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2 moles N
2 moles H
3 moles O
<h3>Further explanation</h3>
Given
The compound is probably NH₂NO₃
Required
The number of moles
Solution
there are 2 N in NH₂NO₃, so moles N = 2moles
there are 2 H in NH₂NO₃, so moles H = 2moles
there are 3 O in NH₂NO₃, so moles O = 3 moles
Answer:
21.344%
Explanation:
For the given chemical reaction, 8 moles of the reactant should produce 4 moles of . However, 195 g of was produced instead. The molar mass of is 61.9789 g/mol.
Thus, the moles of produced = 195/61.9789 = 3.1462 moles
The percent error = [(Actual -Experiment)/Actual]*100%
The percent error = [(4.00 - 3.1462)/4.00]*100% = (0.85376/4.00)*100% = 21.344%
I think the anwer is electrolyte :)... i had it on a test a couple days ago.
NaHCO₃ + H⁺ = Na⁺ + H₂O + CO₂
n(NaHCO₃)=n(H⁺)
m(NaHCO₃)=M(NaHCO₃)n(NaHCO₃)=M(NaHCO₃)n(H⁺)
p=$m(NaHCO₃)=$M(NaHCO₃)n(H⁺)
p=9.20$/kg×84.0kg/kmol×10⁻³kmol=0.7728$
p≈0.77$