Answer:
Mass = 76.176 g
Explanation:
Given data:
Mass of lead(II) chloride produced = 62.9 g
Mass of lead(II) nitrate used = ?
Solution:
Chemical equation:
Pb(NO₃)₂ + 2HCl → PbCl₂ + 2HNO₃
Number of moles of lead(II) chloride:
Number of moles = mass/molar mass
Number of moles = 62.9 g/ 278.1 g/mol
Number of moles = 0.23 mol
Now we will compare the moles of lead(II) chloride with Pb(NO₃)₂ from balance chemical equation:
PbCl₂ : Pb(NO₃)₂
1 : 1
0.23 : 0.23
Mass of Pb(NO₃)₂:
Mass = number of moles × molar mass
Mass = 0.23 mol × 331.2 g/mol
Mass = 76.176 g
The correct answer is BeCl_2(l)+2Cl^-(solvated)→BeCl_4^2-.
Evaluating be behavior to see :
how it differs from the other Group 2A (2) members.
In this reaction Be behaves like other alkaline earth metals
The complete equation can be given as
BeCl_2(l)+2Cl^-(solvated)→BeCl_4^2-
BeCl_2 tends to form a chloro bridged dimer in the vapour state, however at high temperatures of the order of 1200K, this dimer dissociates into the linear monomer.
BeCl_2 has a chain structure in its solid form. Each Be atom in this structure is surrounded by chlorine atoms, two of which are connected by conversion bonds and the remaining two by covalent coordinate connections. This chain structure is displayed.
To know more about BeCl₂(I) + Cl⁻ refer the link:
brainly.com/question/5017059
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Answer:
1. Percentage composition of: Na = 42%; P = 19.0%; O = 39%
2. Simplest formula of compound is PbO₂
3. (i) 2Cu(NO₃) ---> 2CuO + 2NO₂ + 3O₂
(ii) 2C₂H₆ + 7O₂ ---> 4CO₂ + 6H₂O
(iii) Mg₃N₂ + 6H₂O ---> 3Mg(OH)₂ + 2NH₃
4. 48 g of MG will react with 2 moles of Cl₂
5. 0.288 g of SO2 will be produced from the combustion of 0.331 g P₄S₃ in excess O₂
6. 12.8 g of nitric oxide can be produced from the reaction of 8.00 g NH₃ with 17.0 g O₂
7. The stock acid solution should be diluted to 6000 mL or 6.0 L
Explanation:
The full explanation is found in the attachments below
When the order n of the reaction rate is determined by this formula:
rate = K * [A]^n
when we have the unit of rate M S^-1
and the unit of [A] M
So:
1- for k = 8.79 x 10^2 M S^-1 :
when the unit of K here is M S^-1
by substitution in rate formula:
M S^-1 = M S^-1 * M ^n
∴ M^n = 1
∴M^0 = 1
∴ this reaction rate is zero order reaction
2- when k = 4.46 x 10^-1 and has unit M-1 S-1:
by substitution in rate formula:
M S^-1 = M-1 S-1 * M^n
∴M^2 = M^n
∴n = 2 so this is the second order reaction
3- when K 2.35 x 10^6 and it's unit is S-1 :
So by substitution in rate formula:
M S^-1 = S^-1 * M^n
∴ M^n = M
∴ n = 1
So this is the first order reaction
4- when k = 1.88 x 10^-3 and its unit is M-2 S-1
by substitution in rate formula:
M S^-1 = M^-2 S^-1 * M^n
∴ M^n = M^3
∴ n = 3
∴ this is a third order reaction
from 1 & 2 & 3 & 4
so we can see that the correct answer is (4), K= 1.88 x 10^-3 M-2 S-1 is the highest order reaction.
