Answer:
Condensation: 423.3 K
Freezing: 83.96 K
(this is all i could figure out :) hope it helps)
Answer:
An insulated beaker with negligible mass contains liquid water with a mass of 0.205kg and a temperature of 79.9 °C How much ice at a temperature of −17.5 °C must be dropped into the water so that the final temperature of the system will be 31.0 °C? Take the specific heat for liquid water to be 4190J/Kg.K, the specific heat for ice to be 2100J/Kg.K, and the heat of fusion for water to be 334000J/kg.
The answer to the above question is
Therefore 0.1133 kg ice at a temperature of -17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.0 °C
Explanation:
To solve this we proceed by finding the heat reaquired to raise the temperature of the water to 31.0 C from 79.9 C then we use tht to calculate for the mass of ice as follows
ΔH = m×c×ΔT
= 0.205×4190×(79.9 -31.0) = 42002.655 J
Therefore fore the ice, we have
Total heat = mi×L + mi×ci×ΔTi = mi×334000 + mi × 2100 × (0 -−17.5) = 42002.655 J
370750×mi = 42002.655 J
or mi = 0.1133 kg
Therefore 0.1133 kg ice at a temperature of -17.5 ∘C must be dropped into the water so that the final temperature of the system will be 31.0 °C
Explanation:
The given reaction is as follows.
![E + S \rightleftharpoons ES \xrightarrow[]{k_{2}} E + P](https://tex.z-dn.net/?f=E%20%2B%20S%20%5Crightleftharpoons%20ES%20%5Cxrightarrow%5B%5D%7Bk_%7B2%7D%7D%20E%20%2B%20P)
Here, [E] = triose phosphate isomerase = 0.1 
[S] = Dihydroxy acetone phosphate = 5 
[P] = Glyceraldehyde-3-phosphate = 2 
Therefore, velocity of the reaction will be as follows.
v =
= ![\frac{K_{2}[E][S]}{K_{M} + [S]}](https://tex.z-dn.net/?f=%5Cfrac%7BK_%7B2%7D%5BE%5D%5BS%5D%7D%7BK_%7BM%7D%20%2B%20%5BS%5D%7D)
where,
= Michaelic menten constant = 
v = 
= 
or, = 30 nm/s
Hence, we can conclude that the actual velocity of the forward reaction under physiologic conditions if KM = 10 μM is 30 nm/s.
<span>Answer:
Enthalpy Change = (6 x -393.5) + (7 x -285.8) - (-204.6) + (19/2) 0.....???
like.. (6 x Enth CO2) + ( 7 x Enth H2O) - (Enth C6H14) + (19/2) Enth O2</span>