Answer:
1.53 × 10²² atoms Ag
Explanation:
Step 1: Define conversions
3.271 × 10⁻²² g = 1 atom
Step 2: Use Dimensional Analysis
= 1.52858 × 10²² atoms Ag
Step 3: Simplify
We have 3 sig figs.
1.52858 × 10²² atoms Ag ≈ 1.53 × 10²² atoms Ag
Answer:
Here's what I get
Explanation:
Assume the initial concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.
We must calculate the initial concentration of HI.
1. We will need a chemical equation with concentrations, so let's gather all the information in one place.
H₂ + I₂ ⇌ 2HI
I/mol·L⁻¹: 0.30 0.15 x
2. Calculate the concentration of HI
![Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} = 5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}](https://tex.z-dn.net/?f=Q_%7B%5Ctext%7Bc%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BHI%5D%7D%5E%7B2%7D%7D%20%7B%5Ctext%7B%5BH%24_%7B2%7D%24%5D%5BI%24_%7B2%7D%24%5D%7D%7D%20%3D%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.30%20%5Ctimes%200.15%7D%20%3D%20%205.56%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%200.30%20%5Ctimes%200.15%20%5Ctimes%205.56%20%3D%200.250%5C%5Cx%20%3D%20%5Csqrt%7B0.250%7D%20%3D%20%5Ctextbf%7B0.50%20mol%2FL%7D%5C%5C%5Ctext%7BThe%20initial%20concentration%20of%20HI%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.50%20mol%2FL%7D%7D%24%7D)
3. Plot the initial points
The graph below shows the initial concentrations plotted on the vertical axis.
Answer:
The unknown temperature is 304.7K
Explanation:
V1 = 100mL = 100*10^-3L
P1 = 99.10kPa = 99.10*10³Pa
V2 = 74.2mL = 74.2*10^-3L
P2 = 133.7kPa = 133.7*10³Pa
T2 = 305K
T1 = ?
From combined gas equation,
(P1 * V1) / T1 = (P2 * V2) / T2
Solving for T1,
T1 = (P1 * V1 * T2) / (P2 * V2)
T1 = (99.10*10³ * 100*10^-3 * 305) / (133.7*10³ * 74.2*10^-3)
T1 = 3022550 / 9920.54
T1 = 304.67K
T1 = 304.7K
An atom is the smallest part of all matter.
I think it would be B hope not too late