I believe it should be copper (II) sulphide.
Answer : When the Hydrogen chloride gas reacts with oxygen to form chlorine gas and water vapor the reaction can be as follows,


= [
![[Cl_{2}] ^{2} [H_{2}O]^{2} / [HCL]^{4} [O_{2}]](https://tex.z-dn.net/?f=%5BCl_%7B2%7D%5D%20%5E%7B2%7D%20%20%20%20%20%20%5BH_%7B2%7DO%5D%5E%7B2%7D%20%20%2F%20%20%5BHCL%5D%5E%7B4%7D%20%20%5BO_%7B2%7D%5D)
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Answer:
Explanation:
The atomic radius of elements are used to estimate the sizes of elements. The atomic radius is taken as half of the inter-nuclear distance between two covalently bonded atoms of non-metallic elements or half of the distance between two nuclei in the solid state of metals.
To solve this problem we will obtain the atomic radius values of the given elements from a standard atomic radius table;
Si 111 pm
P 98 pm
Cl 79 pm
S 87pm
pm = picometer
We see that chlorine has the least atomic radius
Answer:
<u>Molar</u><u> </u><u>mass</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>diprotic</u><u> </u><u>acid</u><u> </u><u>is</u><u> </u><u>4</u><u>2</u><u>4</u><u> </u><u>grams</u>
Explanation:
[hint: <u>diprotic</u><u> </u><u>acid</u><u> </u><u>only</u><u> </u><u>contains</u><u> </u><u>2</u><u> </u><u>hydrogen</u><u> </u><u>protons</u><u>]</u>
Ionic equation:

first, we get moles of potassium hydroxide in 28.94 ml :

since mole ratio of diprotic acid : base is 2 : 2, moles are the same.
Therefore, <u>m</u><u>o</u><u>l</u><u>e</u><u>s</u><u> </u><u>o</u><u>f</u><u> </u><u>a</u><u>c</u><u>i</u><u>d</u><u> </u><u>t</u><u>h</u><u>a</u><u>t</u><u> </u><u>r</u><u>e</u><u>a</u><u>c</u><u>t</u><u>e</u><u>d</u><u> </u><u>a</u><u>r</u><u>e</u><u> </u><u>0</u><u>.</u><u>0</u><u>0</u><u>8</u><u>7</u><u>4</u><u>3</u><u> </u><u>m</u><u>o</u><u>l</u><u>e</u><u>s</u><u>.</u>

for the molar mass:
I think the correct answer from the choices listed above is option D. the substances cobalt and barium chloride will not react since cobalt os less active than barium so it does not have enough energy to displace barium in the compound. Hope this answers the question.