The final temperature, t₂ = 30.9 °C
<h3>Further explanation</h3>
Given
24.0 kJ of heat = 24,000 J
Mass of calorimeter = 1.3 kg = 1300 g
Cs = 3.41 J/g°C
t₁= 25.5 °C
Required
The final temperature, t₂
Solution
Q = m.Cs.Δt
Q out (combustion of compound) = Q in (calorimeter)
24,000 = 1300 x 3.41 x (t₂-25.5)
t₂ = 30.9 °C
Answer:
one-half
Explanation:
cuz for a first order reaction is a half life independent of concentration and constant over time
<u>Answer:</u> The additional information that is helpful in calculating the mole percent of XCl(s) and ZCl(s) is the molar masses of Z and X
<u>Explanation:</u>
To calculate the mole percent of a substance, we use the equation:

Mass percent means that the mass of a substance is present in 100 grams of mixture
To calculate the number of moles, we use the equation:

We require the molar masses of Z and X to calculate the mole percent of Z and X respectively
Hence, the additional information that is helpful in calculating the mole percent of XCl(s) and ZCl(s) is the molar masses of Z and X
Lithium 6 would have 6 valence electrons in the outer orbital, while lithium 7 would have 7 in the outer orbital.