Answer:
The symbol (l) stands for liquid phase.
Explanation:
Answer:
The answer to your question is 16 g
Explanation:
Data
Percent by mass = 8%
Mass of the solution = 200 g
Mass of solute = ?
Formula
Percent by mass = mass of solute / mass of solution x 100
- Solve for mass of solute
Mass of solute = Percent by mass x mass of solution / 100
- Substitution
Mass of solute = 8 x 200 / 100
- Simplification
Mass of solute = 1600 / 100
- Result
Mass of solute = 16 g
Answer:
C) Covalent bonds come about because of a sharing of electrons; ionic bonds do not.
Explanation:
There are two main types of chemical bonds- covalent and ionic/electrovalent bonds.
Ionic bond: Ionic or electrovalent bonds are characterized by the transfer of electrons from electropositive atoms (metals) to electronegative atoms (non-metals). The metal atoms after donating their electrons become positively charged ions (cations) while the non-metal atoms after accepting electrons become negatively charged ions (anions). Strong electrostatic forces of attraction constitutes ionic bonds.
Covalent bond: Covalent bonds are formed by the sharing of electrons by the atoms involved in the bond; usually between atoms of comparable electronegativities or atoms of the same element. The shared electrons are contributed by each of the atoms involved in the bonding or may be contributed by only one of the atoms. In covalent bonding, molecules rather than ions are formed.
Answer:
It cost Evan $17.70 to send 177 text messages. How many text messages did he send if he spent $19.10?
Explanation:
I cant do this
Answer:
This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M
Explanation:
Step 1: Data given
Molarity of Na2CrO4 = 0.010 M
Molarity of NaBr = 2.5 M
Ksp(PbCrO4) = 1.8 * 10^–14
Ksp(PbBr2) = 6.3 * 10^–6
Step 2: The balanced equation
PbCrO4 →Pb^2+ + CrO4^2-
PbBr2 → Pb^2+ + 2Br-
Step 3: Define Ksp
Ksp PbCrO4 = [Pb^2+]*[CrO4^2-]
1.8*10^-14 = [Pb^2+] * 0.010 M
[Pb^2+] = 1.8*10^-14 /0.010
[Pb^2+] = 1.8*10^-12 M
The minimum [Pb^2+] needed to precipitate PbCrO4 is 1.8*10^-12 M
Ksp PbBr2 = [Pb^2+][Br-]²
6.3 * 10^–6 = [Pb^2+] (2.5)²
[Pb^2+] = 1*10^-6 M
The minimum [Pb^2+] needed to precipitate PbBr2 is 1*10^-6 M
This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M