Hi :)
20 mol NH3 x 6 H2O/4 NH3 = 30 mol H2O
Hope this helped :)
Explanation:
total heat = Heat required to convert 2 kg of ice to 2 kg of water at 0 °C + Heat required to convert 2 kg of water at 0 °C to 2 kg of water at 20 °C.
Heat=mhfg+mCpΔT
Here, m ( mass of ice) = 2 kg
hfg (latent heat of fusion of ice) = 334 KJ
Cp of water (specific heat) = 4.187 KJ/Kg-K
ΔT(Temperature difference) = 20 °C
Therefore, Heat required = 2 x 334 + 2 x 4.187 x (20 - 0 )
Heat reqd= 835.48 KJ
Therefore, to melt 2 kg of ice 835.48 KJ of heat is required.
Answer:
2812.6 g of H₂SO₄
Explanation:
From the question given above, the following data were obtained:
Mole of H₂SO₄ = 28.7 moles
Mass of H₂SO₄ =?
Next, we shall determine the molar mass of H₂SO₄. This can be obtained as follow:
Molar mass of H₂SO₄ = (1×2) + 32 + (16×4)
= 2 + 32 + 64
= 98 g/mol
Finally, we shall determine the mass of H₂SO₄. This can be obtained as follow:
Mole of H₂SO₄ = 28.7 moles
Molar mass of H₂SO₄ =
Mass of H₂SO₄ =?
Mole = mass / Molar mass
28.7 = Mass of H₂SO₄ / 98
Cross multiply
Mass of H₂SO₄ = 28.7 × 98
Mass of H₂SO₄ = 2812.6 g
Thus, 28.7 mole of H₂SO₄ is equivalent to 2812.6 g of H₂SO₄
Answer:
CH₃CH₂CH₂COOH > CH₃CH₂COOH > ClCH₂CH₂COOH > ClCH₂COOH
Explanation:
Electron-withdrawing groups (EWGs) increase acidity by inductive removal of electrons from the carboxyl group.
Electron-donating groups (EDGs) decrease acidity by inductive donation of electrons to the carboxyl group.
- The closer the substituent is to the carboxyl group, the greater is its effect.
- The more substituents, the greater the effect.
- The effect tails off rapidly and is almost zero after about three C-C bonds.
CH₃CH₂-CH₂COOH — EDG — weakest — pKₐ = 4.82
CH₃-CH₂COOH — reference — pKₐ = 4.75
ClCH₂-CH₂COOH — EWG on β-carbon— stronger — pKₐ = 4.00
ClCH₂COOH — EWG on α-carbon — strongest — pKₐ = 2.87
The balanced chemical reaction would be:
FeS(s)+2HCl(aq)→FeCl2(s)+H2S(g)
We are given the amount of the reactants to be used for the reaction. We use these amounts. First, we determine the limiting reactant of the reaction. From the data, we can say that FeS is the limiting ad HCl is the excess reactant. We calculate as follows:
Amount of HCl used: 0.240 mol FeS x 2 mol HCl / 1 mol FeS = 0.48 mol HCl
0.646 - 0.48 = 0.166 mol HCl left
