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1 answer:
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Answer:
6.69%
Explanation:
Given that:
Mass of the fertilizer = 0.568 g
The mass of HCl used in titration (45.2 mL of 0.192 M)
=
= 0.313 g HCl
The amount of NaOH used for the titration of excess HCl (42.4 mL of 0.133M)
=
= 0.0058919 mole of NaOH
From the neutralization reaction; number of moles of HCl is equal with the number of moles of NaOH is needed to complete the neutralization process
Thus; amount of HCl neutralized by 0.0058919 mole of NaOH = 0.0058919 × 36.5 g
= 0.2151 g HCl
From above ; the total amount of HCl used = 0.313 g
The total amount that is used for complete neutralization = 0.2151 g
∴ The amount of HCl reacted with NH₃ = (0.313 - 0.2151)g
= 0.0979 g
We all know that 1 mole of NH₃ = 17.0 g , which requires 1 mole of HCl = 36.5 g
Now; the amount of HCl neutralized by 0.0979 HCl =
= 0.0456 g
Therefore, the mass of nitrogen present in the fertilizer is:
=
= 0.038 g
∴ Mass percentage of Nitrogen in the fertilizer = %
= 6.69%
Number of molecules in 1 dm³ Oxygen = 2.71 x 10²²
<h3>Further explanation</h3>Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters/mol.
The mole is the number of particles contained in a substance
1 mol = 6.02.10²³
1 dm³ of oxygen = 1 L Oxygen
- mol Oxygen :
- molecules of Oxygen :
n=mol=0.045
No = 6.02.10²³
Answer:
The total heat required is 691,026.36 J
Explanation:
Latent heat is the amount of heat that a body receives or gives to produce a phase change. It is calculated as: Q = m. L
Where Q: amount of heat, m: mass and L: latent heat
On the other hand, sensible heat is the amount of heat that a body can receive or give up due to a change in temperature. Its calculation is through the expression:
Q = c * m * ΔT
where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the change in temperature (Tfinal - Tinitial).
In this case, the total heat required is calculated as:
- Q for liquid water. This is, raise 248 g of liquid water from O to 100 Celsius. So you calculate the sensible heat of water from temperature 0 °C to 100° C
Q= c*m*ΔT
Q=103,763.2 J
- Q for phase change from liquid to steam. For this, you calculate the latent heat with the heat of vaporization being 40 and being 248 g = 13.78 moles (the molar mass of water being 18 g / mol, then
)
Q= m*L
Q=562.0862 kJ= 562,086.2 J (being 1 kJ=1,000 J)
- Q for temperature change from 100.0 ∘ C to 154 ∘ C, this is, the sensible heat of steam from 100 °C to 154°C.
Q= c*m*ΔT
Q=25,176.96 J
So, total heat= 103,763.2 J + 562,086.2 J + 25,176.96 J= 691,026.36 J
<u><em>The total heat required is 691,026.36 J</em></u>