6 because all atomic numbers equal the number of electrons
What kind of control like scientific?
Answer:
Yes, the investigations will reach similar conclusions about the reactivity of H2 and Cl2
Explanation:
1. The law of multiple proportions says that when elements form compounds, the proportions of the elements in those chemical compounds can be expressed in small whole number ratios. This means that regardless of whether 1000 times more of the products are used, the reactivity of the products is established by the chemical reaction
2. The law of multiple proportions is an extension of the law of definite composition, which states that compounds will consist of defined ratios of elements.
3. A reaction with more reactants will need more care because more products are produced, which can be toxic
4. H2 and Cl2 reactivity does not depend on the quantities but the chemical properties of each compound
Answer:
the value of equilibrium constant for the reaction is 8.5 * 10⁷
Explanation:
Ti(s) + 2 Cl₂(g) ⇄ TiCl₄(l)
equilibrium constant Kc = ![\frac{1}{[Cl_2]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BCl_2%5D%5E2%7D)
Given that,
We are given:
Equilibrium amount of titanium = 2.93 g
Equilibrium amount of titanium tetrachloride = 2.02 g
Equilibrium amount of chlorine gas = 1.67 g
We calculate the No of mole = mass / molar mass
mass of chlorine gas = 1.67 g
Molar mass of chlorine gas = 71 g/mol
mole of chlorine = 1.67 / 71
= 7.0L
Concentration of chlorine is = no of mole / volume
= 0.024 / 7
= 3.43 * 10⁻³M
equilibrium constant Kc =
= ![\frac{1}{[3.43 * 10^-^3]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5B3.43%20%2A%2010%5E-%5E3%5D%5E2%7D)
= 8.5 * 10⁷
Answer:
The answer to your question is 122.4 g of O₂
Explanation:
Data
mass of O₂ = ?
moles of H₂O = 7.65
Process
1.- Write the balanced chemical reaction
2H₂O ⇒ 2H₂ + O₂
2.- Convert the moles of H₂O to grams
molar mass of H₂O = 2 + 16 = 18 g
18 g of H₂O ---------------- 1 mol
x ----------------- 7.65 moles
x = (7.65 x 18) / 1
x = 137.7 g H₂O
3.- Calculate the grams of O₂
36 g of H₂O -------------------- 32 g of O₂
137.7 g of H₂O ------------------- x
x = (32 x 137.7) / 36
x = 122.4 g of O₂
