Answer:
ΔrxnH = -580.5 kJ
Explanation:
To solve this question we are going to help ourselves with Hess´s law.
Basically the strategy here is to work in an algebraic way with the three first reactions so as to reprduce the desired equation when we add them together, paying particular attention to place the reactants and products in the order that they are in the desired equation.
Notice that in the 3rd reaction we have 2 mol Na₂O (s) which is a reactant but with a coefficient of one, so we will multiply this equation by 1/2-
The 2nd equation has Na₂SO₄ as a reactant and it is a product in our required equation, therefore we will reverse the 2nd . Note the coefficient is 1 so we do not need to multiply.
This leads to the first equation and since we need to cancel 2 NaOH, we will nedd to multiply by 2 the first one.
Taking 1/2 eq 3 + (-) eq 2 + 2 eq 1 should do it.
Na₂O (s) + H₂ (g) ⇒ 2 Na (s) + H₂O(l) ΔrxnHº = 259 / 2 kJ 1/2 eq3
+ 2NaOH(s) + SO₃(g) ⇒ Na₂SO₄ (s) + H₂O (l) ΔrxnHº = -418 kJ - eq 2
+ 2Na (s) + 2 H₂O (l) ⇒ 2 NaOH (s) + H₂ (g) ΔrxnHº = -146 x 2 2 eq 1
<u> </u>
Na₂O (s) + SO₃ (g) ⇒ Na₂SO₄ (s) ΔrxnHº = 259/2 + (-418) + (-146) x 2 kJ
ΔrxnH = -580.5 kJ
Answer:
50.3mL of mercury are in 1.50lb
Explanation:
Punds are an unit of mass. To convert mass to volume we must use density (13.546g/mL). Now, As you can see, density is in grams but the mass of mercury is in pounds. That means we need first, to convert pounds to grams to use density and obtain volume of mercury.
<em>Mass mercury in grams:</em>
1.50lb * (1kg / 2.20lb) = 0.682kg = 682g of mercury.
<em>Volume of mercury:</em>
682g Mercury * (1mL / 13.546g) =
<h3>50.3mL of mercury are in 1.50lb</h3>
<u>Answer:</u> The number of moles of strontium bicarbonate is 
<u>Explanation:</u>
Formula units are defined as lowest whole number ratio of ions in an ionic compound. It is calculate by multiplying the number of moles by Avogadro's number which is 
We are given:
Number of formula units of 
As, number of formula units are contained in 1 mole of a substance.
So, 
number of formula units will be contained in = 
of strontium bicarbonate.
Hence, the number of moles of strontium bicarbonate is
