Answer:
Except where otherwise noted, data are given for materials in their standard state (at 25 °C [77 °F], 100 kPa). Calcium phosphide (CP) is the inorganic compound with the formula Ca3P2. It is one of several phosphides of calcium, being described as the salt-like material composed of Ca2+ and P3−.
Explanation:
Answer:
The answer is 4.28 moles
Explanation:
This is super easy okay, you won't forget this!
Basically mole ratios, we're just looking at the coefficients in front of the compounds, multiplying them, and dividing them as we see fit.
In this example, you can see how you need 2 moles of lithium bromide (LiBr) for the reaction, and 2 moles of lithium chloride (LiCl) will be produced.
Basically, the <u>molar ratio</u> is when you divide numbers and see how much of this do I have for that (if that makes sense).
So if you were to divide the 2 moles of LiBr / 2 moles of LiCl = 1. So we know that the mole ratio for LiBr to LiCl is 1:1 or 2:2, either or, it's the same thing.
SO THE BIG IDEA, if we have 4.28 moles of lithium bromide reacting, we should also have 4.28 moles of lithium chloride produced, BECAUSE the <u>mole ratio</u> is 1:1.
I hope this makes sense please tell me if it doesn't, I will try my best to explain a little more.
Answer:
74 or 74 kPa.
Explanation:
Hello,
In this case, based on the initial information, it is seen that the oxygen and the carbon dioxide form the mixture at 160 kPa, thus, by isolating the oxygen, its pressure will be equal to its initial partial pressure because it gets isolated, hence, we compute its molar fraction as:
Therefore, its initial pressure turns out:
Such pressure will be the oxygen's pressure once it is isolated. Finally, considering the request, the answer will be just 74 (by rounding to the nearest integer and without units).
Best regards.
Answer:
3.97
Explanation:
pH of buffer solution = pKa+Log(Cb/Ca)
pH of buffer solution = -log(Ka)+log(Cb/Ca)............... Equation 1
Where Ca = concentration of acid, Cb = concentration of base.
Given: Ka = 6.5×10⁻⁵, Ca = 0.25 M, Cb = 0.15 M
Substitute into equation 1
pH of buffer solution = -log(6.5×10⁻⁵)+log(0.15/0.25)
pH of buffer solution = 4.19+(0.22)
pH of buffer solution = 3.97.
Hi there
Steps are
Pretreatment
Reductive leaching
Precipitating of Ni
Precipitation of Co
Extraction of Mn^2+
Precipitating of Li
Hope this helped!
