Answer:
Ethane would have a higher boiling point.
Explanation:
In this case, for the lewis structures, we have to keep in mind that all atoms must have <u>8 electrons</u> (except hydrogen). Additionally, each carbon would have <u>4 valence electrons</u>, with this in mind, for methane we have to put the hydrogens around the carbon, and with this structure, we will have 8 electrons for the carbon. In ethane, we will have a bond between the carbons, therefore we have to put three hydrogens around each carbon to obtain 8 electrons for each carbon.
Now, the main difference between methane and ethane is an <u>additional carbon</u>. In ethane, we have an additional carbon, therefore due to this additional carbon, we will have <u>more area of interaction</u> for ethane. If we have more area of interaction we have to give <u>more energy</u> to the molecule to convert from liquid to gas, so, the ethane will have a higher boiling point.
I hope it helps!
Answer: sorry I’m late but it is 11 electrons
Explanation:
The cryosphere can affect the atmosphere by letting the Earth get to warm. The cryosphere protects the Earth from getting to warm.
Answer:- A. 
is not the correct formula.
Explanations:- Charge for Ba is +2 and charge for O is -2, Here the charges are in equal and opposite so their ratio is 1:1 and hence the formula must be BaO and not .
Rest of the formulas are correct as charge for Ca is +2 and charge for N is -3 . On criss cross we get the formula 
.
Charge for K is +1 and charge for Cl is -1 and so KCl is correct.
Charge for Li is +1 and that for S is -2. On criss cross we get the formula 
.
Answer:
% = 76.75%
Explanation:
To solve this problem, we just need to use the expressions of half life and it's relation with the concentration or mass of a compound. That expression is the following:
A = A₀ e^(-kt) (1)
Where:
A and A₀: concentrations or mass of the compounds, (final and initial)
k: constant decay of the compound
t: given time
Now to get the value of k, we should use the following expression:
k = ln2 / t₁/₂ (2)
You should note that this expression is valid when the reaction is of order 1 or first order. In this kind of exercises, we can assume it's a first order because we are not using the isotope for a reaction.
Now, let's calculate k:
k = ln2 / 956.3
k = 7.25x10⁻⁴ d⁻¹
With this value, we just replace it in (1) to get the final mass of the isotope. The given time is 1 year or 365 days so:
A = 250 e^(-7.25x10⁻⁴ * 365)
A = 250 e^(-0.7675)
A = 191.87 g
However, the question is the percentage left after 1 year so:
% = (191.87 / 250) * 100
<h2>
% = 76.75%</h2><h2>
And this is the % of isotope after 1 year</h2>
