The answer is
<span>The density (D) is quotient of mass (m) and
volume (V):
</span>
The unit is g/cm³
It is given:
m = 1.62 kg = 1620 g
V = 205 mL = 205 cm³
D = ?
Thus:
The density of the goblet is 7.90 g/cm³.
The second shell is left at 7, it should be filled to 8 to go to the next shell.
Answer:
40% of the ammonia will take 4.97x10^-5 s to react.
Explanation:
The rate is equal to:
R = k*[NH3]*[HOCl] = 5.1x10^6 * [NH3] * 2x10^-3 = 10200 s^-1 * [NH3]
R = k´ * [NH3]
k´ = 10200 s^-1
Because k´ is the psuedo first-order rate constant, we have the following:
b/(b-x) = 100/(100-40) ; 40% ammonia reacts
b/(b-x) = 1.67
log(b/(b-x)) = log(1.67)
log(b/(b-x)) = 0.22
the time will equal to:
t = (2.303/k) * log(b/(b-x)) = (2.303/10200) * (0.22) = 4.97x10^-5 s
SThe missing coefficient for the skeleton equation below is as follows
skeleton equation
Cr(s) + Fe(No3)2(aq) ------> Fe (s) + Cr(NO3)3 (aq)
the missing coefficient are is as follows
2 Cr(s) + 3 Fe(NO3)2 ---> 3 Fe (s) + 2 Cr(NO3)3
This is obtained by making sure all the molecules are balanced in both sides
Answer:
Following are the answer to this question:
Explanation:
In the given question information is missing, that is equation which can be defined as follows:
- Growing temperatures may change its connection to just the way which consumes thermal energy in accordance with Le chatelier concepts Potential connection is endothermic. Answer: shifts to the right
-
Kc are described as a related to the concentration by the intensity of both the reaction for each phrase which reaches a power equal towards its stoichiometric equation coefficient Kc = \frac{product}{reactant}
It increases [product] but reduces [reactant] Therefore, Kc increases
