Answer: The correct answers are A and B.
Explanation:
Conduction: In the conduction, the heat is
transferred from the hotter body to the colder
body until the temperature on both bodies are
equal.
In thermal equilibrium, there is no heat
transfer as the heat is transferred till the
temperature on the bodies are not same.
In the given problem, an iron bar at 200°C is
placed in thermal contact with an identical iron
bar at 120°C in an isolated system. After 30
minutes, the thermal equilibrium is attained.
Then, the temperature on both iron bars are
equal.Both iron bars are at 160°C in an isolated
system.
But in an open system, the temperatures of the
iron bars after 30 minutes would be less than
160°C. There will be heat lost to the
surrounding. The room temperature is 25°C.
There will be exchange of the heat occur
between the iron bars and the surrounding. But
It would take more than 30 minutes for both iron
bars to reach 160°C because heat would be
transferred less efficiently.
Therefore, the correct options are (A) and (B).
Answer:4
Explanation:To balance the equation you need to make the number of each element equivalent in both sides.
To start add a 2 in front of the MnO2 which balances the Mn.
Then balance the oxygen by adding a 4 in front of H20.
The H then needs a 8 as it’s coefficient.
What that question makes 0 sense and there is no picture
Answer:
6.69 moles
Explanation:
Data obtained from the question include:
Volume = 2230mL
Molarity of NaOH = 3 M
First, we'll convert 2230mL to L. This is illustrated below:
1000mL = 1L
2230mL = 2230/1000 = 2.23L
Now, we can obtain the number of mole of NaOH in the solution as follow:
Molarity = mole /Volume
Mole = Molarity x Volume
Mole of NaOH = 3 x 2.23
Mole of NaOH = 6.69 moles
Therefore, the total number of mole of the solute (NaOH) in the solution is 6.69 moles
Answer:
a) HCN - hybridization sp
b) C(CH₃)₄ - hybridization sp³
c) H₃O⁺ - hybridization sp³
d) - CH₃ - hybridization sp³
Explanation:
Hybridization occurs to allow an atom to make more covalent bonds than the original electronic distribution would allow or to allocate ligands in an energetically stable geometry.
Carbon can have thre hybridization states: sp³ , sp² and sp.
Oxygen usualluy has an sp³ hybridization.
In order to determine the hybridization, we need to consider the number of atoms attached to the central atom and the number of lone pairs.
The figure attached shows the species and the hybridization of their central atoms.