<span>283.89 g/mol is the molar mass of tetraphosphorus decoxide</span>
Answer:
The new pressure is 0.5 atm
Explanation:
Step 1: Data given
Volume of oxygen = 300 mL = 0.300 L
Pressure = 1.00 atm
Temperature = 300 K
The volume increases to 1000mL = 1.00 L
The temperature increases to 500 K
Step 2: Calculate the new pressure
(P1*V1)/T1 = (P2*V2)/T2
⇒with P1 = the initial pressure = 1.00 atm
⇒with V1 = the initial volume = 0.300 L
⇒with T1 = the initial temperature = 300 K
⇒with P2 = the new pressure = TO BE DETERMINED
⇒with V2 = the increased volume = 1.00 L
⇒with T2 = the increased temperature = 500 K
(1.00 atm* 0.300 L)/300 K = (P2 * 1.00L) / 500 K
P2 = (1.00 *0.300 * 500) / (300 *1.00)
P2 = 0.5 atm
The new pressure is 0.5 atm
Answer:
Scandium
Titanium
Vanadium
Chromium
Manganese
Iron
Cobalt
Nickel
Copper
Zinc
Yttrium
Zirconium
Niobium
Molybdenum
Technetium
Ruthenium
Rhodium
Palladium
Silver
Cadmium
Lanthanum
Hafnium
Tantalum
Tungsten
Rhenium
Osmium
Iridium
Platinum
Gold
Mercury
Actinium
Rutherfordium
Dubnium
Seaborgium
Bohrium
Hassium
Meitnerium
Darmstadtium
Roentgenium
Copernicium
Explanation:
all of those are transition metals lol
1) Answer is: c) The reaction will proceed right.
Balanced chemical reaction: N₂(g) + 3H₂(g) ⇄ 2NH₃(g) ΔH = +92 kJ.
Reducing the volume of the system increase the partial pressures of the products and reactants.
With a pressure increase due to a decrease in volume, the side of the equilibrium with fewer moles is more favorable, there are 4 moles at the left side (three moles of hydrogen and one mole of nitrogen) and 2 moles (ammonia) at the right side of the reaction.
2) Answer is: d) The partial pressure of ammonia will increase.
This reaction is endothermic (enthalpy is higher than zero), which means that heat is added.
According to Le Chatelier's principle when the reaction is endothermic heat is included as a reactant and when the temperature increased, the heat of the system increase, so the system consume some of that heat by shifting the equilibrium to the right, producing more ammonia.