Answer:
Root mean squared velocity is different.
Explanation:
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In this case, since we have a mixture of oxygen and nitrogen at STP, which is defined as a condition whereas T = 298 K and P = 1 atm, we can infer that these gases have the same temperature, pressure, volume and moles but a different root mean squared velocity according to the following formula:
Since they both have a different molar mass (MM), nitrogen (28.02 g/mol) and oxygen (32.02 g/mol), thus we infer that nitrogen would have a higher root mean squared velocity as its molar mass is less than that of oxygen.
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Answer:
V = 0.356 L
Explanation:
In this case, we need to use the following expression:
M = n/V (1)
Where:
M: molarity of solution (mol/L or M)
n: moles of solute (moles)
V: Volume of solution (Liters)
From these expression, we can solve for V:
V = n/M (2)
Now, replacing the given data we can solve V:
V = 8.9 / 25
V = 0.356 L
Answer:
CH3CH2CH2Cl
CH3CH2CH2CH2CH2SH
Br2
Explanation:
Dispersion forces increases with increase in relative molecular mass. The specie having the greater relative molecular mass definitely has greater dispersion forces. A rough estimation of the relative molecular masses of the species stated in the answer will reveal this fact.
A strong acid like HCl donates its proton so readily that there is essentially no tendency for the conjugate base Cl– to reaccept a proton. Consequently, Cl– is a very weak base.
Answer:
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Explanation:
