Answer:
22.4L of one mole of any gas
or you can use PV=nRT
3.45*22.4=77.28
Explanation:
The grams of NaCl that are required to make 150.0 ml of a 5.000 M solution is 43.875 g
calculation
Step 1:calculate the number of moles
moles = molarity x volume in L
volume = 150 ml / 1000 = 0.15 L
= 0.15 L x 5.000 M = 0.75 moles
Step 2: calculate mass
mass = moles x molar mass
molar mass of NaCl = 23 + 35.5 = 58.5 mol /L
mass is therefore =0.75 moles x 58.5 mol /l =43.875 g
Answer:
Mass = 18.9 g
Explanation:
Given data:
Mass of Al₂O₃ formed = ?
Mass of Al = 10.0 g
Solution:
Chemical equation:
4Al + 3O₂ → 2Al₂O₃
Number of moles of Al:
Number of moles = mass/molar mass
Number of moles = 10.0 g/ 27 g/mol
Number of moles = 0.37 mol
Now we will compare the moles of Al and Al₂O₃.
Al : Al₂O₃
4 : 2
0.37 : 2/4×0.37 = 0.185 mol
Mass of Al₂O₃:
Mass = number of moles × molar mass
Mass = 0.185 mol × 101.9 g/mol
Mass = 18.9 g
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:
Water
Explanation :-
Higher the intermolecular forces between the liquid particles, higher its boiling point.
