- ---------> 2H2O reduction
adding both equation, gives
Sn2+ + H2O2 + 2H+ ------------> 2 H2O + Sn4+
Sn2+ -----------> Sn4+ + 2e- oxidation
H2O2 + 2H+ + 2e- ---------> 2H2O reduction
adding both equation, gives
Sn2+ + H2O2 + 2H+ ------------> 2 H2O + Sn4+
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