Answer:

Explanation:
The limiting reactant is the reactant that gives the smaller amount of product.
Assemble all the data in one place, with molar masses above the formulas and masses below them.
M_r: 39.10 80.41 2.016
2K + 2HBr ⟶ 2KBr + H₂
m/g: 5.5 4.04
a) Limiting reactant
(i) Calculate the moles of each reactant

(ii) Calculate the moles of H₂ we can obtain from each reactant.
From K:
The molar ratio of H₂:K is 1:2.

From HBr:
The molar ratio of H₂:HBr is 3:2.

(iii) Identify the limiting reactant
HBr is the limiting reactant because it gives the smaller amount of NH₃.
b) Excess reactant
The excess reactant is K.
c) Mass of H₂

Answer:
Kc = 3.1x10²
Explanation:
At equilibrium, the velocity of product formation is equal to the velocity of reactants formation. For a generic reaction, the equilibrium constant (Kc) is:
aA + bB ⇄ cC + dD
![Kc = \frac{[C]^c*[D]^d}{[A]^a*[B]^b}](https://tex.z-dn.net/?f=Kc%20%3D%20%5Cfrac%7B%5BC%5D%5Ec%2A%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%2A%5BB%5D%5Eb%7D)
Where [X] is the molar concentration of X, and the solid substances are not considered (because it's activity is 1, for the other substances, the activity is substituted for the molar concentration, which forms the equation above).
For the reaction given, let's make an equilibrium chart:
Fe³⁺(aq) + SCN⁻(aq) ⇄ FeSCN²⁺(aq)
1.1*10⁻³ 8.2*10⁻⁴ 0 <em> Initial</em>
-x -x +x <em>Reacts</em> (stoichiometry is 1:1:1)
1.1*10⁻³ -x 8.2*10⁻⁴ -x x <em> Equilibrium</em>
x = 1.8*10⁻⁴ M, so the molar concentrations at equilibrium are:
[Fe⁺³] = 1.1*10⁻³ - 1.8*10⁻⁴ = 9.2*10⁻⁴ M
[SCN⁻] = 8.2*10⁻⁴ - 1.8*10⁻⁴ = 6.4*10⁻⁴ M
[FeSCN⁺²] = 1.8*10⁻⁴ M
Kc = [FeSCN⁺²]/([Fe⁺³]*[SCN⁻])
Kc = (1.8*10⁻⁴)/(9.2*10⁻⁴*6.4*10⁻⁴)
Kc = 306 = 3.1x10²
Answer:

Explanation:
The pH of a solution can be found by using the formula
![pH = - log [ {H}^{+} ]](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5B%20%7BH%7D%5E%7B%2B%7D%20%5D)
Since we are finding the H+ ions we find the antilog of the pH
So we have

We have the final answer as

Hope this helps you
The balanced chemical equation for the reaction is
<h3>6Cs(s) + 2CsNO₂(s) —> 4Cs₂O(s) + N₂(g) </h3>
From the question given above, we were told that:
solid cesium reacts with solid cesium nitrite to form solid cesium oxide and nitrogen gas.
The equation for the reaction can be written as follow:
Caesium => Cs
Caesium nitrite => CsNO₂
cesium oxide => Cs₂O
nitrogen gas => N₂
Caesium + Caesium nitrite —> Caesium oxide + Nitrogen gas
<h3>Cs + CsNO₂ —> Cs₂O + N₂</h3>
The above equation can be balance as follow:
Cs + CsNO₂ —> Cs₂O + N₂
There are 2 atoms of N on the right side and 1 atom on the left side. It can be balance by writing 2 before CsNO₂ as shown below:
Cs + 2CsNO₂ —> Cs₂O + N₂
There are 2 atoms of Cs on the right side and a total 3 atoms on the left side. It can be balance by writing 6 before Cs and 4 before Cs₂O as shown below:
6Cs(s) + 2CsNO₂(s) —> 4Cs₂O(s) + N₂(g)
Now the equation is balanced
Learn more: brainly.com/question/11502387
The shape of BeH2 is linear.