Problem One
<em><u>Formula</u></em>
N(t) = No * (1/2)^[t/t_1/2]
<em><u>Givens</u></em>
N(t) = the current mass of the sample = 2.10 grams
No = The original mass of the sample = No [We're trying to find this].
t = time elapsed which is 2.6 billion years or 2.6 * 10^9 years.
t1/2 = the 1/2 life time which is 1.3 billion years of 1.3 *10^9
<em><u>Solution</u></em>
2.10 grams = No (1/2)^(2.6*10^9/1.3 * 10^9)
The 10^9s cancel and you are left with 2.6/1.3 = 2
2.10 grams = No (1/2)^2
2.10 grams = No (1/4) Multiply both sides by 4
2.10 * 4 = No (1/4)*4
8.4 grams = No
which is how many grams you originally had.
Answer B.
Problem Two
Solve for y
2 + 2 = y + 1
4 = y + 1
y = 3
Solve for z
1 + 1 = z + 0
z = 2
The 2 tells you that it is the second member on the periodic table. That's Helium. So the answer looks like this.
The mass of the Helium is 3 and its number is two.
Answer:
a. 0.80V b. 2Ag⁺(aq) + H2(g) ⇄ 2Ag(s) +2H⁺(aq) c. i) 0.88 ii) 1.03 d. Cell is a ph meter with the potential being a function of hydrogen ion concentration
Explanation:
a. The two half cell reactions are
1. 2H⁺(aq) +2e⁻ → H₂(g) Eanode = 0.00V
2. Ag⁺(aq) + e⁻ → Ag(s) Ecathode = 0.80V
The balanced cell reaction is
2Ag(aq)⁺ + H₂(g) ⇄ 2Ag(s) + 2H⁺(aq)
therefore Ecell = Ecathode - Eanode = 0.80 - 0.00 = +0.80V
b. Since the Ecell is positive, the spontaneous cell reaction under standar state conditions is
2Ag(aq)⁺ + H₂(g) ⇄ 2Ag(s) + 2H⁺(aq)
c. Use Nernst Equation
E = Ecell - (0.0592/n)log([H⁺]/[Ag⁺]²[P H₂]), where n is the number of moles of Ag and P H₂= 1.0 atm
i) E = 0.80 - (0.0592/2)log(4.2x10^-2)/(1.0)²(1.0) = 0.88V
ii) E = 0.80 - (0.0592/2)log(9.6x10^-5)/(1.0)²(1.0) = 1.03V
d . From the above calculation we can conclude that the cell acts as a pH meter as a change in hydrogen ion concentration results in a change in the potential of the cell. A change of ph of 2.64 changes the E of cell by 0.15 V.
Answer:
There are 1.8 moles of aluminum.
Explanation:
To solve this problem we need to convert moles of pyrophillite (Al₂Si₄O₁₀(OH)) into moles of aluminum (Al):
- In one mol of Al₂Si₄O₁₀(OH) there are two moles of Al.
With the above information in mind we can now <u>do the conversion</u>:
- 0.900 moles Al₂Si₄O₁₀(OH) *
= 1.8 mol Al
Thus the answer is that there are 1.8 moles of aluminum.
Mulch includes compost, grass clipping, leaves or other organic matter which increaese the fertility of soil by providing the necessary nutrients.
Volume of much used 
To converting into m^3, divide it by 10^6
Thus, riya is applying mulch at rate of 0.25 m^3 per m^2 of garden.
