1.True
2.False
3.B
4.A
5.True
Answer:
3.6 grams of NaCl are needed
Explanation:
Percent solution are solutions whose concentrations are expressed in percentages. The amount(either weight or volume) of a solute is expressed as a percentage of the total weight or volume of solution. Percent solutions can either be expressed as weight/volume % (wt/vol % or w/v %), weight/weight % (wt/wt % or w/w %), or volume/volume % (vol/vol % or v/v %).
A 6.0% wt/wt % solution contains 6 g of solute in 100 g of solution
Therefore, a 100 g solution contains 6.0 g of solute.
60 g of 6.0% solution will contain 60 g solution * 6.0 g solute/ 100 g solution
Mass of NaCl present = 3.6 g of NaCl
A) when the balanced equation of the reaction is:
H2CO3(aq) → HCO3 -(aq) + (H+)
and when we have Ka = 4.3 x10^-7 & PH = 7.4
So first we will get PKa = -㏒ Ka
PKa = -㏒(4.3x10^-7) = 6.37 by substitution with Pka value in the following formula:
PH = Pka + ㏒[salt/acid]
PH= PKa + ㏒[HCO3-]/[H2CO3]
㏒[HCO3-]/[H2CO3] = PH-Pka
[HCO3-] /[H2CO3] = 10^(7.4 - 6.37)
∴[HCO3-]/[H2CO3] = 11.7
∴[H2CO3]/[HCO3-] = 1/11.7 = 0.09
B) when The balanced equation for this reaction is:
H2PO42-(aq) → HPO4-(aq) + H+
and when we have Ka = 6.2x10^-8 & PH = 7.15
So Pka= -㏒Ka = -㏒(6.2x10^-8) = 7.2 by substitution by Pka value in the following formula:
PH = Pka + ㏒[salt/acid]
7.15= 7.2 + ㏒[HPO4]/[H2PO4]
-0.05 = ㏒[HPO4]/[H2PO4]
∴[HPO4]/[H2PO4] = 10^-0.05 = 0.89
∴[H2PO4]/[HPO4] = 1/0.89 = 1.12
c) H3PO4(aq) ↔ H2PO-(aq) + H+
the answer is: because we have Ka =7.5x10^-3 and it is a high value of Ka to make a good buffer, also we need a week acid with th salt of the week acid as H3PO4 is a strong acid so it does'nt make a goof buffer.
Noble gasses have full outer shells giving them 8 electrons
The answer is it is exothermic.
