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:
When the volume increased from 2.00 to 5.25L the new temperature is 808.9 K ( =535.75 °C)
Explanation:
Step 1: Data given
The initial volume of the sample = 2.00 L
The initial temperature = 35 °C = 308 K
The increased volume = 5.25 L
Pressure = constant
Step 2: Calculate the new temperature
V1/T1 = V2/T2
⇒ with V1 = the initial volume = 2.00 L
⇒ with T1 = the initial volume = 308 K
⇒ with V2 = the new volume = 5.25 L
⇒ with T2 = the new temperature
2.00 / 308 = 5.25 / T2
0.00649 = 5.25/T2
T2 = 5.25/ 0.00649
T2 = 808.9 K
When the volume increased from 2.00 to 5.25L the new temperature is 808.9 K ( =535.75 °C)
If a material contains three elements joined in a fixed proportion, it is a(an) _____.
d.) compound
First, we need to get the number of moles of NH3:
no.of moles = mass of NH3 / molar mass
when we have the mass= 155g & the molar mass of NH3 = 14+(1x3) = 17 g/mol
∴ no.of moles = 155g/17mol/g
= 9.12 moles
∴ the heat (in Kj) = no.of moles of NH3 * (ΔHrxn/4moles NH3)
= 9.12 moles * (-906 / 4mol NH3)
= -2065.68 KJ