Answer:
0.1988 J/g°C
Explanation:
-Qmetal = Qwater
Q = mc∆T
Where;
Q = amount of heat
m = mass of substance
c = specific heat of substance
∆T = change in temperature
Hence;
-{mc∆T} of metal = {mc∆T} of water
From the information provided in this question, For water; m= 22.0g, ∆T = (24°C-19°C), c = 4.18J/g°C.
For metal; m= 34.0g, ∆T = (24°C-92°C), c = ?
Note that, the final temperature of water and the metal = 24°C
-{34 × c × (24°C-92°C)} = 22 × 4.18 × (24°C-19°C)
-{34 × c × (-68°C)} = 459.8
-{34 × c × -68} = 459.8
-{-2312c} = 459.8
+2312c = 459.8
c = 459.8/2312
c = 0.1988
The specific heat capacity of the metal is 0.1988 J/g°C
Answer:
Explanation:
Hello there!
In this case, according to the given chemical reaction for this problem about stoichiometry:
Whereas there is a 3:2 mole ratio of oxygen (molar mass = 32.0 g/mol) to iron (III) oxide (molar mass = 159.69 g/mol) and therefore, the correct stoichiometric setup is:
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Answer:hope we can be friends
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Although phlorizin inhibition of Na+-glucose cotransport occurs within a few seconds, 3H-phlorizin binding to the sodium-coupled glucose transport protein(s) requires several minutes to reach equilibrium (the fast-acting slow-binding paradigm). Using kinetic models of arbitrary dimension that can be reduced to a two-state diagram according to Cha’s formalism, we show that three basic mechanisms of inhibitor binding can be identified whereby the inhibitor binding step either (A) represents, (B) precedes, or (C) follows the rate-limiting step in a binding reaction. We demonstrate that each of mechanisms A–C is associated with a set of unique kinetic properties, and that the time scale over which one may expect to observe mechanism C is conditioned by the turnover number of the catalytic cycle. In contrast, mechanisms A and B may be relevant to either fast-acting or slow-binding inhibitors.
Explanation:
There is no way to know which reaction requires which catalyst. However, if you apply copper to a reaction where it does act as a catalyst, the rate of reaction will be much faster as it lowers the activation energy for successful collisions.
Convert 2430j to Kj that is divide by 1000
2430/1000=2.430
delta E =Q+W
q is the heat done
w is the work done and since is work is done is by the system therefore it is negative
w is therefore -5kj
q=-2.430
delta E =-2.430 + -5= -7.43kj/mol
