Answer:
290.82g
Explanation:
The equation for the reaction is given below:
2Al + 3H2SO4 -> Al2(SO4)3 + 3H2 now, let us obtain the masses of H2SO4 and Al2(SO4)3 from the balanced equation. This is illustrated below:
Molar Mass of H2SO4 = (2x1) + 32 + (16x4) = 2 + 32 +64 = 98g/mol
Mass of H2SO4 from the balanced equation = 3 x 98 = 294g
Molar Mass of Al2(SO4)3 = (2x27) + 3[32 + (16x4)]
= 54 + 3[32 + 64]
= 54 + 3[96] = 54 + 288 = 342g
Now, we can obtain the mass of aluminium sulphate formed by doing the following:
From the equation above:
294g of H2SO4 produced 342g of Al2(SO4)3.
Therefore, 250g of H2SO4 will produce = (250 x 342)/294 = 290.82g of Al(SO4)3
Therefore, 290.82g of aluminium sulphate (Al(SO4)3) is formed.
Answer:
1) Endothermic.
2) 
3) 
Explanation:
Hello there!
1) In this case, for these calorimetry problems, we can realize that since the temperature decreases the reaction is endothermic because it is absorbing heat from the solution, that is why the temperature goes from 22.00 °C to 16.0°C.
2) Now, for the total heat released by the reaction, we first need to assume that all of it is released by the solution since it is possible to assume that the calorimeter is perfectly isolated. In such a way, it is also valid to assume that the specific heat of the solution is 4.184 J/(g°C) as it is mostly water, therefore, the heat released by the reaction is:
3) Finally, since the enthalpy of reaction is calculated by dividing the heat released by the reaction over the moles of the solute, in this case NH4Cl, we proceed as follows:
Best regards!
Best regards!
The answer is 'equal'. Hydroxide ions are OH- and Hydrogen ions are H+. Have you noticed they're opposite charges? Positive + negative = neutral. That's all there is to it :)
Answer:
Explanation:
Water is a conductor if it has solutes/ions dissolved in it(tapwater for example)
but pure water is not a good conductor of electricity.
Nails and keychains(made of metal) are conductors of electricity.
Answer:
The standard potential, E cell, for this galvanic cell is 0.5670V
Explanation:
Ni²⁺(aq) + 2e⁻ → Ni(s) E red = - 0.23V ANODE
Cu²⁺(aq) + 2e- → Cu(s) E red = + 0.337V CATHODE
ΔE° = E cathode - E anode
ΔE° = 0.337V - (0.23V) = 0.5670 V
