Answer:Rate of reaction can be determined in terms of concentration of reactants consumed or concentration of product formed per unit time
Explanation: For the reaction below
A ===>B
The reactant is A while the product given is B.
Reaction rate = Δ[B]/Δt = -Δ[A]/Δt
The concentration of A will decrease with time while the concentration of B will increases with time.
The negative sign in -Δ[A]/Δt is to convert the expression to positive since the change will always be negative (decreases)
Explanation:
Elements need a total of eight electrons to gain stability and look like a noble gas. So, they sometimes need sharing of two, four or even six electrons to complete their octate. So, they form double and triple covalent bonds. One more the reason is the interaction between the p orbitals of the combining atoms. for example A double bond, as in ethene H2C=CH2, arises from one combination of the s orbitals and one combination of the p_y orbitals.
Answer:
Radiation is being released from the reactor.
Explanation:
( A P E X )
1) At tne same temperature and with the same volume, initially the chamber 1 has the dobule of moles of gas than the chamber 2, so the pressure in the chamber 1 ( call it p1) is the double of the pressure of chamber 2 (p2)
=> p1 = 2 p2
Which is easy to demonstrate using ideal gas equation:
p1 = nRT/V = 2.0 mol * RT / 1 liter
p2 = nRT/V = 1.0 mol * RT / 1 liter
=> p1 / p2 = 2.0 / 1.0 = 2 => p1 = 2 * p2
2) Assuming that when the valve is opened there is not change in temperature, there will be 1.00 + 2.00 moles of gas in a volumen of 2 liters.
So, the pressure in both chambers (which form one same vessel) is:
p = nRT/V = 3.0 mol * RT / 2liter
which compared to the initial pressure in chamber 1, p1, is:
p / p1 = (3/2) / 2 = 3/4 => p = (3/4)p1
So, the answer is that the pressure in the chamber 1 decreases to 3/4 its original pressure.
You can also see how the pressure in chamber 2 changes:
p / p2 = (3/2) / 1 = 3/2, which means that the pressure in the chamber 2 decreases to 3/2 of its original pressure.
