<h3>Answer:</h3>
Excess Reagent = NBr₃
<h3>Solution:</h3>
The Balance Chemical Equation for the reaction of NBr₃ and NaOH is as follow,
2 NBr₃ + 3 NaOH → N₂ + 3 NaBr + 3 HBrO
Calculating the Limiting Reagent,
According to Balance equation,
2 moles NBr₃ reacts with = 3 moles of NaOH
So,
40 moles of NBr₃ will react with = X moles of NaOH
Solving for X,
X = (40 mol × 3 mol) ÷ 2 mol
X = 60 mol of NaOH
It means 40 moles of NBr₃ requires 60 moles of NaOH, while we are provided with 48 moles of NaOH which is Limited. Therefore, NaOH is the limiting reagent and will control the yield of products. And NBr₃ is in excess as some of it is left due to complete consumption of NaOH.
Answer:
no
Explanation:
"Alkali metals are among the most reactive of all metals, which makes them suitable for specific and limited uses.
Alkali metals include lithium, sodium, potassium, rubidium, cesium and francium. These metals have large atomic radii and generally lose electrons during reactions. "
- Reference
Answer:
Required number is the vertical coordinate of the intersection point of a line at 60°C with the graph of the KNO₃.
Answer:
Both
Explanation:
The combined gas law is also known as the general gas law.
From the ideal gas law we assume that n = 1;
So;
PV = nRT
and then;
= 
If we cross multiply;
P₁V₁T₂ = P₂V₂T₁
So;
T₁ = 
Also;
V₂ = 
So from the choices both are correct
