Answer:
half-life = 3.8 days
total time of decay = 15.2 days
initial amount = 100. g
number of half-lives past: 15.2/3.8 = 4 half-lives
4 half-lives = 1/16 remains
100. g x 1/16 = 6.25 g
Answer:
Q = 3937.56 J
Explanation:
Heat transferred due to change in temperature is given by :
c is the specific heat of water, c=4.18 J/g-°C
We have, m = 15 g,
So,
Hence, 3937.56 J of heat is transferred.
Answer:
k = 0.0306 min-1
Explanation:
The table is given as;
Time, Concentration
0 1.48
5 1.27
10 0.98
15 0.84
The integrated rate law for a first order reaction is given as;
ln [A] = -kt + ln [Ao]
where;
[A] = Final Concentration
[Ao] = Initial Concentration
k = rate constant
t = time
In the table, taking the first two sets of values;
t = 5
k = ?
[Ao] = 1.48
[A] = 1.27
Inserting into the equation;
ln(1.27) = - k (5) + ln(1.48)
ln(1.27) - ln(1.48) = -5k
-0.1530 = -5k
k = -0.1530 / -5
k = 0.0306 min-1
The answer for the following problem is mentioned below.
<u><em>Therefore volume occupied by methane gas is 184.78 × 10^-3 liters </em></u>
Explanation:
Given:
mass of methane() = 272 grams
pressure (P) = 250 k Pa =250×10^3 Pa
temperature(t) = 54°C =54 + 273 = 327 K
Also given:
R = 8.31JK-1 mol-1 ,
Molar mass of methane() = 16.0 grams
We know;
According to the ideal gas equation,
<u><em>P × V = n × R × T</em></u>
here,
n = m÷M
n =272 ÷ 16
<u><em>n = 17 moles</em></u>
Therefore,
250×10^3 × V = 17 × 8.31 × 327
V = ( 17 × 8.31 × 327 ) ÷ ( 250×10^3 )
V = 184.78 × 10^-3 liters
<u><em>Therefore volume occupied by methane gas is 184.78 × 10^-3 liters </em></u>
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Answer:C. The value of n for H+(H2O)n can be calculated for almost all solutions.
Explanation:
An hydrate can be described as a substance that contains water or with an hydrogen bonded water molecule group.
The hydrate group doesn't necessarily have a fixed formula.