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Lesechka [4]
3 years ago
15

Just wanted to make sure I have the right idea.

Chemistry
1 answer:
Rasek [7]3 years ago
4 0

Explanation:

yes you got the right idea

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Which of the following are true statements?
Digiron [165]

Answer:

The first is true

The second is true IF the statement means that the human body contains energy

The third is true

The fourth is false

8 0
2 years ago
What did ancient Indians create using metallurgy?<br> texts<br> plows<br> sitars<br> paintings
SVETLANKA909090 [29]
I’m guessing the answer is plows or sitars

7 0
2 years ago
Read 2 more answers
What minimum frequency of light is required to ionize boron?
Nadusha1986 [10]

Answer:

frequency  = 8.22 x 10¹⁴ s⁻¹

Explanation:

An electron's positional potential energy while in a given principle quantum energy level is given by Eₙ = - A/n² and A = constant = 2.18 x 10⁻¹⁸j. So to remove an electron from the valence level of Boron (₅B), energy need be added to promote the electron from n = 2 to n = ∞. That is, ΔE(ionization) = E(n=∞) - E(n=2) = (-A/(∞)²) - (-A/(2)²) = [2.18 x 10⁻¹⁸j/4] joules = 5.45 x 10⁻¹⁹ joules.

The frequency (f) of the wave ionization energy can then be determined from the expression ΔE(izn) = h·f; h = Planck's Constant = 6.63 x 10⁻³⁴j·s. That is:

ΔE(izn) = h·f => f = ΔE(izn)/h = 5.45 x 10⁻¹⁹ j/6.63 x 10⁻³⁴ j·s = 8.22 x 10¹⁴ s⁻¹

3 0
2 years ago
How many grams are in 4.65 mol of AI(NO2)3
nirvana33 [79]
To find out how many grams are in 4.65 moles of Al(NO₂)₃
Find out what the molar mass of Al(NO₂)₃ is
Al = 26.98 g/mol Al
N = 14 g/mol N
O = 16 g/mol O
Next, you have to look at the subscripts and figure out which they belong to, in this case:
Al = 26.98 g/mol Al
N₃ = 42 g/mol N₃
O₆ = 96 g/mol O₆
Finally, add the numbers together, so:
26.98 g/mol Al + 42 g/mol N₃ + 96 g/mol O₆ = 
164.98 g/mol Al(NO₂)₃
Now, you have 4.65 mol Al(NO₂)₃ so
164.98 g/mol Al(NO₂)₃ × 4.65 mol Al(NO₂)₃ =
767.157 grams of Al(NO₂)₃

6 0
2 years ago
The barium isotope 133ba has a half-life of 10.5 years. a sample begins with 1.1×1010 133ba atoms. how many are left after (a) 5
Sladkaya [172]
<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>
4 0
3 years ago
Read 2 more answers
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