Answer:
0.296 J/g°C
Explanation:
Step 1:
Data obtained from the question.
Mass (M) =35g
Heat Absorbed (Q) = 1606 J
Initial temperature (T1) = 10°C
Final temperature (T2) = 165°C
Change in temperature (ΔT) = T2 – T1 = 165°C – 10°C = 155°C
Specific heat capacity (C) =..?
Step 2:
Determination of the specific heat capacity of iron.
Q = MCΔT
C = Q/MΔT
C = 1606 / (35 x 155)
C = 0.296 J/g°C
Therefore, the specific heat capacity of iron is 0.296 J/g°C
Find the density of the plastic. if the density is less than that of water, it's going to float. If it's more than the density of water, then it's going to sink.
I am not sure but this is what I think
This question uses the formula connecting mass, density and volume
Which is Density= Mass/Volume
Convert the mass in g
92.5 kg = 92,500g
7.87g/ml = 92,500/ Volume
Volume= 92,500/7.87
= 11,753.5 ml
Now since we have to give the answer in liters we can just divide by 1000 and get
11.75 litres
Answer:
4 g OF IODINE-131 WILL REMAIN AFTER 32 DAYS.
Explanation:
Half life (t1/2) = 8 days
Original mass (No) = 64 g
Elapsed time (t) = 32 days
Mass remaining (Nt) = ?
Using the half life equation we can obtain the mass remaining (Nt)
Nt = No (1/2) ^t/t1/2
Substituting the values, we have;
Nt = 64 * ( 1/2 ) ^32/8
Nt = 64 * (1/2) ^4
Nt = 64 * 0.0625
Nt = 4 g
So therefore, 4 g of the iodine-131 sample will remain after 32 days with its half life of 8 days.
