Answer:
The work done and heat absorbed are both -8,1 kJ
Explanation:
The work done in an isobaric process is defined as:
W = -P (Vf - Vi)
Where P is pressure ( 10 atm)
Vf = 10 L
Vi = 2 L
Thus, <em>W = -80 atm×L ≡ -8,1 kJ</em>
This is the work done in expansion of the gas. As the gas remains at the same temperature, there is no change in internal energy doing that all work was absorbed as heat.
I hope it helps!
In this problem, you’re given a mass (m) and a force (F) applied to an object and you’re asked to calculate the acceleration (a). If we plug this into the provided equation:
F = ma
(7.3 N) = (3.2 kg)a
a = 2.3 m/s/s
The acceleration of the object is 2.3 m/s/s.
Hope this helps!
Mg(OH)₂ ⇄ Mg²⁺ + 2 OH⁻
Ksp = [Mg²⁺] [OH⁻]²
6.0 x 10⁻¹⁰ = 0.10 x [OH⁻]²
[OH⁻] = 7.746 x 10⁻⁵ M
when Mg(OH)₂ 1st precipitates, [OH⁻] = 7.746 * 10⁻⁵ M
Fe(OH)₂ <—> Fe²⁺ + 2OH⁻
Ksp = [Fe²⁺] [OH⁻]²
7.9 x 10⁻¹⁶ = [Fe²⁺] x (7.746 x 10⁻⁵)²
[Fe²⁺] = 1.32 x 10⁻⁷ M
Answer: 1.32 x 10⁻⁷ M
Answer:
potassium.
Explanation:
located at start of period 3, will have larger radius
