Answer:
Lead(II) sulfate
Explanation:
This looks like a double displacement reaction, in which the cations change partners with the anions.
The possible products are
Pb(NO₃)₂ (aq)+ Na₂SO₄(aq) ⟶PbSO₄(?) + 2NaNO₃(?)
To predict the product, we must use the solubility rules. Two important ones for this question are:
- Salts containing Group 1 elements are soluble.
- Most sulfates are soluble, but PbSO₄ is an important exception.
Thus, NaNO₃ is soluble and PbSO₄ is the precipitate.
Answer is: -601,2 kJ/mol
Chemical reaction: Mg(OH)₂ → MgO + H₂O.
ΔHrxn = 37,5 kJ/mol.
ΔHf(Mg(OH)₂) = <span>−924,5 kJ/mol.
</span>ΔHf(H₂O) = <span>−285,8 kJ/mol.
</span>ΔHrxn -enthalpy of reaction.
ΔHf - enthalpy of formation.
<span>ΔHrxn=∑productsΔHf−∑reactantsΔHf.
</span>ΔHf(MgO) = -924,5 kJ/mol - (-285,8 kJ/mol) + 37,5 kj/mol.
ΔHf(MgO) = -601,2 kJ/mol.
Answer:
The particles in the neutral paper can shift, causing the paper to become polarized and attracted to the rod.
Explanation:
The neutral paper has an even distribution of its electrons throughout the paper. If a charged rod is brought near the neutral paper, this can cause the electrons in the paper to shift. If the rod is negative, the electrons will be repelled from the rod and cause the molecules in the paper to have a slight positive charge on the part of the paper closest to the rod. If the rod is positive, the electrons will be attracted to the rod and cause a slight negative charge on the side of the paper closest to the rod.