Answer:
2.77 mol N
Explanation:
M(N2O) = 2*14 + 16 = 44 g/mol
61.0 g * 1 mol/44g = (61/44) mol N2O
N2O ---- 2N
1 mol 2 mol
(61/44) mol x mol
x = (61/44)*2/1 = 2.77 mol N
Answer:
a) 2.01 g
Explanation:
- Na₂CO₃ (s) + 2AgNO₃ (aq) → Ag₂CO₃ (s) + 2NaNO₃
First we <u>convert 0.0302 mol AgNO₃ to Na₂CO₃ moles</u>, in order to <em>calculate how many Na₂CO₃ moles reacted</em>:
- 0.0302 mol AgNO₃ *
= 0.0151 mol Na₂CO₃
So the remaining Na₂CO₃ moles are:
- 0.0340 - 0.0151 = 0.0189 moles Na₂CO₃
Finally we <u>convert Na₂CO₃ moles into grams</u>, using its <em>molar mass</em>:
- 0.0189 moles Na₂CO₃ * 106 g/mol = 2.003 g Na₂CO₃
The closest answer is option a).
Let us start with the total area of the lawn. Area= width x length, ie, 21 x 20 = 420 sq. ft. Snow flakes per square foot per minute = 1350 So Snow flakes for 420 sq.feet per minute = 420 x 1350 = 567000. Snow flakes for 1 hour = 567000 x 60 = 34020000 (60 minutes) Weight of 34020000 snow flakes = 34020000 x 1.60 = 54432000mg. To convert it into kilograms, divide this number by 1000000 (1 kilogram = 1000000 milligrams) Thus 54432000/1000000 = 54.432 kilograms or 54 kilograms and 432 grams.
Answer:
b. The splitting of the d-orbitals is smaller in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.
Explanation:
The spectrochemical series is an arrangement of ligands in increasing order of their magnitude of crystal field splitting.
Ligands that occurs towards the right in the series are called strong field ligands and they tend to cause a greater magnitude of crystal field splitting. Ligands that occur towards the left hand side in the series are called weak field ligands and they tend to cause a lesser magnitude of crystal field splitting.
Since Cl^- is a weak field ligand, it causes a lesser magnitude of d orbital splitting compared to ethylenediammine (en) which causes a greater magnitude of d orbital splitting.
Hence; the splitting of the d-orbitals is smaller in the [Ni(Cl)6]4- complex than in the [Ni(en)3]2+ complex.
Answer:
Explanation:
concentration unit relates moles of solute to volume of solution. ... First find the number of moles of KCl in the 25.00 mL of 0.500 M solution: ... 13) What volume of 1.25 M sulfuric acid is needed to dissolve 0.750 g of ... mL of 0.0962 M hydrochloric acid is titrated with a calcium hydroxide solution, and ...
