Answer:
The answer to your question is: letter C) 5730
Explanation:
Data
Originally Carbon - 14 254 g
Currently Carbon - 14 127 g
Half-life of Carbon -14 is when half of the original amount of Carbon disintegrates.
Then, if originally there were 254 and now there are 127, only one half life passed.
Half life of carbon -14 is 5730 years old.
Answer:
The net force is 2 N in the direction the book is being pushed.
Explanation:
Answer:
T2 = 29°C
Explanation:
Given data:
Heat added = 420 j
Mass of water = 25 g
Initial temperature = 25°C
Final temperature = ?
Solution;
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Specific heat capacity of water = 4.18 j/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
Now we will put the values.
420 j = 25 g ×4.18 j/g.°C × (Final temperature - initial temperature)
420 j = 25 g ×4.18 j/g.°C × (T2 - 25°C)
420 j = 104.5 j/°C × (T2 - 25°C)
420 j /104.5 j/°C = T2 - 25°C
4°C + 25°C = T2
T2 = 29°C
Answer:
T2 = 36.38°C
Explanation:
Given data:
Mass of water = 75 g
Initial temperature = 30 °C
Final temperature = ?
Heat absorbed = 2000 J
Solution:
Specific heat capacity of water is 4.18 J/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = Final temperature - initial temperature
2000 J = 75 g×4.18 J/g.°C × T2- T1
2000 J = 313.5 J/°C × T2- T1
2000 J = 313.5 J/°C × T2 - 30 °C
2000 J / 313.5 J/°C = T2 - 30 °C
6.38 °C = T2 - 30 °C
T2 = 6.38 °C + 30°C
T2 = 36.38°C
Answer:
for given question is 2.79 and 
is 0.52
{i- vant hoff’s constant ; Kb- constant ; m molarity }
M = no. of moles of the solute present in one kg of solution
Let the weight of amount of solute be “w” and its molecular mass be “M”
Let the mass of the solvent in the given question be “x”
