The molar mass of B(NO₃)₃ - Boron nitrate : 196.822 g/mol
<h3>Further explanation</h3>
In stochiometry therein includes
<em>Relative atomic mass (Ar) and relative molecular mass / molar mass (M) </em>
So the molar mass of a compound is given by the sum of the relative atomic mass of Ar
M AxBy = (x.Ar A + y. Ar B)
The molar mass of B(NO₃)₃ - Boron nitrate :
M B(NO₃)₃ = Ar B + 3. Ar N + 9.Ar O
M B(NO₃)₃ = 10.811 + 3. 14,0067 + 9. 15,999
M B(NO₃)₃ = 196.822 g/mol
The limiting reagent when 5 g of NaOH and 4.4 g CO₂ allowed to react will be NaOH
<h3>What is Limiting reagent ?</h3>
The limiting reactant (or limiting reagent) is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed.
Given chemical equation in balanced form ;
2NaOH(s) + CO₂(g) → Na₂CO₃(s) + H₂O(l).
According to the Chemical equation ;
- The limiting reagent when 5 g of NaOH and 4.4 g CO₂ allowed to react will be NaOH
If 44 g CO₂ requires 80 g of NaOH, therefore, 4.4 g CO₂ will require atleast 8 g of NaOH.
But the available quantity is 5 g NaOH. thus, NaOH is the Limiting reagent.
- 6.625 g of Na₂CO₃ are expected to be produced 5.0 g of NaOH and 4.4 g of CO₂ are allowed to react
As 80 g NaOH produces 106 g of Na₂CO₃.
Therefore 5 g NaoH will produce ;
106 / 80 x 5 = 6.625 g
Learn more about limiting reagent here ;
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Answer:
Explanation:
mole of HCl remaining after reaction with CaCO₃
= .3 M of NaOH of 32.47 mL
= .3 x .03247 moles
= .009741 moles
Initial HCl taken = .3 x .005 moles = .0015 moles
Moles of HCl reacted with CaCO₃
= .009741 - .0015 = .008241 moles
CaCO₃ + 2HCl = CaCl₂ + CO₂ + H₂O .
1 mole 2 moles
2 moles of HCl reacts with 1 mole of CaCO₃
.008241 moles of HCl reacts with .5 x .008241 moles of CaCO₃
CaCO₃ reacted with HCl = .5 x .008241 = .00412 moles
the mass (in grams) of calcium carbonate in the tablet
= .00412 x 100 = .412 grams . ( molar mass of calcium carbonate = 100 )
Explanation:
P1V1 = nRT1
P2V2 = nRT2
Divide one by the other:
P1V1/P2V2 = nRT1/nRT2
From which:
P1V1/P2V2 = T1/T2
(Or P1V1 = P2V2 under isothermal conditions)
Inverting and isolating T2 (final temp)
(P2V2/P1V1)T1 = T2 (Temp in K).
Now P1/P2 = 1
V1/V2 = 1/2
T1 = 273 K, the initial temp.
Therefore, inserting these values into above:
2 x 273 K = T2 = 546 K, or 273 C.
Thus, increasing the temperature to 273 C from 0C doubles its volume, assuming ideal gas behaviour. This result could have been inferred from the fact that the the volume vs temperature line above the boiling temperature of the gas would theoretically have passed through the origin (0 K) which means that a doubling of temperature at any temperature above the bp of the gas, doubles the volume.
From the ideal gas equation:
V = nRT/P or at constant pressure:
V = kT where the constant k = nR/P. Therefore, theoretically, at 0 K the volume is zero. Of course, in practice that would not happen since a very small percentage of the volume would be taken up by the solidified gas.
Answer: 2N2O3 -> 2N2 + 3O2
Explanation: