Answer:
Orbitals
Shells
Energy Levels
Explanation:
Elements in the same period will have the same energy levels which is the same as orbitals and shells.
- The period of an element indicates the energy level in an atom.
- Elements in period 1 will have one energy level or shell or orbitals.
- Down a group this is not the case.
- The energy level increases from top to bottom.
- As we go down the group, we are transiting from one period to another.
Answer: Product
Explanation:
A Chemical reaction shows how substances, written in symbols known as reactants, are converted to different substances called products. An arrow from the left hand side (reactants) points to right hand side ( products )shows the direction of flow of reaction.
In the equation. C + O2 --------> CO2
reactant------> product
CO2 is the product
Answer:
sulfur
Explanation:
In oxygen family sulfur has yellow color and also having stinky smell. Thus given statements are about sulfur.
It is present in oxygen family.
It has six valance electrons.
Its atomic number is 16.
Its atomic weight is 32 amu.
The electronic configuration of sulfur is given below,
S₁₆ = 1s² 2s² 2p⁶ 3s² 3p⁴
We can see the valance shell is third shell and it have six electrons thus sulfur have six valance electrons. (3s² 3p⁴ )
Sulfur is used in vulcanisation process.
It is used in bleach and also as a preservative for many food.
it is used to making gun powder.
Answer:
1. V₁ = 2.0 mL
2. V₁ = 2.5 mL
Explanation:
<em>You are provided with a stock solution with a concentration of 1.0 × 10⁻⁵ M. You will be using this to make two standard solutions via serial dilution.</em>
To calculate the volume required (V₁) in each dilution we will use the dilution rule.
C₁ . V₁ = C₂ . V₂
where,
C are the concentrations
V are the volumes
1 refers to the initial state
2 refers to the final state
<em>1. Perform calculations to determine the volume of the 1.0 × 10⁻⁵ M stock solution needed to prepare 10.0 mL of a 2.0 × 10⁻⁶ M solution.</em>
C₁ . V₁ = C₂ . V₂
(1.0 × 10⁻⁵ M) . V₁ = (2.0 × 10⁻⁶ M) . 10.0 mL
V₁ = 2.0 mL
<em>2. Perform calculations to determine the volume of the 2.0 × 10⁻⁶ M solution needed to prepare 10.0 mL of a 5.0 × 10⁻⁷ M solution.</em>
C₁ . V₁ = C₂ . V₂
(2.0 × 10⁻⁶ M) . V₁ = (5.0 × 10⁻⁷ M) . 10.0 mL
V₁ = 2.5 mL
V ( H2SO4) = 35 mL / 1000 => 0.035 L
M ( H2SO4) = ?
V ( NaOH ) = 25 mL / 1000 => 0.025 L
M ( NaOH ) = 0.320 M
number of moles NaOH:
n = M x V
n = 0.025 x 0.320 => 0.008 moles of NaOH
Mole ratio:
<span>2 NaOH + H2SO4 = Na2SO4 + 2 H2O
</span>
2 moles NaOH ---------------------- 1 mole H2SO4
0.008 moles moles NaOH ---------- ??
0.008 x 1 / 2 => 0.004 moles of H2SO4 :
Therefore:
M ( H2SO4) = n / V
M = 0.004 / 0.035
= 0.114 M
hope this helps!
