the branch of science that deals with the identification of the substances of which matter is composed; the investigation of their properties and the ways in which they interact, combine, and change; and the use of these processes to form new substances.the branch of science that deals with the identification of the substances of which matter is composed; the investigation of their properties and the ways in which they interact, combine, and change; and the use of these processes to form new substances.
Answer:
Bin 1 points to a carbon bonded to a double bonded carbon and single bonded to two hydrogens. --- trigonal planar, tetrahedral
Bin 2 points to a carbon double bonded to a carbon and single bonded to a carbon and one hydrogen.------- trigonal planar, tetrahedral
Bin 3 is a carbon single bonded to two carbons and single bonded to two hydrogens. ----- tetrahedral, tetrahedral
Bin 4 is the same as bin 3.--------tetrahedral, tetrahedral
Bin 5 is a carbon triple bonded to a carbon and single bonded to a carbon.---- linear, tetrahedral
Bin 6 is triple bonded to a carbon and single bonded to a hydrogen.---linear, tetrahedral
Explanation:
A single C-C or C-H bond is in a tetrahedral geometry, the carbon atom is bonded to four species with a bond angle of 109°.
A C=C bond is trigonal planar with a bond angle of 120°.
Lastly, a C≡C bond has a linear geometry with a bond angle of 180° between the atoms of the bond.
Answer:
[NH₃] = 14.7 mol/L
Explanation:
28 wt% is a type of concentration that indicates that 28 g of ammonia is contained in 100 g of solution.
Let's determine the amount of ammonia:
28 g . 1 mol / 17.03g = 1.64 moles of NH₃
You need to consider that, when you have density's data it is always referred to solution:
Mass of solution is 100 g, let's find out the volume
0.90 g/mL = 100 g /V
V = 100 g / 0.90mL/g → 111.1 mL
We convert the volume to L → 111.1 mL . 1 L/1000mL = 0.1111 L
mol/L = 1.64 mol/0.1111L → 14.7 M
mol/L = M → molarity a sort of concentration that indicates the moles of solute in 1L of solution
Answer: Edge length of the unit cell = 628pm
Explanation: For a body centred cubic structured system, the relationship between the edge length of the unit cell and radius of the atoms in the structure is
Edge length of Unit cell (a) = (4R)/(√3)
R = 272pm = (272 × (10^-12))m = (2.72 × (10^-10))m
a = (4 × (2.72 × (10^-10)))/(√3)
a = (6.28157 × (10^-10))m = 628pm
I think it’s 44.6 J, but I’m not to sure so hoped this helped /:).
