Answer: It will take 6.93 sec for tracer concentration to drop by 50% and 13.9 sec for tracer concentration to drop by 75%
Explanation:
Expression for rate law for first order kinetics is given by:
where,
k = rate constant =
t = age of sample
a = let initial amount of the reactant = 100 g
a - x = amount left after decay process
a) for tracer concentration to drop by 50%
a - x = amount left after decay process
= 50
It will take 6.93 sec for tracer concentration to drop by 50%
b) for tracer concentration to drop by 75 %
a - x = amount left after decay process
= 25
It will take 13.9 sec for tracer concentration to drop by 75%.
I'm certain that the answer is Sr
Answer:
1429.32 mmHg
Explanation:
Initial Pressure P1 = 3000.0mmHg
Initial Temperature T1 = 500.0°C + 273 = 573 K ( Converting to kelvin temperature)
Final Temperature T2 = 0.00°C + 273 = 273 K ( Converting to kelvin temperature)
Final Pressure P2 = ?
The pressure of a given amount of gas is directly proportional to the absolute temperature provided volume remains constant.
This is given by the mathematical expression;
P1 / T1 = P2 / T2
Inserting the values;
3000 / 573 = P2 / 273
P2 = 273 * 3000 / 573
P2 = 1429.32 mmHg
Answer:
Explanation:
Atom X,
Neutron num = 4, Mass num = 7
Therefore, proton number = Mass num - Neutron num
= 7 - 4
= 3
Atom Y,
Neutron num = 5, Mass num = 9
Therefore, proton number = Mass num - Neutron num
= 9 - 5
= 4
1) False. Atom X belongs to group 3 while atom Y belongs to group 4
2) False. Atom X is in a column on the left of Atom Y
3) True. Atom X and atom Y belong to the same period two on the periodic table
4) Fasle. Atom X and Y are not isotopes because they do not have the same atomic number.
Isotopy is a phenomenon whereby atom of same element have same atomic number but different mass number
Answer:
Q = -33.6kcal .
Explanation:
Hello there!
In this case, according to the equation for the calculation of the total heat of reaction when a fixed mass of a fuel like ethane is burnt, we can write:
Whereas n stands for the moles and the other term for the enthalpy of combustion. Thus, for the required total heat of reaction, we first compute the moles of ethane in 3 g as shown below:
Next, we understand that -337.0kcal is the heat released by the combustion of 1 mole of ethane, therefore, to compute Q, we proceed as follows:
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