Answer: The rate of disappearance ( -Δ[B]/Δt) under the same conditions is 0.200 M/s.
Explanation:
Initial Rate of the reaction = 0.100 M/s
Rate of disappearance of B:
![-\frac{1}{2}\frac{\Delta [B]}{\Delta t}=0.100 M/s](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B%5CDelta%20%5BB%5D%7D%7B%5CDelta%20t%7D%3D0.100%20M%2Fs)
![-\frac{\Delta [B]}{\Delta t}=2\times 0.100 M/s=0.200 M/s](https://tex.z-dn.net/?f=-%5Cfrac%7B%5CDelta%20%5BB%5D%7D%7B%5CDelta%20t%7D%3D2%5Ctimes%200.100%20M%2Fs%3D0.200%20M%2Fs)
The rate of disappearance ( -Δ[B]/Δt) under the same conditions is 0.200 M/s.
Answer:
The catabolic process of converting carbohydrates to CO2 requires<u> oxidation</u> of carbon.
Explanation:
There are multiple definitions of reduction-oxidation. There is one that explains it with respect to oxygen, the other with respect to hydrogen and another with respect to electrons. The relevant one here is the one that explains it in terms of hydrogen. Oxidation is the removal of hydrogen while reduction is the addition of hydrogen. During repiration, the carbon loses the hydrogens attached to it and is therefore oxidized. These hydrogens attach themselves to oxygen which means oxygen is reduced.
Answer:
Kc = 3.90
Explanation:
CO reacts with 
to form 
and 
. balanced reaction is:
No. of moles of CO = 0.800 mol
No. of moles of = 2.40 mol
Volume = 8.00 L
Concentration = 
Concentration of CO = 
Concentration of = 
Initial 0.100 0.300 0 0
equi. 0.100 -x 0.300 - 3x x x
It is given that,
at equilibrium 
= 0.309/8.00 = 0.0386 M
So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M
At equilibrium = 0.300 - 0.0386 × 3 = 0.184 M
At equilibrium = 0.0386 M
![Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BH_2O%5D%5BCH_4%5D%7D%7B%5BCO%5D%5BH_2%5D%5E3%7D)
Answer:
a) CaF₂.
b) 7.81g of CaF₂ are present.
Explanation:
a) The calcium ion has as charge Ca²⁺ and fluoride ion is F⁻, that means formula unit is:
CaF₂
b) 1 molecule of CaF₂ contains 2 anions, F⁻. Thus, the moles of CaF₂ is:
1.2x10²³ anions * (1 moleculeCaF₂ / 2 anions) = 6x10²² molecules of CaF₂
6.022x10²³ molecules = 1mol:
6x10²² molecules of CaF₂ * (1mol / 6.022x10²³molecules) = 0.10 moles CaF₂.
1 mole of CaF₂ has a mass of 78.07g:
0.10 moles CaF₂ * (78.07g / mol) =
7.81g of CaF₂ are present
Super saturated solution is formed.
<u>Explanation:</u>
Solubility is the property of any substance's capacity, that is the solute of the substance is dissolved in the given solvent to form the solution. We have three different types of solution, unsaturated, saturated and supersaturated solution.
- Unsaturated solution is a solution with lesser amount of solute than its solubility at equilibrium.
- Saturated solution is a solution with the maximum solute dissolved in the solvent.
- Super saturated solution is a solution with more solute than it is required.
The solubility of KI at 30°C is 153 g / 100 ml. Here 180 g of KI in 100 ml of water at 30°C is given, which has more solute than required, so it is super saturated solution.
