I hope this helped♡ I drew the realationship of variables
____NaNO3 + ___PbO --> ___Pb(NO3)2 + ___Na[2]O
To balace the eqaution, you need to have the same number of atoms for each element on both the reactant (left) and product (right) side.
To start off, you wanna know the number of atoms in each element on both sides, so take it apart:
[reactants] [product]
Na- 1 Na- 2
N- 1 N- 2(it's 2 because the the subscript [2] is outside of the parenthesis)
O- 4 O- 7 (same reason as above)
Pb- 1 Pb- 1
Na is not balanced out, so add a coefficient to make it the same on both sides.In this case, multiply by 2:
2NaNO3
Now Na is balanced, but the N and O are also effected by this, so they also have to be multiplied by 2 and they become:
Na- 2 Na- 2
N- 2 N- 2 (it balanced out)
O- 7 (coefficient times subscript, plus lone O) O- 7 (balanced out)
Pb was already balanced so no need to mess with it, just put a 1 where needed (it doesn't change anything).
Now to put it back together, it will look like this:
2NaNO3 + 1PbO --> 1Pb(NO3)2 + 1Na[2]O
Answer:
-255.4 kJ
Explanation:
The free energy of a reversible reaction can be calculated by:
ΔG = (ΔG° + RTlnQ)*n
Where R is the gas constant (8.314x10⁻³ kJ/mol.K), T is the temperature in K, n is the number of moles of the products (n =1), and Q is the reaction quotient, which is calculated based on the multiplication of partial pressures by the partial pressure of the products elevated by their coefficient divide by the multiplication of the partial pressure of the reactants elevated by their coefficients.
C₂H₂(g) + 2H₂(g) ⇄ C₂H₆(g)
Q = pC₂H₆/[pC₂H₂ * (pH₂)²]
Q = 0.261/[8.58*(3.06)²]
Q = 3.2487x10⁻³
ΔG = -241.2 + 8.314x10⁻³x298*ln(3.2487x10⁻³)
ΔG = -255.4 kJ
The answer <span>is <span>8.9 g/mL</span>.</span>
The density (D) is <span>equal to mass (m) divided by volume (V): D = m/V
Let's find the mass of the object:
m = 156 g - 105.5 g = 50.5 g
Let's find the volume of the volume:
V = 30.7 mL - 25 mL = 5.7 mL
The density is:
D = m/V = 50.5 g / 5.7 mL = 8.9 g/mL</span>
First. let's write the reaction formula: HBr +LiOH ----> LiBr + H₂O
let's get the moles of LiOH first
moles= Molarity x Liters
moles= 0.253 M x 0.01673 Liter= 0.00423 moles LiOH
using the balanced equation, you can see that 1 mol LiOH is equal to 1 mol HBr. so:
0.00423 mol LiOH = 0.00423 mol HBr
now let's find the concentration
molarity= mol/ Liters
0.00423 mol/ 0.01000 Liters= 0.423 M
