Answer:
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the vapor pressure of pure toluene by using the Raoult's law as shown below:
Thus, we solve for the mole fraction of benzene in the vapor phase first:
Which means that the mole fraction of toluene in the vapor phase is 0.24, and therefore, the vapor pressure of pure toluene turns out to be:
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Answer:
37.15 h.
Explanation:
- The decay of radioactive elements obeys first-order kinetics.
- For first-order reaction: <em>kt = lna/(a-x).</em>
where, k is the rate constant of the reaction <em>(k = 2.4 x 10⁻² h⁻¹)</em>.
t is the time of the reaction <em>(t = ??? day)</em>.
a is the initial concentration of nuclides <em>(a = 100 %)</em>.
(a-x) is the remaining concentration of nuclides <em>(a - x = 100% - 59% = 41.0 %)</em>.
<em>∴ kt = lna/(a-x)</em>
(2.4 x 10⁻² h⁻¹)(t) = ln(100.0%)/(41.0%).
(2.4 x 10⁻² h⁻¹)(t) = 0.8916.
<em>∴ t </em>= (0.8916)/(2.4 x 10⁻² h⁻¹) = <em>37.15 h.</em>
Answer:
894 deg K
Explanation:
The computation is shown below:
Given that
V1 denotes the initial volume of gas = 2.00 L
T1 denotes the initial temperature of gas = 25 + 273 = 298 K
V2 denotes the final volume of gas = 6.00 L
T2 = ?
Based on the above information
Here we assume that the pressure is remain constant,
So,
V1 ÷ T1 = V2 ÷ T2
T2 = T1 × V2 ÷ V1
= (298)(6) ÷ (2)
= 894 deg K
Because they have the same number of electrons in the outer shell, they have the same reactivity (they all want to lose/gain a certain amount of electrons to have a full shell of 8), so they have the similar melting, boiling points etc because they all undergo very similar reactions to certain factors so have similar outcomes
Answer:
a theory for the hydrogen atom based on quantum theory that energy is transferred only in certain well defined quantities. Electrons should move around the nucleus but only in prescribed orbits.
Explanation:
