Answer and Explanation:
The reaction is in the gas phase, so the equilibrium constant is expressed in terms of the partial pressures (P) of the products and reactants, as follows:
We have the following data:
P(SO₃) = 2.6 atm
P(O₂) = 0.43 atm
We need Kp for this reaction. We can assume that in Appendix 4 we found that Kp = 7 x 10²⁴.
Then, we introduce the data in the equilibrium constant expression to calculate the partial pressure f SO₂ (PSO₂), as follows:
Therefore, the partial pressure of SO₂ is 1.5 x 10⁻¹² atm (for the given Kp).
Answer:
Bar graphs have an x-axis and a y-axis like most bar graphs, like the one above, the x-axis runs horizontally (flat).
Sometimes bar graphs are made so that the bars are sidewise like in the graph below. Then the x-axis has numbers representing different time periods or names of things being compared.
have a good day and be safe
-Hops
Answer:
3 g/cm³
Explanation:
Density= Mass divided by Volume
45÷15= 3
The balanced chemical reaction is written as:
<span>Zn + 2AgNO3 = Zn(NO3)2 + 2Ag
To determine the grams of silver metal that is being produced, it is important to first determine which is the limiting reactant and the excess reactant from the given initial amounts. We do as follows:
4.35 g Zn ( 1 mol / 65.38 g ) ( 2 mol AgNO3 / 1 mol Zn ) = 0.1331 mol AgNO3 needed
35.8 g AgNO3 ( 1 mol / 169.87 g ) ( 1 mol Zn / 2 mol AgNO3 ) = 0.1054 mol Zn needed
Therefore, the limiting reactant would be the zinc metal since it would be consumed completely in the reaction. The excess amount of AgNO3 would be:
0.2107 mol AgNO3 - 0.1331 mol AgNO3 = 0.0776 mol AgNO3 left ( 169.87 g / 1 mol ) = 13.19 g AgNO3 left
0.0665 mol Zn ( 2 mol Ag / 1 mol Zn) ( 107.9 g / 1 mol) = 14.3581 g Ag produced</span>
V1 = 5L = 5000ml V2 = ?
T1 = 25°C = 25+273 T2 = -120°C
= 297K = -120+273
= 153K
By Charles Law,
V1/T1 = V2/T2
5000/297 = V2/153
By solving,
V2 = 2.576L
