First, we write the reaction equation:
3Pb(NO₃)₂ + 2Na₃PO₄ → 6NaNO₃ + 3Pb₃(PO₄)₂
Moles of Pb ions present:
moles = concentration x volume
= 0.15 x 0.25
= 0.0375
From the equation,
moles Pb : moles Na₃PO₄
= 3 : 2
Moles of Na₃PO₄:
2/3 x 0.0375
= 0.025
volume = moles / concentration
= 0.025 / 0.1
= 0.25 L
= 250 ml
Answer: -
6
Explanation: -
The given unbalanced chemical equation is As + NaOH -- > Na3AsO3 + H2
We see there 3 sodium on the right side from Na3AsO3.
But there are only 1 sodium on the left from NaOH.
So we multiply NaOH by 3.
As + 3 NaOH -- > Na3AsO3 + H2
Now we see the number of Hydrogen on the left is 3.
But the number of hydrogens is 2 on the left.
So, we multiply to get both sides 6 hydrogen.
As + 6NaOH -- > Na3AsO3 + 3 H2
Rebalancing for Na,
As + 6NaOH -- > 2Na3AsO3 + 3 H2.
Finally balancing As,
2 As + 6 NaOH -- > 2Na3AsO3 + 3H2
The coefficient of the NaOH molecule in the balanced reaction is thus 6
Answer:
In order to react with 45 g of water 1.25 moles of CaC₂ are required.
Explanation:
Given data:
Moles of CaC₂ needed = ?
Mass of water = 45.0 g
Solution:
Chemical equation:
CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂
Number of moles of water:
Number of moles = mass/ molar mass
Number of moles = 45 g/ 18 g/mol
Number of moles = 2.5 mol
Now we will compare the moles of water and CaC₂ from balance chemical equation:
H₂O : CaC₂
2 : 1
2.5 : 1/2×2.5 =1.25 mol
In order to react with 45 g of water 1.25 moles of CaC₂ are required.
Answer:
<h3>
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Explanation:
First balance the chemical equation:
⇄ 
two components are solid so these two will not exert any kind of pressure in the container so at equilibrium only CO2 will apply pressure on the container
Therefore only partial pressure of CO2 will be taken for the calculation of equilibrium pressure constant i.e. Kp
![K_p=[CO_2]](https://tex.z-dn.net/?f=K_p%3D%5BCO_2%5D)
![[CO_2]=p](https://tex.z-dn.net/?f=%5BCO_2%5D%3Dp)
Answer:
Empirical formula is Li₂CO₃.
Explanation:
Percentage of oxygen= 65.0%
Percentage of lithium = 18.7%
Percentage of carbon= 16.3%
Empirical formula = ?
Solution:
Number of gram atoms of C = 16.3/12 = 1.4
Number of gram atoms of Li = 18.7/6.94 = 2.7
Number of gram atoms of O = 65.0/ 16 = 4.1
Atomic ratio:
Li : C : O
2.7/1.4 : 1.4/1.4 : 4.1/1.4
2 : 1 : 3
Li : C : O = 2 : 1 : 3
Empirical formula is Li₂CO₃.
