second compound
Let molar mass of x is = X
Let molar mass of y is = Y
Moles of x in second compound = Mass / molar mass = 7 / X
Moles of y in second compound = Mass / molar mass = 4.5 / Y
For second compound
7 / X : 4.5/ Y = 1:1
Therefore
X / Y = 7/4.5
Y / X = 4.5/ 7
The mass of x in first compound = 14g
moles of x in first compound = 14/X
Mass of y in first compound = 3
moles of y in first compound = 3 / Y
14 / X : 3/ Y = 14Y / 3X = 14 X 4.5 / 3 X 7 = 3 :1
Thus molar ratio in first compound = moles of x / Moles of y = 3:2
Formula = x3y
Answer:
2Al(s) +3Ni²⁺(aq) ⟶ 2Al³⁺(aq) + 3Ni(s)
Explanation:
The unbalanced equation is
Al(s) + Ni²⁺(aq) ⟶ Ni(s) + Al³⁺(aq)
(i) Half-reactions
Al(s) ⟶ Al³⁺(aq) + 3e⁻
Ni²⁺(aq) + 2e⁻ ⟶ Ni(s)
(ii) Balance charges
2 × [Al(s) ⟶ Al³⁺(aq) + 3e⁻]
3 × [Ni²⁺(aq) + 2e⁻ ⟶ Ni(s)]
gives
2Al(s) ⟶ 2Al³⁺(aq) + 6e⁻
3Ni²⁺(aq) + 6e⁻ ⟶ 3Ni(s)
(iii) Add equations
2Al(s) ⟶ 2Al³⁺(aq) + 6e⁻
<u>3Ni²⁺(aq) + 6e⁻ ⟶ 3Ni(s) </u>
2Al(s) +3Ni²⁺(aq) + <em>6e</em>⁻ ⟶ 2Al³⁺(aq) + 3Ni(s) + <em>6e⁻
</em>
Simplify (cancel electrons)
2Al(s) +3Ni²⁺(aq) ⟶ 2Al³⁺(aq) + 3Ni(s)
Answer:
Percent yield = 57.8 %
Theoretical yield = 11.781 g
Explanation:
Given data:
Mass of CaO produced = 6.81 g
Mass of CaCO₃ react = 20.7 g
Theoretical yield = ?
Percent yield = ?
Solution:
Chemical equation:
CaCO₃ → CaO + CO₂
Number of moles of CaCO₃ :
Number of moles = mass/molar mass
Number of moles = 20.7 g/ 100.1 g/mol
Number of moles = 0.21 mol
Now we will compare the moles of CaCO₃ with CaO.
CaCO₃ : CaO
1 : 1
0.21 : 0.21
Theoretical yield of CaO:
Mass = number of moles × molar mass
Mass = 0.21 mol × 56.1 g/mol
Mass = 11.781 g
Percent yield:
Percent yield = ( actual yield / theoretical yield ) × 100
Percent yield = (6.81 g/ 11.781 g) × 100
Percent yield = 57.8 %
Answer:
Explanation:
Inertia is the reluctancy of a moving body to come to rest or change it direction or a body at rest to start moving
Answer:
V = 177.4 L.
Explanation:
Hello there!
In this case, since this gas can be assumed as ideal due to the given data, we can use the following equation:
Thus, by solving for volume we obtain:
So we can plug in the temperature in Kelvins (537 K), the pressure in atmospheres (0.404 atm) and the molar mass (54 g/mol) to obtain:
Best regards!
