Answer:
the second answer its science behind it
Answer:
b
Explanation:
1. What 2 subatomic particles have charges? List the particle name and its charge.
Answer: Proton - positive charge (+)
Neutron - neutral charge (0)
Electron - negative charge (-)
Explanation:
2 2 6 2 6 2 10 3
1s 2s 2p 3s 3p 4s 3d 4p
=
Answer:
ARSENIC
Explanation:
It has an atomic number of 33
Solid diarsenic trioxide reacts with fluorine gas (F2) to produce liquid arsenic pentafluoride and oxygen gas (O2). Write the Qc for this reaction.
Answer:
QC= [O2]^3/[F2]^10
Explanation:
In the laboratory you dissolve 18.7 g of copper(II) bromide in a volumetric flask and add water to a total volume of 375mL.
Required:
a. What is the molarity of the solution?
b. What is the concentration of the copper(II) cation?
c. What is the concentration of the acetate anion?
Answer:
a) - 0.2 M
b) - 0.2 M
c)- 0
Explanation:
The chemical formula of copper (II) bromide is CuBr₂. Its molar mass (MM) is calculated as follows:
MM(CuBr₂)= MM(Cu) + (2 x MM(Br) = 63.5 g/mol + (2 x 80 g/mol)= 223.5 g/mol
a). Molarity = moles CuBr₂/1 L solution
moles CuBr₂ = mass/MM = 18.7 g x 1 mol/223.5 g = 0.084 mol
Volume in L = 375 mL x 1 L/1000 mL = 0.375 L
M = 0.084 mol/(0.375 L) = 0.223 M ≅ 0.2 M
b). When is added to water, CuBr₂ dissociates into ions as follows:
CuBr₂ ⇒ Cu²⁺ + 2 Br⁻
We have 1 mol Cu²⁺ (copper (II) cation) per mol of CuBr₂. Thus, the concentration of copper (II) cation is:
0.2 mol CuBr₂ x 1 mol Cu²⁺/mol CuBr₂ = 0.2 M
c). The concentration of acetate anion is 0. There is no acetate anion in the solution (the anion from CuBr₂ is bromide Br⁻).
The molar mass of gallium (Ga) is 69.72 g/mol.
Calculate the number of atoms in a 27.2 mg sample of Ga.
Write your answer in scientific notation using three significant figures.
atoms Ga
Answer:
2.35 x 10²⁰ atoms Ga
Explanation:
After converting from mg to g, use the molar mass as the unit converter to convert to moles. Then using Avogadro's number, 6.022 x 10²³ convert from moles to atoms of Ga.
[tex]27.2mgGa*\frac{1g}{1000mg} *\frac{1 mol Ga}{69.72gGa} *\frac{6.022*10^2^3 atoms Ga}{1 molGa} = 2.349 * 10^2^0 atoms Ga[/tex]
Then round to 3 significant figures = 2.35 x 10²⁰ atoms Ga.
The number of atoms in 27.2 mg sample of Ga is 2.35 × 10²⁰ atoms
StoichiometryFrom the question, we are to calculate the number of atoms in a 27.2 mg sample of Ga.
First, we will determine the number of moles of Ga present
Using the formula,
[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass} [/tex]
Mass = 27.2 mg = 0.0272 g
Molar mass = 69.72 g/mol
Then,
[tex]Number\ of\ moles \ of\ Ga = \frac{0.0272}{69.72} [/tex]
[tex]Number\ of\ moles \ of\ Ga = [/tex] 0.000390132 moles
Now, for the number of atoms present
From the formula
Number of atoms = Number of moles × Avogadro's constant
Then,
Number of Ga atoms = 0.000390132 × 6.022×10²³
Number of Ga atoms = 2.35 × 10²⁰ atoms
Hence, the number of atoms in 27.2 mg sample of Ga is 2.35 × 10²⁰ atoms
Learn more on stoichiometry here: https://brainly.com/question/14464650