Answer:
See explanation
Explanation:
The elements in group form univalent positive ions and element in group 17 form univalent negative ions. Hence, when a group 1 element reacts with a group 17 element, a compound of the sort MX is formed. Hence, when a group 1 element reacts with bromine, a salt is formed with the general formula MBr.
Elements of group 1 are highly electro positive metals. They react with water to form the metal hydroxide and release hydrogen gas. Hence, when group 1 elements react with water, hydrogen gas is released.
A group 1 element forms a univalent positive ion while a group 16 element forms a divalent negative ion. Hence, when a groups 1 element reacts with oxygen, the compound formed must have the general formula M2O.
The reactivity of group 1 metal increases down the group hence Cs is the most reactive group 1 element.
Lithium displays a slightly different chemistry from other group 1 elements because of its small size.
To
determine the percent ionization of the acid given, we make use of the acid
equilibrium constant (Ka) given. It is the ration of the equilibrium
concentrations of the dissociated ions and the acid. The dissociation reaction
of the HF acid would be as follows:<span>
HF = H+ + F-
The acid equilibrum constant would be expressed as follows:
Ka = [H+][F-] / [HF] = 3.5 x 10-4
To determine the equilibrium concentrations we use the ICE table,
HF
H+ F-
I 0.337 0
0
C -x +x
+x
---------------------------------------------
E 0.337-x x
x
3.5 x 10-4 = [H+][F-] / [HF]
3.5 x 10-4 = [x][x] / [0.337-x] </span>
Solving for x,
x = 0.01069 = [H+] = [F-]
percent ionization = 0.01069 / 0.337 x 100 = 3.17%
Answer:
1.89 nol Cu(NO3)2
Explanation:
if you calculate it it will be 1.89
<h3>
Answer:</h3>
78.34 g
<h3>
Explanation:</h3>
From the question we are given;
Moles of Nitrogen gas as 2.3 moles
we are required to calculate the mass of NH₃ that may be reproduced.
<h3>Step 1: Writing the balanced equation for the reaction </h3>
The Balanced equation for the reaction is;
N₂(g) + 3H₂(g) → 2NH₃(g)
<h3>Step 2: Calculating the number of moles of NH₃</h3>
From the equation 1 mole of nitrogen gas reacts to produce 2 moles of NH₃
Therefore, the mole ratio of N₂ to NH₃ is 1 : 2
Thus, Moles of NH₃ = Moles of N₂ × 2
= 2.3 moles × 2
= 4.6 moles
<h3>Step 3: Calculating the mass of ammonia produced </h3>
Mass = Moles × molar mass
Molar mass of ammonia gas = 17.031 g/mol
Therefore;
Mass = 4.6 moles × 17.031 g/mol
= 78.3426 g
= 78.34 g
Thus, the mass of NH₃ produced is 78.34 g
