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:
A chemical equation is balanced when the number of each kind of atom is the same on both sides of the reaction.
Explanation:
The law of conservation of matter (except in nuclear reactions) indicates that atoms can neither be created or destroyed.
The number of atoms that are in the reactants must be the same as the number of the atoms that are in the product.
The number and types of molecules can (and will) change. The atoms that make up the molecules are rearranged but the number and kinds of atoms stay the same.
<span>To find the mass of 3.00 moles of magnesium chloride (MgCl2), first record the atomic mass of magnesium (Mg) and chloride (Cl), which are both listed on the periodic table as follows:
Mg=24 g/mole
Cl=38 g/mole
Now, double the Cl mass since there are 2 Cl moles in MgCl2 and then add it to the Mg mass like so:
(38 g/mole*2 moles)+24 g/mole=100 g/mole
Finally, to calculate the mass of 3.00 moles of MgCl2, convert the combined atomic mass to grams as follows:
3.00 moles * 100 g/mole = 300 g</span>
Answer:
b) C = 0.50 J/(g°C)
Explanation:
∴ Q = 50 J
∴ m = 10.0 g
∴ ΔT = 35 - 25 = 10 °C
specific heat (C) :
⇒ C = Q / mΔT
⇒ C = 50 J / (10.0 g)(10 °C)
⇒ C = 0.50 J/(g°C)
Orbitals am only hold two electrons each, so 3 orbitals can hold 6 electrons
