Answer:
262 ppm of Na₃PO₄
Explanation:
In a dilution, the concentration of the initial solution is decreased. When you take 5.00mL of the solution that is diluted to 25.0mL The solution is diluted 25/5 = <em>5 times</em>
If you make another two serial dilutions the final solution wil decrease its concentration 5*5*5 = 125 times
As original solution containing 0.200 M of Na3PO4, the final solution will have a concentration of:
0.200M / 125 = <em>1.6x10⁻³M</em>
Molarity is defined as the ratio between moles and liters. 1.6x10⁻³ moles of Na3PO4 in 1L are:
1.6x10⁻³mol ₓ (164g/mol) = 0.262g Na₃PO₄ / L
Assuming density of Na3PO4 as 1g/mL the concentration of the solution is:
0.262mL Na₃PO₄ / L
As 1mL = 1000μL:
262μL Na₃PO₄ / L
μL of solute per L of solution is equal to ppm, that means the solution has:
<h3>262 ppm of Na₃PO₄</h3>
Answer:
melting point will be higher than that of pure ethyl acetate
Explanation:
Answer:
132g
Explanation:
The reaction equation is given as:
CaCO₃ → CaO + CO₂
Given;
Number of moles of CaCO₃ = 3 moles
Unknown:
Mass of CO₂ produced = ?
Solution:
From the balanced reaction equation;
1 mole of CaCO₃ will produce 1 mole of CO₂;
3 mole of CaCO₃ will then produce 3 moles of CO₂
To find the mass of CO₂;
Mass of CO₂ = number of moles x molar mass
Molar mass of CO₂ = 12 + 2(16) = 44g/mol
Mass of CO₂ = 3 x 44 = 132g
Considering the first reaction there is no NO on the left and NO2 on the right we start with adding equation 1. To remove O3 we subtract equation 2, then here is 1/2O2 left which we remove by subtracting 1/2 equation 3.
That is; Eqn 1 - Eqn 2 - 1/2(Eqn 3)
NO(g) + O3(g) + 3/2 O2(g) + O(g) = NO2(g) + O2(g) + O3(g) + 1/2 O2(g)
This gives;
NO(g) + O(g) = NO2(g) as required, since O3(g) + 3/2O2(g) is on both sides and thus subtracts out.
Hence; ΔH = ΔH1-ΔH2-1/2 ΔH3
= (-198.9 + 142.3 - 0.5×495.0) kJ = -304.1 kJ