Answer:
The mass of PbSO4 formed 15.163 gram
Explanation:
mole of Pb(NO₃)₂ = 1.25 x 0.05 = 0.0625
mole of Na₂SO₄ = 2 x 0.025 = 0.05
Pb(NO₃)₂ + Na₂SO₄ → PbSO₄ + 2 NaNO₃
( Mole/Stoichiometry ) 
= 0.0625 = 0.05
From (Mole/ Stoichiometry ) we can conclude that Na₂SO₄ is limiting reagent.
Mass of PbSO₄ precipitate = 0.05 x Molecular mass of PbSO₄
= 0.05 x 303.26 g
= 15.163 g
Answer:
Rb
Alkali Metals are Group 1 so
Rb it isnt Hydrogen because it is a gas
Explanation:
Answer:
C) hydrogen bonding
Explanation:
All atoms and molecules have London Dispersion Forces between them, but they are usually overshadowed but the much stronger forces. In this scenario the major attractive force in HF molecules are hydrogen bonds. Hydrogen bonds are electrostatic forces of attraction found when Hydrogen is bonded to a more electronegative atom such as Oxygen, Chlorine and Fluorine.
Answer: 94.13 L
Explanation: In STP in an ideal gas there is a standard value for both temperature and pressure. At STP,pressure is equal to 1atm and the temperature at 0°C is equal to 273.15K. This problem is an ideal gas so we use PV=nRT where R is a constant R= 0.08205 L.atm/mol.K.
To find volume, derive the equation, it becomes V=nRT/P. Substitute the values. V= 4.20 mol( 0.08205L.atm/mol.K)(273.15K) / 1 atm = 94.13 L. The mole units, atm and K will be cancelled out and L will be the remaining unit which is for volume.
Answer:
balanced in ACID not BASE
Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O
Answer
Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O
Explanation:
Cr2O7^2-(aq) + Hg(l) ----> Hg^2+(aqH) + Cr^3+(aq)
add H^1+ (acid) to capture the O and make 7 water molecules
Cr2O7^2-(aq) + Hg(l) + H^1+ ----> Hg^2+(aqH) + Cr^3+(aq) + 7H2O
Cr goes from +6 to +3 by gaining 3 e
Hg goes from 0 to +2 by losing 2 e
we need 3 Hg for every 2 Cr
so
Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O
2 Cr on the right and left
Net 12 positive charges on the right and the left
3 Hg on the right and left
14 H on the right and left
the equation is balanced
we cannot balance the equation in a basic solution with OH^1-
we have plenty of O in the dichromate ion. we need to convert it to water which take free H^1+ from the acid
