2 meters per second. You do 10 divided by 5 to find your answer, which is 2.
The bond dipole moment<span> uses the idea of </span>electric dipole moment<span> to measure the </span>polarity<span> of a chemical bond within a </span>molecule<span>. It occurs whenever there is a separation of positive and negative charges. In the diagram above, option B exhibited a bond dipole moment. I hope this helps.</span>
Answer:
By increasing the pressure, the molar concentration of N2O4 will increase
Explanation:
We have the equation 2NO2 ⇔ N2O4
This equation is reversible and exotherm. By <u>decreasing the temperature</u>, the reaction will produce more energy, so the reaction will move to the right. But a lower temperature also lowers the rate of the process, so, the temperature is set at a compromise value that allows N2O4 to be made at a reasonable rate with an equilibrium concentration that is not too unfavorable
So <u>increasing the temperature</u> will shift the equilibrium to the left. The equilibrium shifts in the direction that consumes energy.
If we d<u>ecrease the concentration of NO2</u>, the equilibrium will shift to the left, resulting in forming more reactants.
To increase the molar concentration of the product N2O4, we have to <u>increase the pressure</u> of the system.
NO2 takes up more space than N2O4, so increasing the pressure would allow the reactant to collide more form more product.
By increasing the pressure, the molar concentration of N2O4 will increase
Answer:
A) Ca(s) + C(s) + 3/2 O₂(g) → CaCO₃(s)
Explanation:
Standard enthalpy of formation of a chemical is defined as the change in enthalpy durin the formation of 1 mole of the substance from its constituent elements in their standard states.
The consituent elements of calcium carbonate, CaCO₃, in their standard states (States you will find this pure elements in nature), are:
Ca(s), C(s) and O₂(g)
That means, the equation that represents standard enthalpy of CaCO₃ is:
<h3>A) Ca(s) + C(s) + 3/2 O₂(g) → CaCO₃(s)</h3><h3 />
<em>Is the equation that has ΔH° = -1207kJ/mol</em>
