The amount of heat released when 12.0g of helium gas condense at 2.17 K is; -250 J
The latent heat of vapourization of a substance is the amount of heat required to effect a change of state of the substance from liquid to gaseous state.
However, since we are required to determine heat released when the helium gas condenses.
The heat of condensation per gram is; -21 J/g.
Therefore, for 12grams, the heat of condensation released is; 12 × -21 = -252 J.
Approximately, -250J.
Read more on latent heat:
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Answer:
I thank it is C
Step-by-step explanation:
hope this helps if not please let me now
Step-by-step explanation:
$ 3750 = truck rental; $125 per ton of sugar transported
C is cost; S is number of tons transported
Equation relating C to S would be a linear equation like y = mx + b
C = 125S + $3750
This equation would be graphed in the first quadrant only
you would start with your y-intercept at (0, 3750)
As x increases by 1, your y increases by 125 yielding these points:
(1, 3875) (2, 4000) (3, 4125) etc.
This shows that for each increase by one ton of sugar, the cost goes up $125
Answer:
mean = 0.76
median = 0.75
range = 1.79
Step-by-step explanation:
mean = sum/no of items
for media they need to be sorted: 0.05,0.06,0.75,1.10,1.84
median is the middle item, so 0.75
range = largest item - smallest item = 1.84 - 0.05 = 1.79
Answer:
The overview of the given problem is outlined in the following segment on the explanation.
Step-by-step explanation:
The proportion of slots or positions that have been missed due to numerous concurrent transmission incidents can be estimated as follows:
Checking a probability of transmitting becomes "p".
After considering two or even more attempts, we get
Slot fraction wasted,
= ![[1-no \ attempt \ probability-first \ attempt \ probability-second \ attempt \ probability+...]](https://tex.z-dn.net/?f=%5B1-no%20%5C%20attempt%20%5C%20probability-first%20%5C%20attempt%20%5C%20probability-second%20%5C%20attempt%20%5C%20probability%2B...%5D)
On putting the values, we get
= ![1-no \ attempt \ probability-[N\times P\times probability \ of \ attempts]](https://tex.z-dn.net/?f=1-no%20%5C%20attempt%20%5C%20probability-%5BN%5Ctimes%20P%5Ctimes%20probability%20%5C%20of%20%5C%20attempts%5D)
= ![1-(1-P)^{N}-N[P(1-P)^{N}]](https://tex.z-dn.net/?f=1-%281-P%29%5E%7BN%7D-N%5BP%281-P%29%5E%7BN%7D%5D)
So that the above seems to be the right answer.
