Answer:
0.01235 moles
Explanation:
<em>Given data:</em>
volume = 47.5 ml = 0.0475 L
concentration = 0.260 M
<em>To find:</em>
number of moles: ?
<em>Formula:</em>
No of moles = concentration x volume (in litres)
M is equal to moles per liter, so if you multiply the volume in liters times the value of M, the result will tell you the number of moles.
<em>Solution: </em>
no. of moles = 0.260 M x 0.0475 L
= 0.01235 moles
The Balanced Chemical Equation is as follow;
4 KO₂ + 2 CO₂ → 2 K₂CO₃ + 3 O₂
First find out the Limiting Reagent,
According to equation,
284 g (4 moles) KO₂ reacted with = 44.8 L (2 moles) of CO₂
So,
27.9 g of KO₂ will react with = X L of CO₂
Solving for X,
X = (44.8 L × 27.9 g) ÷ 284 g
X = 4.40 L of CO₂
Hence, to consume 27.9 g of KO₂ only 4.40 L CO₂ is required, while, we are provided with 29 L of CO₂, it means CO₂ is in excess and KO₂ is is limited amount, Therefore, KO₂ will control the yield of K₂CO₃. So,
According to eq.
284 g (4 moles) KO₂ formed = 138.2 g of K₂CO₃
So,
27.9 g of KO₂ will form = X g of K₂CO₃
Solving for X,
X = (138.2 g × 27.9 g) ÷ 284 g
X = 13.57 g of K₂CO₃
So, 13.57 g of K₂CO₃ formed is the theoretical yield.
%age Yield = 13.57 / 21.8 × 100
%age Yield = 62.24 %
Answer:
The answer to your question is: D) 8
Explanation:
We can know the valence of an element if we know in which is located.
For example if an element is in group IA it would have 1 valence electron,
if an element is located in group IIIA it would have 3 valence electrons, if an element is located in group VIIA, it would have 7 valence electron and so on.
Neon is located in group VIIIA, so it would have 8 valence electrons.
Answer:
Neutral
Explanation:
They have the same # of protons and electrons
Answer:
k = 6.31 x 10⁻³ min⁻¹
Explanation:
The equation required to solve this question is:
k = 0693 / t half-life
This equation is derived from the the equation from the radioctive first order reactions:
ln At/A₀ = -kt
where At is the number of isoopes after a time t , and A₀ is the number of of isotopes initially. The half-life is when the number of isotopes has decayed by a half, so
ln(1/2) = -kt half-life
-0.693 = - k t half-life
t half-life = 109.8 min
⇒ k = 0.693 / t half-life = 0.693 / 109.8 min = 6.31 x 10⁻³ min⁻¹