E=MC(delta)T
=20.0g x 9.00J/g x (94.4-22.8) C
= 12,924.0 J
Answer:
4.5 moles
Explanation:
One mole is equal to 6.022 x 10^23 atoms
2.71 x 10^24 atoms * 1 mol/ 6.022 x 10^23 atoms = 4.5 moles
Answer:
1.07 g
Explanation:
Half-life of Pu-234 = 4.98 hours
Initially present = 45 g
mass remains after 27 hours = ?
Solution:
Formula
mass remains = 1/ 2ⁿ (original mass) ……… (1)
Where “n” is the number of half lives
To find "n" for 27 hours
n = time passed / half-life . . . . . . . .(2)
put values in equation 2
n = 27 hr / 4.98 hr
n = 5.4
Mass after 27 hr
Put values in equation 1
mass remains = 1/ 2ⁿ (original mass)
mass remains = 1/ 2^5.4 (45 g)
mass remains = 1/ 42.2 (45 g)
mass remains = 0.0237 x 45 g
mass remains = 1.07 g
Answer:
Keqq = 310
Note: Some parts of the question were missing. The missing values are used in the explanation below.
Explanation:
<em>Given values: ΔH° = -178.8 kJ/mol = -178800 J/mol; T = 25°C = 298.15 K; ΔS° = -552 J/mol.K; R = 8.3145 J/mol.K</em>
Using the formula ΔG° = -RT㏑Keq
㏑Keq = ΔG°/(-RT)
where ΔG° = ΔH° - TΔS°
㏑Keq = ΔH° - TΔS°/(-RT)
㏑Keq = {-178800 - (-552 * 298.15)} / -(8.3145 * 298.15)
㏑Keq = -14221.2/-2478.968175
㏑Keq = 5.73674166
Keq = e⁵°⁷³⁶⁷⁴¹⁶⁶
Keq = 310.05
