Answer:
1 × 10²⁴ atoms O
General Formulas and Concepts:
<u>Chemistry - Atomic Structure</u>
- Reading a Periodic Table
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
Explanation:
<u>Step 1: Define</u>
30 g O
<u>Step 2: Identify Conversions</u>
Avogadro's Number
Molar Mass of O - 16.00 g/mol
<u>Step 3: Convert</u>
<u />
= 1.12913 × 10²⁴ atoms O
<u>Step 4: Check</u>
<em>We are given 1 sig fig. Follow sig fig rules and round.</em>
1.12913 × 10²⁴ atoms O ≈ 1 × 10²⁴ atoms O
Answer:
I. Changing the pressure:
Increasing the pressure: the amount of H₂S(g) will increase.
Decreasing the pressure: the amount of H₂S(g) will decrease.
II. Changing the temperature:
Increasing the temperature: the amount of H₂S(g) will decrease.
Decreasing the temperature: the amount of H₂S(g) will increase.
III. Changing the H₂ concentration:
Increasing the H₂ concentration: the amount of H₂S(g) will increase.
Decreasing the H₂ concentration: the amount of H₂S(g) will decrease.
Explanation:
Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.
I. Changing the pressure:
When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.
For the reaction: CH₄(g) + 2H₂S(g) ⇄ CS₂(g) + 4H₂(g),
The reactants side (left) has 3.0 moles of gases and the products side (right) has 5.0 moles of gases.
Increasing the pressure: will shift the reaction to the side with lower moles of gas (left side), amount of H₂S(g) will increase.
Decreasing the pressure: will shift the reaction to the side with lower moles of gas (right side), amount of H₂S(g) will decrease.
II. Changing the temperature
The reaction is endothermic since the sign of ΔH is positive.
So the reaction can be represented as:
CH₄(g) + 2H₂S(g) + heat ⇄ CS₂(g) + 4H₂(g).
Increasing the temperature:
The T is a part of the reactants, increasing the T increases the amount of the reactants. So, the reaction will be shifted to the right to suppress the effect of increasing T and the amount of H₂S(g) will decrease.
Decreasing the temperature:
The T is a part of the reactants, increasing the T decreases the amount of the reactants. So, the reaction will be shifted to the left to suppress the effect of decreasing T and the amount of H₂S(g) will increase.
III. Changing the H₂ concentration:
H₂ is a part of the products.
Increasing the H₂ concentration:
H₂ is a part of the products, increasing H₂ increases the amount of the products. So, the reaction will be shifted to the left to suppress the effect of increasing H₂ and the amount of H₂S(g) will increase.
Decreasing the H₂ concentration:
H₂ is a part of the products, decreasing H₂ decreases the amount of the products. So, the reaction will be shifted to the right to suppress the effect of decreasing H₂ and the amount of H₂S(g) will decrease.
The electron domain geometry is trigonal bipyramidal while the molecular geometry of the compound is seesaw.
The shapes of molecules is determined by the number of electron pairs on the valence shell of the central atom in the molecule. These electron domains include lone pairs and bond pairs.
The lone pairs only contribute towards the electron domain geometry and not the molecular geometry. SCl4 has five electron domains hence its electron domain geometry is trigonal bipyramidal. The molecular geometry of the compound is seesaw.
Learn more: brainly.com/question/6505878
Answer:
Chemical reaction A, because the reactant is a compound
Explanation:
In a decomposition reaction, a compound is broken down into its components, so the number of products is greater than the number of reactants
Answer:
given that
....the mass of the metal is 20g(0.02kg)
....specific heat capacity(c) is 0.4J/g°C
....ΔT=??
heat(Q)=3.9KJ(3900J)
Q=mcΔT
ΔT= Q/mc
.....=3900÷(20g x 0.4J/g°C)
.....=487.5°C
