If you have 58.93g of Co it means that you only have 1 mol (use a periodic table to find the answer, if you had more find it by proportion, it's easier).
There's 6.022 x 10^23 atoms per mol so you have 6.022 x 10^23 atoms of Co.
(once again if you had more mol, you could find the answer by proportions).
The correct answer for the question that is being presented above is this one: "<span>16.728 g."</span>
Given that
ΔHsolid = -5.66 kJ/mol.
This means that 5.66 kJ of heat is released when 1 mole of NH3 solidifies
When 5.57 kJ of heat is released
amount of NH3 solidifies = 5.57/5.66 = 0.984 moles
<span>molar mass of NH3 = 17 g/mole </span>
<span>1 mole of NH3 = 17 g </span>
So, 0.984 moles of NH3 = 17 X 0.984 = 16.728 g
<span>are
in random, constant, straight-line motion</span>
<span>
</span>
<span>your welcome :)</span>
<span>all I Ask is for a thank u peaceeee</span>
The answer is C. Since aluminum reacts with chloride displacing only Copper.
Answer:
3.07 × 10⁻⁴
Explanation:
Step 1: Calculate the concentration of H⁺
We will use the definition of pH.
![pH = -log [H^{+} ]\\\[ [H^{+} ] = antilog -pH = antilog -2.37 = 4.27 \times 10^{-3} M](https://tex.z-dn.net/?f=pH%20%3D%20-log%20%5BH%5E%7B%2B%7D%20%5D%5C%5C%5C%5B%20%5BH%5E%7B%2B%7D%20%5D%20%3D%20antilog%20-pH%20%3D%20antilog%20-2.37%20%3D%204.27%20%5Ctimes%2010%5E%7B-3%7D%20M)
Step 2: Calculate the concentration of HY
5.22 × 10⁻³ mol of HY are dissolved in 0.088 L. The concentration of the acid (Ca) is:
Step 3: Calculate the acid dissociation constant (Ka)
We will use the following expression.
![Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(4.27 \times 10^{-3} )^{2} }{0.0593} = 3.07 \times 10^{-4}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5E%7B2%7D%20%7D%7BCa%7D%20%3D%20%5Cfrac%7B%284.27%20%5Ctimes%2010%5E%7B-3%7D%20%29%5E%7B2%7D%20%7D%7B0.0593%7D%20%3D%203.07%20%5Ctimes%2010%5E%7B-4%7D)
