4.05 kg + 567.95 g + 100.1 g add the correct value using the correct sig figs and or least precise degree of precision. When i d
1 answer:
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Answer:
The total change in enthalpy for the reaction is - 81533.6 J/mol
Explanation:
Given the data in the question;
Reaction;
HCl + NaOH → NaCl + H₂O
Where initial temperature is 21.2 °C and final temperature is 28.0 °C. Ccal is 1234.28 J
Moles of NaOH = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol
Moles of HCl = 50.mL × 1.00 M = 50.0 mmol = 0.0500 mol
so, 0.0500 moles of H₂O produced
Volume of solution = 50.mL + 50.mL = 100.0 mL
Mass of solution m = volume × density = 100.0mL × 1.0 g/mL = 100 g
now ,
Heat energy of Solution q= (mass × specific heat capacity × temp Δ) + Cal
we know that; The specific heat of water(H₂O) is 4.18 J/g°C.
so we substitute
q_soln = (100g × 4.18 × ( 28.0 °C - 21.2 °C) ) + 1234.28
q_soln = 2842.4 + 1234.28
q_soln = 4076.68 J
Enthalpy change for the neutralization is ΔH
ΔH = -q_soln / mole of water produced
so we substitute
ΔH = -( 4076.68 J ) / 0.0500 mol
ΔH = - 81533.6 J/mol
Therefore, the total change in enthalpy for the reaction is - 81533.6 J/mol
Answer:
Second Law
Explanation:
Newton's second law states that the acceleration caused in a body is directly proportional to the force applied and inversely proportion to the mass of the body.
This is given by :
In this case the suggestion given to reduce the aircraft's cargo load is the right move as reducing the load on the aircraft will decrease the mass of the whole aircraft. This in turn will help the aircraft to accelerate more as acceleration inversely varies with mass. Thus the aircraft will be able to reach its flying speed even on a short run way.
Hence, Newton's second law is applied.