Answer:
Increasing pressure shifts rxn right (lower molar volume side of equation).
Explanation:
In general, if a stress is applied to a gas phase reaction, the reaction will shift away from the applied stress and establish a new equilibrium for the given reaction. The applied stress factors include ...
changes in masses of reactant or product,
changes in applied temperature values, and/or
changes in applied pressure values for a reaction confined in a reaction vessel.
For this problem, if pressure is increased on the reaction N₂(g) + 3H₂(g) ⇄ 2NH₃(g), the reaction will shift away from the applied stress, that is, in the direction of the product side of the reaction in order to relieve the applied stress as there are fewer number of moles of gas on the product side of the equation.
Answer:
The correct answer is 199.66 grams per mole.
Explanation:
Based on law of effusion given by Graham, a gas rate of effusion is contrariwise proportionate to the square root of molecular mass, that is, rate of effusion of gas is inversely proportional to the square root of mass. Therefore,
R1/R2 = √ M2/√ M1
Here rate is the rate of effusion of the gas expressed in terms of number of mole per uni time or volume, and M is the molecular mass of the gas.
Rate Q/Rate N2 = √M of N2/ √M of Q
The molecular mass of N2 or nitrogen gas is 28 grams per mole and M of Q is molecular mass of Q and based on the question Q needs 2.67 times more to effuse in comparison to nitrogen gas, therefore, rate of Q = rate of N2/2.67
Now putting the values we get,
rate of N2/2.67/rate of N2 = √28/ √M of Q
√M of Q = √ 28 × 2.67
M of Q = (√ 28 × 2.67)²
M of Q = 199.66 grams per mole
A dichromate ion has 56 valence electrons. Valence electrons are the electrons that are present in an atom which are responsible or allows the formation of a chemical bond. When you draw the lewis structure of a dichromate ions, you will have to calculate the valence electrons by (6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6 ) – (-2) = 56.
Answer:
Explanation:
Hello,
In this case, we have a second order kinetics given the second power of the concentration of chlorine (V) oxide in the rate expression, thus, the integrated equation for the concentration decay is:
Thus, the final concentration for a 94% decrease is:
Therefore, we compute the time for such decrease:
Regards.
<span>Answer:
To determine hybridization you count the # of un-bonded PAIR of electrons. Theny you count bonded domains (a double bond still counts as one bonded domain, so does a triple bond).
For the first Carbon on top, you see it bonded to 2 oxygens and 1 carbon.However, when you count up the valence electrons that would be present, there is supposed to be 2 more electrons (ONE electron pair) on carbon. To do hybridization you must also be familiar with drawing lewis structures. So when you draw the pair of un-bonded electrons on carbon, you see that there is ONE un-bonded electron pair, and THREE bonded domains.
ONE plus THREE = FOUR.
so that would be sp3.</span>