Remembering that
d = m ÷ v
d = ?
m = 89 g
v = 10 cm³
Therefore:
d = 89 ÷ 10
d = 8,9 g÷cm³
Answer :
121.5 <span>
μCi
Explanation : We have Ce-141 half life given as 32.5 days so if the activity is 3.8 </span><span>μci after 162.5 days of time elapsed we have to find the initial activity.
We can use this formula;
</span>

3.8 /

=

((0.693 X 162.5 ) / 32.5) = 121.5
<span>
On solving we get, The initial activity as 121.5 </span>μci
Answer:
K8S4O16 or K8(SO4)4 depending on if the SO4 is supposed to represent sulfate or not
Explanation:
Find the molar mass of K2SO4 first:
2K + S + 4O ≈ 174 g/mol
Divide the goal molar mass of 696 by the molar mass of the empirical formula:
696 / 174 = 4
This means you need to multiply everything in the empirical formula by 4:
K2SO4 --> K8S4O16 or K8(SO4)4 depending on if the SO4 is for sulfate or not
I think the answer would be nitrogen
Answer: The final temperature of both the weight and the water at thermal equilibrium is
.
Explanation:
The given data is as follows.
mass = 7.62 g, 
Let us assume that T be the final temperature. Therefore, heat lost by water is calculated as follows.
q =
= 
Now, heat gained by lead will be calculated as follows.
q =
=
According to the given situation,
Heat lost = Heat gained
= 
T = 
Thus, we can conclude that the final temperature of both the weight and the water at thermal equilibrium is
.