The volume of a gas is the same as its CONTAINER.
Gases generally has no shape and no definite volume. When a gas is placed in a container, the gas usually takes the shape and the volume of the container, that is, the gas fills up all the available spaces in the container. Thus, the volume of a gas will always be the same as its container. This is in contrast with solids, which have definite shape and volume and liquids, which have definite volume but no fixed shape.
1A: The legs can be a adjusted, as well as the sand can be swapped out. It’s a very good design for running multiple tests.
1B: He could add books or something under the front or back legs in order to increase/decrease the incline, therefore imitating the hypothesis.
1C: He can change out the sand grains to finer ones, or coarser ones, and record his results of each test.
2: If he sets the model at a steep incline and tests it with coarse sand and fine sand, seeing which one makes a narrower, deeper hole.
Answer:
∆H° rxn = - 93 kJ
Explanation:
Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds broken minus bonds formed (H according to Hess Law.
We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.
N₂ (g) + 3H₂ (g) ⇒ 2NH₃ (g)
1 N≡N = 1(945 kJ/mol) 3 H-H = 3 (432 kJ/mol) 6 N-H = 6 ( 389 kJ/mol)
∆H° rxn = ∑ H bonds broken - ∑ H bonds formed
∆H° rxn = [ 1(945 kJ) + 3 (432 kJ) ] - [ 6 (389 k J]
∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ
be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond, N≡N, we have to use this one .
Answer:
1.811 g
Explanation:
The computation of the mass need to use to make the solution is shown below:
We know that molarity is
So,
= 0.031 moles
Now
where,
The Molecular weight of NaCl is 58.44 g/mole
And, the moles are 0.031 moles
So, the mass of NaCL is
= 1.811 g
We simply applied the above formulas
