Answer:
4.90 M
Explanation:
In case of titration , the following formula can be used -
M₁V₁ = M₂V₂
where ,
M₁ = concentration of acid ,
V₁ = volume of acid ,
M₂ = concentration of base,
V₂ = volume of base .
from , the question ,
M₁ = ? M
V₁ = 125.0 mL
M₂ = 4.56 M
V₂ = 134.1 mL
Using the above formula , the molarity of acid , can be calculated as ,
M₁V₁ = M₂V₂
Substituting the respective values ,
M₁ * 125.0 mL = 4.56 M * 134.1 mL
M₁ = 4.90 M
Answer:
This addition reaction yields 3-BromoPentane and 2-BromoPentane.
Explanation: The reaction is an addition reaction that follows the Markonikoff's principle engaging the electrophillic addition mechnism with electrophile having no lone pair so rearrangement of carbonation is possible. It yields two possible products.
Answer:
3.5 × 10⁵ g of salt
Explanation:
<em>What is the mass (grams) of salt in 10.0 m³ of ocean water?</em>
We have this data:
- 1.000 mol salt is equal to 58.44 g salt
- 1.0 L of ocean water contains 0.60 mol of salt
We will need the following relations:
We can use proportions:
The H+ concentration in the lake has increased as a result of the acid rain.
The original pH of the lake was 7, which mean the water is neutral, but due to the acid rain, it drops to 5. This means, that the water has become acidic and how have more hydrogen ion. The H+ concentration in the lake has 100 times compares with its original pH.
To solve this question you need to calculate the number of the gas molecule. The calculation would be:
PV=nRT
n=PV/RT
n= 1 atm * 40 L/ (0.082 L atm mol-1K-<span>1 * 298.15K)
</span>n= 1.636 moles
The volume at bottom of the lake would be:
PV=nRT
V= nRT/P
V= (1.636 mol * 277.15K* 0.082 L atm mol-1K-1 )/ 11 atm= <span>3.38 L</span>
