Technically yes but no because you have to fill the 3s orbital before the 5s orbital.
It is a plant so it is considered living.
Answer: it C
Explanation: Because is the only that makes sense and the only one that seems right
Answer:
a. 86,942 = 8.69x10⁴
f. 56,018 = 5.60x10⁴
b. 0.00621875 = 6.22x10⁻³
g. 0.003870 = 3.87x10⁻³
c. 107062.03 = 1.07x10⁵
h. 15.9994 = 16.0
d. 4.00049 = 4.00
i. 12.011 = 12.0
e. 96 = 9.60x10¹
j. 2,000 = 2.00x10³
Explanation:
Hello,
In this case, rounding off each measurement to three significant figures, considering that if the fourth figure is five or more, the third one is rounded, we obtain:
a. 86,942 = 8.69x10⁴ (four places until the last digit and third digit is not rounded).
f. 56,018 = 5.60x10⁴ (four places until the last digit and third digit is not rounded).
b. 0.00621875 = 6.22x10⁻³ (three places until the first digit and third digit is rounded).
g. 0.003870 = 3.87x10⁻³ (three places until the first digit and third digit is not rounded).
c. 107062.03 = 1.07x10⁵ (five places until the last non-decimal digit and third digit is not rounded).
h. 15.9994 = 16.0 (those consecutive nines round the number to 16.0).
d. 4.00049 = 4.00
i. 12.011 = 12.0
e. 96 = 9.60x10¹ (exponential notation is needed since the number initially has two significant figures so an additional zero is added)
j. 2,000 = 2.00x10³ (three places until the last digit and third digit is not rounded).
Best regards.
This question is asking for the pH of a buffer solution between ammonia and nitric acid, with given volumes and concentrations. At the end, the result turns out to be 10.488.
<h3>Buffers</h3>
In chemistry, buffers are known as substances attempting to hold a relatively constant pH by mixing and acid and a base (weak and strong). In such a way, for the substances given, the first step will be to calculate the consumed moles as they are mixed:
Now, since ammonia is in a greater proportion, one can calculate how much of it is left after being consumed by the nitric acid:
And its new concentration:
Next, with ammonia's ionization:
We set up the equilibrium expression based on ammonia's Kb:
Which can be solved by introducing x and using ammonia's Kb:
Then, we solve for x which is also equal to the concentration of ammonium and hydroxide ions in the solution:
Ultimately, we calculate the pOH and then turn it into pH with:
Learn more about buffers: brainly.com/question/24188850