The answer is (4) Add enough solvent to 30.0 g of solute to make 1.0 L solution. The molarity is calculated using volume of the solution. When solute dissolving, the total volume will change. So the final volume of solution need to be 1.0 L.
Answer:
The hydroxide ions decrease.
Explanation:
I got it right on the quiz. This is what I saw. Read this, "Adding water to an acid or base will change its pH. Water is mostly water molecules so adding water to an acid or base reduces the concentration of ions in the solution. When an acidic solution is diluted with water the concentration of H + ions decreases and the pH of the solution increases towards 7."
Hope this helps! Tell me if this is wrong just incase.
Answer:C.He should locate the chloroform stored in a dark container in chemical storage and should take it to the fume hood to pour.
Explanation:
The statement of the question clearly states that chloroform is sensitive to light and it's vapour is toxic.
If a substance is sensitive to light, then it must be stored in a dark bottle. This is because. If a substance that is sensitive to light is stored in a transparent container, it may be decomposed by light.
Being a substance whose fumes are toxic, Malik should pour the liquid in a fumes hood so that he does not inhale the fumes.
A. N₂ (g) + 3 H₂ (g) --> 2 NH₃ (g)
B. The value for standard enthalpy of formation is empirical given that the reactants involved were pure elements. So, you can search this on the internet or in any textbook. The Hf for NH₃ is -46.0 kJ/mol.
C. C (s) + O₂ (g) --> CO₂ (g)
D. The Hf for CO₂ is <span>-393.5 kJ/mol
E. 4 Fe (s) + 3 O</span>₂ (g) --> 2 Fe₂O₃ (s)
F. The Hf for solid Fe₂O₃ is -826.0 kJ/mol.
G. C (s) + 2 H₂ (g) --> CH₄ (g)
H. The Hf for methane gas is -74.9 kJ/mol.
Answer:
A 3s orbital is at a greater average distance from the nucleus than a 2s orbital
Explanation:
As the principal quantum number n increases, the distance of the orbital from the nucleus increases. Hence if we consider the 2s and 3s orbitals, it is easy to see that the 3s orbital is at a greater distance from the nucleus than the 2s orbitals.
This is clearly seen when we plot the radial distribution against the distance from the nucleus. This enables us to visualize the region in space in which an electron may be found.
