Answer:
n = 3.0 moles
V = 60.0 L
T = 400 K
From PV = nRT, you can find P
P = nRT/V = (3.0 mol)(0.0821 L-atm/K-mol)(400 K)/60.0L
P = 1.642 atm = 1.6 atm (to 2 significant figures)
Explanation:
Answer:
How many grams of K2Cr2O7, are soluble in 100 g of water at 95 ºC? Solubility Curve DRAFT. 10th - 12th grade. 3326 times. Chemistry. 63% average accuracy. 3 years ago ... When 20 grams of potassium chlorate, KClO 3, is dissolved in 100 grams of water at 80 ºC, the solution can be correctly described as: answer choices . supersaturated. saturated.
Explanation:
ABFO scales, calculator, string, tape, laser pointer, permanent markers, and a protractor
Answer 1)

+

----> 2 HCl
This is a pure
chemical reaction that is taking place.
In this reaction the two chemical species which are Chlorine and Hydrogen undergo addition reaction and form two moles of hydrochloric acid. No new species is formed except the combination of reactants which gives the product.
Answer 2) H + H ----> He + n
This is a nuclear reaction.
As the reactant species is undergoing a reaction which is resulting in formation of new species completely different than the elements in reactants.
Also there is a neutron which is generated after the reaction this is clear indication of nuclear fusion reaction. Hence, it can be called as nuclear reaction.
Answer:
B. - 210 kJ
Explanation:
<em>∵ ΔHrxn = ∑(bond energies)products - ∑(bond energies)reactants.</em>
- The bond formation in the products releases energy (exothermic).
- The bond breaking in the reactants requires energy (endothermic).
The products:
- H₂O contains 2 O-H (- 459 kJ/mol) bonds.
- O₂ contain 1 O=O (- 494 kJ/mol) bond.
The reactants:
- H₂O₂ contain 2 O–H (459 kJ/mol) bonds and 1 O–O (142 kJ/mol) bond.
∵ ΔHrxn = ∑(bond energies)products - ∑(bond energies)reactants.
<em>∴ ΔHrxn = [2 (2 x (O–H bond energy) + (1 x (O=O bond energy)] - 2 [(2 x (O–H bond energy) + (1 x (O–O bond energy)] </em>= [2 (2 x - 459 kJ/mol) + (1 x - 494 kJ/mol)] - 2 [(2 x 459 kJ/mol) + (1 x 142 kJ/mol)] = (- 2330 kJ) + (2120 kJ) = <em>- 210 kJ.</em>