298 g of calcium carbonate CaCO₃
Explanation:
We have the following chemical reaction:
CaCN₂ (s) + 3 H₂O (l) → CaCO₃ (s)+ 2 NH₃ (g)
number of moles = mass / molar weight
number of moles of H₂O = 161 / 18 = 8.94 moles
Knowing the chemical reaction we devise the following reasoning:
if 3 moles of H₂O produces 1 mole of CaCO₃
then 8.94 moles of H₂O produces X moles of CaCO₃
X = (8.94 × 1) / 3 = 2.98 moles of CaCO₃
mass = number of moles × molar weight
mass of CaCO₃ = 2.98 × 100 = 298 g
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number of moles
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The balanced chemical reaction is:
<span>3N2H4(l)→4NH3(g)+N2(g)
</span>
The amounts given for the N2H4 reactant will be the starting point for our calculations.
2.6mol N2H4 ( 4 mol NH3 / 3 mol N2H4 ) = 3.47 mol NH3
4.05mol N2H4 ( 4 mol NH3 / 3 mol N2H4 ) = 5.4 mol NH3
63.8g N2H4 <span>( 4 mol NH3 / 3 mol N2H4 ) = 85.07 mol NH3</span>
Answer:
Explanation:
the chemical equilibrium constant can be easily calculated since the concentrations at equilibrium are given.the calculation shows the value of Kc for the reversible reaction and forward reaction
Answer: The approximate molecular mass of the polypeptide is 856 g/mol
Explanation:
To calculate the concentration of solute, we use the equation for osmotic pressure, which is:
Or,
where,
= osmotic pressure of the solution = 4.19 torr
i = Van't hoff factor = 1 (for non-electrolytes)
Mass of solute (polypeptide) = 0.327 g
Volume of solution = 1.70 L
R = Gas constant = 
T = temperature of the solution = ![26^oC=[273+26]K=299K](https://tex.z-dn.net/?f=26%5EoC%3D%5B273%2B26%5DK%3D299K)
Putting values in above equation, we get:
Hence, the molar mass of the polypeptide is 856 g/mol
