An element that exist as a discreet atom has only one atom which can stand alone on its own. Example of this is Neon, which is a noble gas. The chemical symbol for neon is Ne.
Elements are able to exist as discreet atom because they are chemical stable and inert.
Answer:
The answer to your question is: 83.9 %
Explanation:
Data
Cu = 31.8 g
S = 50 g
CuS = 40 g
yield = ?
Equation
Cu + S ⇒ CuS
MW Cu = 64 g
MW S = 32 g
MW CuS = 96 g
Ratio (theoretical/experimental)
Experimental 50/31.8 = 1.57
Theoretical 32/64 = 0.5 limiting reactant Cu
64 g of Cu ------------------ 96 g of CuS
31.8 g ------------------- x
x = (31.8 x 96) / 64
x = 47.7 g of CuS
% yield = (40/47.7) x 100
= 83.9 %
Answer:
The standard reduction potential E°cell (Cr6+/Cr3+) is -0.13V
Explanation:
<u>Step 1</u>: Data given
3 Ni^2+(aq) + 2 Cr(OH)3(s) + 10 OH− (aq) → 3 Ni(s) + 2 CrO4^2−(aq) + 8 H2O(l) ΔG∘ = +87000 J/mol
Ni2+(aq) + 2 e− → Ni(s) E∘red = -0.28 V
<u>Step 2:</u> The half reactions:
Cathode: Ni2+(aq) + 2 e− → Ni(s) E° = -0.28 V
Anode: CrO4^2-(aq) + 4H2O(l) +3e- → Cr(OH)3(s) + 5OH- (aq) E°= unknown
<u>Step 3:</u> Calculate E°cell
ΔG° = -n*F*E°cell
⇒ with ΔG° = the gibbs free energy
⇒ n = the number of electrons in the net reaction = 6
⇒ F = the Faraday constant = 96485 C
⇒ E°cell= the standard cell potential
<u>Step 4:</u> Calculate E°(Cr6+/Cr3+
E°cell= ΔG°/(-n*F)
E°cell = 87000 /(-6*96485)
E°cell = -0.150 V
E°cell = E°(Ni2+/Ni) - E°(Cr6+/Cr3+)
E°(Cr6+/Cr3+) = -0.13V
The standard reduction potential E°cell (Cr6+/Cr3+) is -0.13V
Answer:
Option D is correct.
H₂O + CO₂ → H₂CO₃
Explanation:
First of all we will get to know what law of conservation of mass states.
According to this law, mass can neither be created nor destroyed in a chemical equation.
This law was given by French chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.
Example:
6CO₂ + 6H₂O + energy → C₆H₁₂O₆ + 6O₂
there are six carbon atoms, eighteen oxygen atoms and twelve hydrogen atoms on the both side of equation so this reaction followed the law of conservation of mass.
Now we will apply this law to given chemical equations:
A) H₂ + O₂ → H₂O
There are two hydrogen and two oxygen atoms present on left side while on right side only one oxygen and two hydrogen atoms are present so mass in not conserved. This equation not follow the law of conservation of mass.
B) Mg + HCl → H₂ + MgCl₂
In this equation one Mg, one H and one Cl atoms are present on left side while on right side two hydrogen, one Mg and two chlorine atoms are present. This equation also not follow the law of conservation of mass.
C) KClO₃ → KCl + O₂
There are one K, one Cl and three O atoms are present on left side of chemical equation while on right side one K one Cl and two oxygen atoms are present. This equation also not following the law of conservation of mass.
D) H₂O + CO₂ → H₂CO₃
There are two hydrogen, one carbon and three oxygen atoms are present on both side of equation thus, mass remain conserved. Thus is correct option.
We have to know the number of Na⁺ ions in the unit cell.
The number of Na⁺ ions in the unit cell is (D) 8.
Sodium oxide (Na2O) crystallizes in a structure in which the O2– ions are in a face - centered cubic lattice and the Na+ ions are in tetrahedral holes.
O²⁻ ions are in a face centred cubic lattice, so the number of O²⁻ ions per unit cell is equal to 4. The number of tetrahedral hole= 2 X 4=8. Na+ ions are present in tetrahedral holes, which indicates there are 8 number of Na+ ions in the unit cell.
