An example is when on a seesaw if a kid is on one side and an adult is on the other then it is unbalanced because the adult weighs more
Answer:
reee, here is your answer.
Explanation: CH3OH(l) + 3O2(g) rightarrow CO2(g) + 3H2O(g) CH3OH(l) + O2(g) rightarrow CO2(g) + 2H20(g) CH3OH(l) + 2O2(g) rightarrow 2CO2(g) + 4H20(g) 2CH3OH(l) + 3O2(g) rightarrow 2CO2(g) + 4H20(g) Correct Calculate Delta H degree rxn at 25 degree C.
Answer:
Given,
P1=760 torr
T1=21°C
T2=250°C
P2=?
Solution,
According to Gay-Lussac's Law or Third Gas Law,
P1T2=P2T1
760×250=21×P2
P2=9047 torr
9047 is the pressure of the gas at 250 degree Celsius.
Answer:
Oxidation by FAD
Explanation:
1. Oxidation by NAD⁺
Succinate ⇌ Fumarate + <u>2H⁺ + 2e⁻</u>; E°´ = -0.031 V
<u>NAD⁺ + </u><u>2H⁺ + 2e⁻</u><u> ⇌ NADH + H⁺; </u> E°´ = <u> -0.320 V</u>
Succinate + NAD⁺ ⇌ Fumarate + NADH + H⁺; E°' = -0.351 V
2. Oxidation by FAD
Succinate ⇌ Fumarate + 2H⁺ + 2e⁻; E°´ = -0.031 V
<u>FAD + 2H⁺ + 2e⁻ ⇌ FADH₂; </u> E°´ = <u>-0.219 V
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Succinate + FADH₂ ⇌ Fumarate + FAD; E°' = -0.250 V
Neither reaction is energetically favourable, but FAD has a more positive half-cell potential.
FAD is the stronger oxidizing agent.
The oxidation by FAD has a more positive cell potential, so it is more favourable energetically.
<span>I'm going to assume that the other reactant is sodium phosphate(you did not indicate this)
so i'm guessing that the reaction did this
CaCL2 + Na3(PO4) ------------> Ca3(PO4)2 + NaCl
(since this reaction was done in water(aqueous solutions involved) the NaCl would be in the solution and thus the filtrate of the product so it won't influence the mass of the calcium phosphate product)
so then you need to write the balanced equation for this reaction(lets assume you can do this)
3 CaCL2 + 2 Na3(PO4) ------------> Ca3(PO4)2 + 6 NaCl
so in theoretical analysis 3 moles of calcium chloride is involved in the production of 1 mole of calcium phosphate
in terms of molar mass equivalents 3 molesw of calcium chloride weigh = 3 x (40 + 71)g/mol = 333g
1 mole of calcium phosphate = 310 g
so if you only have 0.513 g of calcium chloride then this would calculate to get the mass of calcium phosphate
mass = [0.513 / 333] x 310 g = 0.477grams
your actual yield is what you massed when you performed the lab. hope that helped.
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