Answer: Its option "A" Both A and R are true and R is the correct explanation of A.
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Answer:
3.925 mol.
Explanation:
- From the balanced equation:
<em>2 Na₂O₂(s) + 2 H₂O(l) → 4 NaOH(s) + O₂(g)
,</em>
It is clear that 2 moles of Na₂O₂ react with 2 moles of H₂O to produce 4 moles of NaOH and 1 mole of O₂
.
<em>Using cross multiplication:</em>
4 moles of NaOH produced with → 1 mole of O₂
.
15.7 moles of NaOH produced with → ??? mole of O₂
.
<em>∴ The no. of moles of O₂ made =</em> (1 mole)(15.7 mole)/(4 mole) = <em>3.925 mol.</em>
A chemical element that has an atomic number less than 58 and an atomic mass greater than 135.6m is barium (atomic no. 56 and atomic mass137.13 ) and lanthanum (atomic no. 57 and atomic mass 135.6).
<h3>Give a brief introduction about Barium and Lanthanum.</h3>
Barium is an element with the symbol Ba and atomic number 56. It is an alkaline earth metal that is soft and silvery, and it is the fifth element in group 2. Barium is never found in nature as a free element due to its extreme chemical reactivity. Oil well drilling fluid uses barium sulfate as an insoluble ingredient. It is employed as an X-ray radiocontrast agent in a purer form to image the human gastrointestinal tract. Barium compounds that dissolve in water have been employed as rodenticides despite being hazardous.
Chemical element lanthanum has the atomic number 57 and the symbol La. It is a silvery-white, ductile, soft metal that slowly tarnishes when exposed to air. It serves as the eponym for the group of 15 related elements in the periodic table between lanthanum and lutetium, of which lanthanum is the first and prototype. The rare earth elements traditionally include lanthanum.
Learn more about elements here:-
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<span>The molecular formula that describes the problem is
2CH3COOH (aq) + Ca(OH)2 (s) ---> Ca(CH3COO)2 (aq) + 2H2O (l)
The net equation is written as follows:
2CH3COOH- (aq) + 2H+ (aq) + Ca(OH)2 (s) ---> Ca2+ (aq) + 2 CH3COO- (aq) + 2H2O (l)
canceling out spectator ions
2H+ (aq) + Ca(OH)2 (s) ---> Ca2+ (aq) + 2 H2O (l)</span>
