Answer:
M KIO3 = 1.254 mol/L
Explanation:
∴ w KIO3 = 553 g
∴ mm KIO3 = 214.001 g/mol
∴ volumen sln = 2.10 L
⇒ mol KIO3 = (553 g)×(mol/210.001 g) = 2.633 mol
⇒ M KIO3 = (2.633 mol KIO3 / (2.10 L sln)
⇒ M KIO3 = 1.254 mol/L
The quick deployment of the bag which I think is quite similar to a cushion and a balloon. It holds your head/face away from any hard surfaces. Although it does not protect your legs or other extremities that are out of the bags range.
The heat lost by the metal should be equal to the heat
gained by the water. We know that the heat capacity of water is simply 4.186 J
/ g °C. Therefore:
100 g * 4.186 J / g °C * (31°C – 25.1°C) = 28.2 g * Cp *
(95.2°C - 31°C)
<span>Cp = 1.36 J / g °C</span>
Answer:
20.5torr
Explanation:
Given parameters:
V₁ = 15L
P₁ = 8.2 x 10⁴torr
V₂ = 6 x 10⁴L
Unknown:
P₂ = ?
Solution:
To solve this problem we have to apply the claims of Boyle's law.
Boyle's law is given mathematically as;
P₁ V₁ = P₂V₂
where P₁ is the initial pressure
V₁ is the initial volume
P₂ is final pressure
V₂ is final volume
8.2 x 10⁴ x 15 = P₂ x 6 x 10⁴
P₂ = 20.5torr
