Answer:
The answer is C. The partial pressure of hydrogen will be unchanged.
Explanation:
⇒ 
Argon with electronic configuration
(that is atomic number 18) is an inert gas making it unreactive and it's addition to the reaction has no effect on the partial pressure of either the reactant or production or the state of the system.
The partial pressure of hydrogen will remain unchanged on the addition of Argon.
Answer:
all cells are produced from other preexisting cells through cell division
Both of you are overlooking a pretty big component of the question...the Group I cation isn't being dissociated into water. We're testing the solubility of the cation when mixed with HCl. And this IS a legitimate question, seeing as our lab manual is the one asking.
<span>By the way, the answer you're looking for is "Because Group I cations have insoluble chlorides". </span>
<span>"In order...to distinguish cation Group I, one adds HCl to a sample. If a Group I cation is present in the sample, a precipitate will form." </span>
The Zn that is 1.33 g is used at the start of the reaction where f is 520 ml and h2 collected over water is 28oc and the atmospheric pressure is 1.0 atm.
Given If 520 ml of H2 is gathered over Wate at 28 diploma Celsius and the atmospheric strain is 1 ATM if vapour strain of wate at 28 diploma celsius is 28.three mmhg then the quantity of zn in grams taken at begin of the response is.
We recognise that
h * 2 = PT - P * h * 20 = 1atm - 0.037atm
= 0.963 atm
1 * h * 2 = Ph * 2V / R * T
= 0.963 atm x 0.520 L / 0.0821 L atm/
molK * 301
= 0.02 mol h2
= 0.02molZn
So 0.02 mol Zn x 65.39 g/mol
= 1.33 g Zn
Read more about zinc;
brainly.com/question/28880469
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Answer:
<h2>0.44 </h2>
Explanation:
The pH of a solution can be found by using the formula
![pH = - log [ { H_3O}^{+}]](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%20%5B%20%7B%20H_3O%7D%5E%7B%2B%7D%5D)
From the question we have

We have the final answer as
<h3>0.44</h3>
Hope this helps you