The empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.
<h3>How to calculate empirical formula?</h3>
The empirical formula of a compound is a notation indicating the ratios of the various elements present in a compound, without regard to the actual numbers.
The empirical formula of the given compound can be calculated as follows:
- Hafnium = 55.7% = 55.7g
- Chlorine = 44.3% = 44.3g
First, we convert mass values to moles by dividing by the molar mass of each element
- Hafnium = 55.7g ÷ 178.49g/mol = 0.312mol
- Chlorine = 44.3g ÷ 35.5g/mol = 1.25mol
Next, we divide each mole value by the smallest
- Hafnium = 0.312 ÷ 0.312 = 1
- Chlorine = 1.25 ÷ 0.312 = 4
Therefore, the empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.
Learn more about empirical formula at: brainly.com/question/14044066
#SPJ1
Answer:
1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹
2. 0.58 mol
Explanation:
1.Given ΔO₂/Δt…
2H₂O₂ ⟶ 2H₂O + O₂
-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt
d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹
d[H₂O]/dt = 2d[O₂]/dt = 2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = 6.6 × 10⁻³mol·L⁻¹s⁻¹
2. Moles of O₂
(a) Initial moles of H₂O₂
(b) Final moles of H₂O₂
The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.
(c) Moles of H₂O₂ reacted
Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol
(d) Moles of O₂ formed
A ratio of 0.89% w / v or 0.0089 in fraction means that
there is 0.89 weight of NaCl pero volume of solution. Therefore the mass of
NaCl is:
mass NaCl = 0.0089 * 200
<span>mass NaCl = 1.78 grams</span>
Answer:
See explanation
Explanation:
The particular reactants in the Fischer esterification reaction were not stated.
Generally, a Fischer esterification is a reaction that proceeds as follows;
RCOOH + R'OH ⇄RCOOR' + H2O
This reaction occurs in the presence of an acid catalyst.
We can shift the equilibrium of this reaction towards the products side in two ways;
I) use of a large excess of either of the reactants
ii) removal of one of the products as it is formed.
Any of these methods shifts the equilibrium of the Fischer esterification reaction towards the products side.
The answer I think it’s C I think.
