B. It has a negative charge and much less mass than a proton.
Answer:
20 g Ag
General Formulas and Concepts:
<u>Chemistry - Stoichiometry</u>
- Using Dimensional Analysis
<u>Chemistry - Atomic Structure</u>
Explanation:
<u>Step 1: Define</u>
[RxN] Cu (s) + AgNO₃ (aq) → CuNO₃ (aq) + Ag (s)
[Given] 10 g Cu
<u>Step 2: Identify Conversions</u>
[RxN] 1 mol Cu = 1 mol Ag
Molar Mass of Cu - 63.55 g/mol
Molar Mass of Ag - 197.87 g/mol
<u>Step 3: Stoichiometry</u>
<u />
= 16.974 g Ag
<u>Step 4: Check</u>
<em>We are given 1 sig fig. Follow sig fig rules and round.</em>
16.974 g Ag ≈ 20 g Ag
Answer:
The answer is
<h2>13.84 %</h2>
Explanation:
The percentage error of a certain measurement can be found by using the formula

From the question
actual volume = 22.4 L
error = 22.4 - 19.3 = 3.1
The percentage error is

We have the final answer as
<h3>13.84 %</h3>
Hope this helps you
Answer:
The new equilibrium concentration of HI: <u>[HI] = 3.589 M</u>
Explanation:
Given: Initial concentrations at original equilibrium- [H₂] = 0.106 M; [I₂] = 0.022 M; [HI] = 1.29 M
Final concentrations at new equilibrium- [H₂] = 0.95 M; [I₂] = 0.019 M; [HI] = ? M
<em>Given chemical reaction:</em> H₂(g) + I₂(g) → 2 HI(g)
The equilibrium constant (
) for the given chemical reaction, is given by the equation:
![K_{c} = \frac {[HI]^{2}}{[H_{2}]\: [I_{2}]}](https://tex.z-dn.net/?f=K_%7Bc%7D%20%3D%20%5Cfrac%20%7B%5BHI%5D%5E%7B2%7D%7D%7B%5BH_%7B2%7D%5D%5C%3A%20%5BI_%7B2%7D%5D%7D)
<u><em>At the original equilibrium state:</em></u>

<u><em>Therefore, at the new equilibrium state:</em></u>
![\Rightarrow [HI]^{2} = 713.59 \times 0.01805 = 12.88](https://tex.z-dn.net/?f=%5CRightarrow%20%5BHI%5D%5E%7B2%7D%20%3D%20713.59%20%5Ctimes%200.01805%20%3D%2012.88)
![\Rightarrow [HI] = \sqrt {12.88} = 3.589 M](https://tex.z-dn.net/?f=%5CRightarrow%20%5BHI%5D%20%3D%20%5Csqrt%20%7B12.88%7D%20%3D%203.589%20M)
<u>Therefore, the new equilibrium concentration of HI: [HI] = 3.589 M</u>