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
The material which requires the most heat to raise its temperature from 10°C to 30°C is oil.
<h3>What is the formula to calculate absorbed heat?</h3>
The formula which we used to calculate the amount of involved heat in relation with specific heat is:
Q = mcΔT, where
- Q = absorbed heat
- m = mass
- c = specific heat
- ΔT = change in temperature
Among the given materials, specific heat of oil is highest than other materials so will require maximum absorbed heat.
Hence, oil requires the most heat.
To know more about specific heat, visit the below link:
brainly.com/question/6198647
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The equation is: C+O2=>CO2
Since we got 10 molecules of CO2 new balanced equation would be 10C+10O2=>10CO2
from this equation we can see that we have 10 molecules of oxygen, however ,we need to find atoms. There are 2 atoms in the oxygen molecule so we need to multiply 10 by 2 which gives us 20 atoms.
The answer: there are 20 atoms of oxygen
Answer:
4.56 X 10^ -4 g/mL
Explanation:
A solution is prepared by diluting 6.0 mL of a 7.6 x 10-4 g/mL solution to a total volume of 10.0 mL. Calculate the concentration of the dilute solution.
(7.6 X10^-4 gm/m L) x( 6.0 m L ) = 45.6 X 10^-4 g
this is dissolved )in 10 m L=45.6 X 10^-4 g/ 10
4.56 X 10^ -4 g/mL
check
6/10 =0.6
4.56/7.6 = o.,6