Answer:
64.52 mg.
Explanation:
The following data were obtained from the question:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Final amount (N) =.?
Next, we shall determine the rate constant (K).
This is illustrated below:
Half life (t½) = 1590 years
Rate/decay constant (K) =?
K = 0.693 / t½
K = 0.693/1590
K = 4.36×10¯⁴ / year.
Finally, we shall determine the amount that will remain after 1000 years as follow:
Half life (t½) = 1590 years
Initial amount (N₀) = 100 mg
Time (t) = 1000 years.
Rate constant = 4.36×10¯⁴ / year.
Final amount (N) =.?
Log (N₀/N) = kt/2.3
Log (100/N) = 4.36×10¯⁴ × 1000/2.3
Log (100/N) = 0.436/2.3
Log (100/N) = 0.1896
Take the antilog
100/N = antilog (0.1896)
100/N = 1.55
Cross multiply
N x 1.55 = 100
Divide both side by 1.55
N = 100/1.55
N = 64.52 mg
Therefore, the amount that remained after 1000 years is 64.52 mg
CH4 + 2 O2 ---> CO2 + 2 H2O Q = 891,6 kJ / mol CH4
1 mol CH4 = 16 g
16 g ---- 891,6 kJ
x g ----- 272 kJ
x = 272 kJ × 16 g / 891,6 kJ = 4,88 g
You must burn 4,88 g of CH4.
:-) ;-)
Answer:
= 15.51 mL
Explanation:
Here's is the reaction:
2HgO(s) ⇒ 2 Hg(s)+O₂(g)
In this reaction 2mol HgO = 1mol O₂
The molecular weight of HgO = 216.59g
so, 3.0g HgO = 3.0g x 1.00molHgO/216.59gHgO
= 0.0138511 molHgO
The amount of Oxygen follows:
0.0138511 molHgOx1/2= 0.00692555 mol O₂
Now, volume of 1 any gas = 22400mL
so, 0.00692555 mol O₂ x22400mLO₂/1mol O₂
= 15.513232mL O₂
Answer:
That is the membrane of the cell
Explanation:
The cell membrane is made up of layers of protein and fat. It is made with a <em>phospholipid bilayer</em>. It allows some substances to pass into the cell while blocking others. This is known as <em>semipermeability</em>.
