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tatyana61 [14]
2 years ago
8

. The electron affinity for an oxygen atom is negative. What statement is probably true regarding

Chemistry
1 answer:
kramer2 years ago
7 0
The best and most correct answer among the choices provided by your question is the second choice or letter B.

<span>Regarding the second electron affinity for an oxygen, i. e., the electron affinity for O-, it is much larger and negative.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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3 years ago
Please Help!
Whitepunk [10]

We can use two equations for this problem.<span>

t1/2 = ln 2 / λ = 0.693 / λ
Where t1/2 is the half-life of the element and λ is decay constant.

20 days = 0.693 / λ 
λ   = 0.693 / 20 days        (1) 

Nt = Nο eΛ(-λt)                (2)

Where Nt is atoms at t time, No is the initial amount of substance, λ is decay constant and t is the time taken.
t = 40 days</span>

<span>No = 200 g

From (1) and (2),
Nt =  200 g eΛ(-(0.693 / 20 days) 40 days)
<span>Nt = 50.01 g</span></span><span>

</span>Hence, 50.01 grams of isotope will remain after 40 days.

<span>
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3 0
3 years ago
Can someone please help
oksian1 [2.3K]

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Even ur best Friend Can you to have red flag feeling

6 0
2 years ago
What does a graph representing Charles’s law show?
Juliette [100K]
<span>Volume increases at the same rate as temperature.</span>
5 0
3 years ago
Read 2 more answers
1. The manufacturer of the vinegar used in the experiment stated that the vinegar contained 5.0% acetic acid. What is the percen
Naddik [55]

Answer:

72.8 % (But verify explanation).

Explanation:

Hello,

In this case, with the following obtained results, the percent error is computed as follows:

Volume of vinegar= 7.0 mL

Volume of NaOH= (7+6.6+6.4)/3= 6.7 mL

Used concentration of NaOH= 1.5M

Concentration of acetic acid= (concentration of NaOH*volume of NaOH)/volume of vinegar= (6.7mL*1.5M)/7.0M= 1.44M

Assuming we have 100 mL (0.100L) of vinegar, moles of acetic acid in vinegar = 1.44M x 0.100 L= 0.144 mol

Mass of acetic acid in 100g of vinegar = 0.144 mol x 60.0g/mol= 8.64 g

% of acetic acid in vinegar=8.64 %

% error in percentage of acetic acid = [(8.64% - 5.0%)/5.0] x 100=72.8 %

Clearly, this result depend on your own measurements, anyway, you can change any value wherever you need it.

Regards.

8 0
3 years ago
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