Answer:
Explanation:
Cesium Lewis dot structure would look like this:
·Cs, because it only has one valence electron. But, since it has a plus, that means we lost an electron. So, we have to get rid of the dot and you have:
Answer:
17 ohms
Explanation:
Given that,
In a series circuit there are 17 light bulbs, each providing 1 ohm of resistance.
We need to find the total resistance of the circuit.
The equivalent resistance for series combination is given by :
For 17 light bulbs, the equivalent resistance is given by :
Hence, the total resistance is 17 ohms. Hence, the correct option is (d).
Answer:
See explanation below
Explanation:
In order to calculate this, we need to use the following expression to get the concentration of the base:
MaVa = MbVb (1)
We already know the volume of NaOH used which is 13.4473 mL. We do not have the concentration of KHP, but we can use the moles. We have the mass of KHP which is 0.5053 g and the molecular formula. Let's calculate the molecular mass of KHP:
Atomic weights of the elements to be used:
K = 39.0983 g/mol; H = 1.0078 g/mol; C = 12.0107 g/mol; O = 15.999 g/mol
MM KHP = (1.0078*5) + (39.0983) + (8*12.0107) + (4*15.999) = 204.2189 g/mol
Now, let's calculate the mole of KHP:
moles = 0.5053 / 204.2189 = 0.00247 moles
With the moles, we also know that:
n = M*V (2)
Replacing in (1):
n = MbVb
Now, solving for Mb:
Mb = n/Vb (3)
Finally, replacing the data:
Mb = 0.00247 / (13.4473/1000)
Mb = 0.184 M
This would be the concentration of NaOH
T is amount after time t
<span>Ao is initial amount </span>
<span>t is time </span>
<span>HL is half life </span>
<span>log (At) = log [ Ao x (1/2)^(t/HL) ] </span>
<span>log (At) = log Ao + log (1/2)^(t/HL) </span>
<span>log (At) = log Ao + (t/HL) x log (1/2) </span>
<span>( log At - log Ao) / log (1/2) = t / HL </span>
<span>log (At/Ao) / log (1/2) = t / HL </span>
<span>HL = t / [( log (At / Ao)) / log (1/2) ] </span>
<span>HL = 14.4 s / [ ( log (12.5 / 50) / log (1/2) ] </span>
<span>HL = 14.4 s / 2 = 7.2 seconds </span>
Answer:
The work done and heat absorbed are both -8,1 kJ
Explanation:
The work done in an isobaric process is defined as:
W = -P (Vf - Vi)
Where P is pressure ( 10 atm)
Vf = 10 L
Vi = 2 L
Thus, <em>W = -80 atm×L ≡ -8,1 kJ</em>
This is the work done in expansion of the gas. As the gas remains at the same temperature, there is no change in internal energy doing that all work was absorbed as heat.
I hope it helps!
