Answer:
4.5 g/L.
Explanation:
- To solve this problem, we must mention Henry's law.
- Henry's law states that at a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid.
- It can be expressed as: P = KS,
P is the partial pressure of the gas above the solution.
K is the Henry's law constant,
S is the solubility of the gas.
- At two different pressures, we have two different solubilities of the gas.
<em>∴ P₁S₂ = P₂S₁.</em>
P₁ = 525.0 kPa & S₁ = 10.5 g/L.
P₂ = 225.0 kPa & S₂ = ??? g/L.
∴ S₂ = P₂S₁/P₁ = (225.0 kPa)(10.5 g/L) / (525.0 kPa) = 4.5 g/L.
Answer:
the normality of the given solution is 0.0755 N
Explanation:
The computation of the normality of the given solution is shown below:
Here we have to realize the two sodiums ions per carbonate ion i.e.
N = 0.321g Na_2CO_3 × (1mol ÷ 105.99g)×(2eq ÷ 1mol)
= 0.1886eq ÷ 0.2500L
= 0.0755 N
Hence, the normality of the given solution is 0.0755 N
Total number of atoms = 7
Total number of H atom = 5
% of H in ammonium hydroxide = 5/7 ×100 = 71.4 %
Pitch is sometimes defined as the fundamental frequency of a sound wave. For most practical purposes, this is fine, and pitch and frequency can be thought of as equivalent. On the other hand, for most practical purposes, amplitude can be thought of as volume.
However, technically, pitch and volume are human perceptions. Thus, our perception of pitch and volume are not solely based on frequency and amplitude respectively, but are based on a combination of both. Frequency overwhelming dictates perceived pitch, but amplitude also does have some small, small effect on our pitch perception, especially when it is very large. For example, a very loud sound can have a different perceived pitch than you would predict from its frequency alone.
Hope that helps!
<span>LiOH+HBr---> LiBr +h20. Moles of LiOH = 10/24 = 0.41moles. According to stoichiometry, moles of LiOH = moles of LiBr = 0.41moles. Therefore mass of LiBr =moles of LiBr x molecular weight of LiBr = o.41 x 87 = 35.67g. Hope it helps </span>
