Answer:
Minimum volume of H₂SO₄ required for H₂SO₄ to be in excess = 0.0556 mL
Explanation:
Pb(NO₃)₂ + H₂SO₄ -----> PbSO₄ + 2HNO₃
For this reaction, we know that the max concentration of Pb(NO₃) according to the bottle is 0.999M and to ensure the other reactant in the reaction is in excess, we'll do the calculation with a Pb(NO₃) that's a bit higher, that is, 1.0M.
Knowing that Concentration in mol/L = (number of moles)/(volume in L)
Number of moles of Pb(NO₃) added = concentration in mol/L × volume in L = 1 × 0.001 = 0.001 mole
According to the reaction,
1 mole of Pb(NO₃) reacts with 1 mole of H₂SO₄
0.001 mole of Pb(NO₃) will react with 0.001×1/1 mole of H₂SO₄
Therefore number of H₂SO₄ required for the reaction and for the H₂SO₄ to be in excess is 0.001 mole of H₂SO₄
So, the concentration of commercial H₂SO₄ is usually 18.0M, using this as the assumed value.
Volume of H₂SO₄ = (number of H₂SO₄ required for it to be in excess)/(concentration of H₂SO₄)
Volume of H₂SO₄ = 0.001/18 = 0.0000556 L = 0.0556 mL.
QED!!!
FeSO₄*7H₂O(s) = FeSO₄(s) + 7H₂O(g)
M(FeSO₄*7H₂O)=278.0 g/mol
M(FeSO₄)=151.9 g/mol
m(FeSO₄*7H₂O)/M(FeSO₄*7H₂O)=m(FeSO₄)/M(FeSO₄)
m(FeSO₄)=M(FeSO₄)m(FeSO₄*7H₂O)/M(FeSO₄*7H₂O)
m(FeSO₄)=151.9*100.0/278.0=54.6 g
m(FeSO₄)=54.6 g
Using the combined gas law, where PV/T = constant, we first solve for PV/T for the initial conditions: (4.50 atm)(36.0 mL)/(10.0 + 273.15 K) = 0.57213.
Remember to use absolute temperature.
For the final conditions: (3.50 atm)(85.0 mL)/T = 297.5/T
Since these must equal, 0.57213 = 297.5/T
T = 519.98 K
Subtracting 273.15 gives 246.83 degC.
Answer:
Option D is correct.
H₂O + CO₂ → H₂CO₃
Explanation:
First of all we will get to know what law of conservation of mass states.
According to this law, mass can neither be created nor destroyed in a chemical equation.
This law was given by French chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.
Example:
6CO₂ + 6H₂O + energy → C₆H₁₂O₆ + 6O₂
there are six carbon atoms, eighteen oxygen atoms and twelve hydrogen atoms on the both side of equation so this reaction followed the law of conservation of mass.
Now we will apply this law to given chemical equations:
A) H₂ + O₂ → H₂O
There are two hydrogen and two oxygen atoms present on left side while on right side only one oxygen and two hydrogen atoms are present so mass in not conserved. This equation not follow the law of conservation of mass.
B) Mg + HCl → H₂ + MgCl₂
In this equation one Mg, one H and one Cl atoms are present on left side while on right side two hydrogen, one Mg and two chlorine atoms are present. This equation also not follow the law of conservation of mass.
C) KClO₃ → KCl + O₂
There are one K, one Cl and three O atoms are present on left side of chemical equation while on right side one K one Cl and two oxygen atoms are present. This equation also not following the law of conservation of mass.
D) H₂O + CO₂ → H₂CO₃
There are two hydrogen, one carbon and three oxygen atoms are present on both side of equation thus, mass remain conserved. Thus is correct option.
Answer: 178.9 g
Explanation:
Density = 
find volume of the cube: (5.80 cm) (5.80 cm) (5.80cm) = 195.112 cm³
1.0 cm³ = 1.0 mL
so 195.112 cm³ = 195.112 mL
plug value into density equation:
0.917 g/mL = (mass) / (195.112 mL)
and solve for mass!
