We are given with the total mass of fertilizer which is 14.1 oz. This is equivalent to 399.73 grams. 15% of the total is the amount of nitrogen. Thus, the nitrogen amount is 59.96 grams.
Answer:
c = 0.898 J/g.°C
Explanation:
1) Given data:
Mass of water = 23.0 g
Initial temperature = 25.4°C
Final temperature = 42.8° C
Heat absorbed = ?
Solution:
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
Specific heat capacity of water is 4.18 J/g°C
ΔT = 42.8°C - 25.4°C
ΔT = 17.4°C
Q = 23.0 g × × 4.18 J/g°C × 17.4°C
Q = 1672.84 j
2) Given data:
Mass of metal = 120.7 g
Initial temperature = 90.5°C
Final temperature = 25.7 ° C
Heat released = 7020 J
Specific heat capacity of metal = ?
Solution:
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 = 25.7°C - 90.5°C
ΔT = -64.8°C
7020 J = 120.7 g × c × -64.8°C
7020 J = -7821.36 g.°C × c
c = 7020 J / -7821.36 g.°C
c = 0.898 J/g.°C
Negative sign shows heat is released.
Cellular respiration is the process by which the chemicalenergy of "food" molecules is released and partially captured in the form of ATP. Carbohydrates, fats, and proteins can all be used as fuels in cellular respiration, but glucose is most commonly used as an example to examine the reactions and pathways involved.
The reaction is actually endothermic because delta H is positive, indicating that it absorbing heat.
Buffer solution resist the change in pH upon addition of small amount of strong acid or strong base.
Buffer consists of weak acid as HF / and its conjugate base NaF
When strong acid as HCl is added to buffer, it respond with its conjugate base to convert the strong acid to weak acid like this:
HCl (S.A) + NaF → NaCl + HF (W.A)
moles of HF we already have = M * V(in liters)
= 0.0955 M * 0.033 L = 3.15 x 10⁻³ mole
moles of HCl added = 8.00 x 10⁻⁵ mole
one mole HCl reacts with 1 mole NaF to give 1 mole HF
so the amount added to HF = 8.00 x 10⁻⁵
Total moles of HF present = (3.15 x 10⁻³) + (8.00 x 10⁻⁵) = 3.23 x 10⁻³ mole
