Phosphate group , pentose sugar , nitrogen base nucleotide
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
We need the reading for this I think
Answer: The correct answer to the question is option A. HBr + NaOH –> NaBr + H₂O
Explanation: An acid-base reaction is otherwise known as neutralization reaction. This is a reaction in which an acid reacts with a base to produce salt and water only.
