For what? what is your question
Answer:
0.676 grams of manganese (IV) oxide should be added.
Explanation:
Moles of chlorine gas = n
Volume of the chlorine gas = V = 205 mL = 0.205 L
Pressure of the chlorine gas = 705 Torr = 
1 atm = 760 Torr
Temperature of the chlorine gas = T = 25°C = 25 + 273 K = 298 K
( ideal gs equation)


According to reaction, 1 mole of chlorine gas is obtained from 1 mole of manganese(IV) oxide,then 0.00777 moles of chlorine gas will be obtained from :
of manganese (IV) oxide
Mass of 0.00777 moles of manganese (IV) oxide:
0.00777 mol × 87 g/mol = 0.676 g
0.676 grams of manganese (IV) oxide should be added.
<h3>Answer:</h3>
Have a <em>strong</em><em> </em><em><u>polar</u></em><em><u> </u></em><em><u>attractio</u></em><em><u>ns</u></em><em><u> </u></em>between <em><u>molecule</u></em><em><u>'s</u></em><em><u> </u></em><em><u> </u></em><em><u>involvin</u></em><em><u>g</u></em><em><u> </u></em><em><u>H</u></em><em><u>,</u></em><em><u> </u></em><em><u>F</u></em><em><u>,</u></em><em><u> </u></em><em><u>O</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>N</u></em><em><u>.</u></em>
SUBMIT!
<em><u>hope</u></em><em><u> </u></em><em><u>it</u></em><em><u> </u></em><em><u>help</u></em><em><u>s</u></em>
Answer:
See explanation.
Explanation:
Hello,
In this case, for the described chemical reaction:
2 HCl(aq) + Mg(OH)2(aq) → MgCl2(aq) + 2 H2O(l)
We can notice there is a 2:1 molar ratio between the moles of hydrochloric acid and magnesium hydroxide, therefore, at the equivalence point:

And in terms of volumes and concentrations we verify:

So we use the given data to proof it:

Therefore, we can conclude the data is wrong by means of the 2:1 mole ratio that for sure was not taken into account. This is also supported by the fact that normalities are actually the same, but the nomality of magnesium hydroxide is the half of the hydrochloric acid normality since the acid is monoprotic and the base has two hydroxyl ions.
Best regards.