Answer : The volume of 
produced at standard conditions of temperature and pressure is 0.2422 L
Explanation :
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
where,
= initial pressure of 
gas = (740-22.4) torr = 717.6 torr
= final pressure of gas at STP= 760 torr
= initial volume of gas = 280 mL
= final volume of gas at STP = ?
= initial temperature of gas = 
= final temperature of gas = 
Now put all the given values in the above equation, we get:
Therefore, the volume of produced at standard conditions of temperature and pressure is 0.2422 L
The organism that would have the most variation in the DNA of its offspring is the cat (Option C). Meiosis is a type of cell division that generates more genetic variability than asexual types of reproduction.
Meiosis is a type of reductional cell division by which a parental cell produces 4 daughter cells (gametes), each containing half of the genetic material.
Animals (e.g., cats) generate gametes by meiosis which fuse during fertilization to produce new offspring.
Both amoeba and bacteria reproduce by a type of asexual reproduction called binary fission. Moreover, yeasts also reproduce asexually by a process called budding and fission.
Both asexual and sexual types of reproduction generate genetic variability by the emergence of new mutations in daughter cells.
Meiosis generates much more genetic variability than asexual types of reproduction due to two different processes:
- Random assortment of chromosomes, which produces new allele combinations.
- Recombination, i.e., by the exchange of genetic material (DNA) between non-sister chromatids during Prophase I.
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In finding the molarity of a solution, we use the following formula:
What is Molarity?
The number of moles of the solute is calculated by dividing the mass of the solute by its molar mass.
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The molar mass of NH4NO3 and (NH4)3PO4 are 80.043 g/mol and 149.0867 g/mol, respectively.
![[NH+4]=0.1596 mol20.0 L=7.98×10−3 M NH+4](https://tex.z-dn.net/?f=%5BNH%2B4%5D%3D0.1596%20mol20.0%20L%3D7.98%C3%9710%E2%88%923%20M%20NH%2B4)
![[PO3−4]=0.0296 mol20.0 L=1.48×10−3 M PO3−4](https://tex.z-dn.net/?f=%5BPO3%E2%88%924%5D%3D0.0296%20mol20.0%20L%3D1.48%C3%9710%E2%88%923%20M%20PO3%E2%88%924)
Therefore, ![[PO3−4]\\](https://tex.z-dn.net/?f=%5BPO3%E2%88%924%5D%5C%5C)
has a molarity of 
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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
Greater rainfall has an greater increase on the rate of chemical weathering. Rain is a form of precipitation.
