<span>a) 7.9x10^9
b) 1.5x10^9
c) 3.9x10^4
To determine what percentage of an isotope remains after a given length of time, you can use the formula
p = 2^(-x)
where
p = percentage remaining
x = number of half lives expired.
The number of half lives expired is simply
x = t/h
where
x = number of half lives expired
t = time spent
h = length of half life.
So the overall formula becomes
p = 2^(-t/h)
And since we're starting with 1.1x10^10 atoms, we can simply multiply that by the percentage. So, the answers rounding to 2 significant figures are:
a) 1.1x10^10 * 2^(-5/10.5) = 1.1x10^10 * 0.718873349 = 7.9x10^9
b) 1.1x10^10 * 2^(-30/10.5) = 1.1x10^10 * 0.138011189 = 1.5x10^9
c) 1.1x10^10 * 2^(-190/10.5) = 1.1x10^10 * 3.57101x10^-6 = 3.9x10^4</span>
Answer:
a) 965,1 lbf
b) 4,5 kg
c) 1,33 * 10^6 dynes
Explanation:
Mass of an object refers to the amount of mattter it cotains, it can be expressed it gr, kg, lbm, ton, etc.
Weight of an object refers to a force, and is the measurement of the pull of gravitiy on an object. It may be definide as the mass times the acceleration of gravity.
w=mg
In Planet Earth, the nominal "average" value for gravity is 9,8 m/s² (in the International System) or 32,17 ft/s² (in the FPS system).
To solve this problem we'll use the following conversion factors:
1 lbf = 1 lbm*ft/s²
1 N = 1 kg*m/s²
1 dyne = 1 gr*cm/s² and 1 N =10^5 dynes
1 ton = 907,18 kg
1 k = 1000 gr
a) m = 30 lbm
![w = 30 lbm * 32,17 ft/s^{2} = 965, 1 \frac{lmb*ft}{s^{2} } = 965,1 lbf](https://tex.z-dn.net/?f=w%20%3D%2030%20lbm%20%2A%2032%2C17%20ft%2Fs%5E%7B2%7D%20%3D%20965%2C%201%20%5Cfrac%7Blmb%2Aft%7D%7Bs%5E%7B2%7D%20%7D%20%3D%20965%2C1%20lbf)
b) w = 44 N
First, we clear m of the weight equation and then we replace our data.
![m = \frac{w}{g} = \frac{44 N}{9,8 \frac{m}{{s}^{2}} } = 4,5 kg](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7Bw%7D%7Bg%7D%20%3D%20%5Cfrac%7B44%20N%7D%7B9%2C8%20%5Cfrac%7Bm%7D%7B%7Bs%7D%5E%7B2%7D%7D%20%7D%20%3D%204%2C5%20kg)
c) m = 15 ton
here is an attached photo with a detailed explanation, good luck!
The atomic number of the undiscovered element is 168
Element 118 will have just filled its 7p orbitals. therefore the predicted element to fill completely up to its 8 p orbital would have to filled a whole set of s, p, d, f and g orbitals
That's another 2 + 6 + 10 +14 + 18 = 50 electrons
To determine the total number of quantum numbers we have to find
Nml × Nms
we have Nml × Nms = ( 2 + 1 ) × 2
8s + 8P + 7d + 6f + 5g = 2 + 6 + 10 + 14 + 18 = 50
The element right below should be
Z = 118 + 50
= 168
Hence the atomic number of the undiscovered element is 168
Learn more about the atomic number on
brainly.com/question/14514242
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Answer:
Na ·
Explanation:
Hello!
In this case, since the Lewis dot structure of any element illustrates the number of valence electrons at the outer shell; we can set up the electron configuration of sodium to obtain:
![Na^1^1\rightarrow 1s^2,2s^2,2p^6,3s^1](https://tex.z-dn.net/?f=Na%5E1%5E1%5Crightarrow%201s%5E2%2C2s%5E2%2C2p%5E6%2C3s%5E1)
We can see there is only one valence electron, for that reason the Lewis dot structure is:
Na ·
Whereas the dot represents the valence electron.
Best regards!