Answer:
Kc = 1.09x10⁻⁴
Explanation:
<em>HF = 1.62g</em>
<em>H₂O = 516g</em>
<em>F⁻ = 0.163g</em>
<em>H₃O⁺ = 0.110g</em>
<em />
To solve this question we need to find the moles of each reactant in order to solve the molar concentration of each reactan and replacing in the Kc expression. For the reaction, the Kc is:
Kc = [H₃O⁺] [F⁻] / [HF]
<em>Because Kc is defined as the ratio between concentrations of products over reactants powered to its reaction coefficient. Pure liquids as water are not taken into account in Kc expression:</em>
<em />
[H₃O⁺] = 0.110g * (1mol /19.01g) = 0.00579moles / 5.6L = 1.03x10⁻³M
[F⁻] = 0.163g * (1mol /19.0g) = 0.00858moles / 5.6L = 1.53x10⁻³M
[HF] = 1.62g * (1mol /20g) = 0.081moles / 5.6L = 0.0145M
Kc = [1.03x10⁻³M] [1.53x10⁻³M] / [0.0145M]
<h3>Kc = 1.09x10⁻⁴</h3>
Answer: Option (E) is the correct answer.
Explanation:
When we move from top to bottom in a group then there occurs an increase in atomic size of the atoms due to increase in the number of electrons.
For example, in group 2A elements beryllium is the smallest in size whereas radium being at the bottom is the largest in size.
Also, atomic number of beryllium is 4 and atomic number of radium is 88.
Thus, we can conclude that out of the given options radium is the 2A element which has the largest atomic radius.
A molecule that can h-bond will not always necessarily and does not have guarantee to have a higher boiling point than one than cannot have h-bond.
we can take an example of Pentan-2-one that cannot h-bond but instead of this it has a high boiling point that is 102.3 °C, while propan-1-ol can h-bond but it has a boiling point of 97.2°C, that is lower than the boiling point of Pentan-2-one.
Answer:
14.4g
Explanation:
First, we need to write a balanced equation for the reaction between Fe and O2 to produce Fe2O3. This is illustrated below:
4Fe + 3O2 —> 2Fe2O3
From the balanced equation,
4moles of Fe produced 2moles of Fe2O3.
Therefore, 0.18mol of Fe will produce = (0.18x2) /4 = 0.09mol of Fe2O3.
Now we need to find the mass present in 0.09mol of Fe2O3. This can be achieved by doing the following:
Molar Mass of Fe2O3 = (56x2) + (16x3) = 112 + 48 = 160g/mol
Number of mole of Fe2O3 = 0.09mol
Number of mole = Mass /Molar Mass
Mass = number of mole x molar Mass
Mass of Fe2O3 = 0.09 x 160 = 14.4g
Answer:
24 mol Cu
General Formulas and Concepts:
<u>Chemistry</u>
Explanation:
<u>Step 1: Define</u>
RxN: 2Cu (s) + O₂ (g) → 2CuO (s)
Given: 12 moles O₂
<u>Step 2: Stoichiometry</u>
<u />
= 24 mol Cu
<u>Step 3: Check</u>
<em>We are given 2 sig figs.</em>
Our final answer is in 2 sig figs, so no need to round.
