Answer:
0.057 M
Explanation:
Step 1: Given data
Solubility product constant (Ksp) for HgBr₂: 2.8 × 10⁻⁴
Concentration of mercury (II) ion: 0.085 M
Step 2: Write the reaction for the solution of HgBr₂
HgBr₂(s) ⇄ Hg²⁺(aq) + 2 Br⁻
Step 3: Calculate the bromide concentration needed for a precipitate to occur
The Ksp is:
Ksp = 2.8 × 10⁻⁴ = [Hg²⁺] × [Br⁻]²
[Br⁻] = √(2.8 × 10⁻⁴/0.085) = 0.057 M
For the same amount of energy, the number of photons in red light will be greater than the number of photons in blue light.
This is because the energy carried by a photon is inversely proportional to the wavelength of the photon. A longer wavelength means there is a lower energy in the photons and a shorter wavelength means that there is a higher energy. Therefore, in order for the photons to deliver one joule of energy, more of the red light photons will be required.
Hey there!:
absorbance = log 100 - log Transmitance
absorbace = 0.85
log 100 = 2
- log transmitance = absorbace / log 100
0.85 / 2= 0.425
transmitance = 10 ^ ( - 0.425 )
transmitance = 0.376
Answer:
1. First
2. Third
3. Fourth
4.remain the same as
Explanation:
Given the reaction equation;
Rate= k[A] [B]^3
We can see that the order of reaction is first order with respect to reactant A and third order with respect to reactant B. This gives an overall fourth order reaction.
If the concentration of A is doubled and that of B is halved. The rate of reaction remains the same.
The balanced chemical reaction is:
<span>N2 + 3H2 = 2NH3 </span>
We are given the amount of H2 being reacted. This will be our starting point.
26.3 g H2 (1 mol H2 / 2.02 g H2) 2 mol
O2/3 mol H2) ( 17.04 g NH3 / 1mol NH3) = 147.90 g O2
Percent yield = actual yield / theoretical
yield x 100
Percent yield = 79.0 g / 147.90 g x 100
Percent yield = 53.4%
