Hello!
At
Standard Pressure and Temperature, an ideal gas has a molar density of
0,04464 mol/L.So, we need to apply a simple conversion factor to calculate the density of Sulfur Dioxide using the molar mass of Sulfur Dioxide.
So, the Density of Sulfur Dioxide (SO₂) at STP is
2,8599 g/LHave a nice day!
There are several information's already given in the question. Based on those information's, the answer can be easily deduced.
Amount of gasoline required by Harry's car to travel 25 miles = 1 gallon
Then
amount of gasoline required
by Harry's car to travel 15000 miles = 15000/25
= 600 gallons
So
Amount of CO2 released by burning 1 gallon of gasoline = 20 pounds
Then
Amount of CO2 released
by burning 600 gallon of gasoline = 600 * 20
= 12000 pounds
From the above deduction, it can be concluded that the amount of CO2 that will be added by Harry's car to the atmosphere is 12000 pounds.
Answer:
Sodium hydroxide is a highly caustic base and alkali that decomposes proteins at ordinary ambient temperatures and may cause severe chemical burns. It is highly soluble in water, and readily absorbs moisture and carbon dioxide from the air. It forms a series of hydrates NaOH·nH
2O.[11] The monohydrate NaOH·H
2O crystallizes from water solutions between 12.3 and 61.8 °C. The commercially available "sodium hydroxide" is often this monohydrate, and published data may refer to it instead of the anhydrous compound.
As one of the simplest hydroxides, sodium hydroxide is frequently utilized alongside neutral water and acidic hydrochloric acid to demonstrate the pH scale to chemistry students.[12]
Sodium hydroxide is used in many industries: in the manufacture of pulp and paper, textiles, drinking water, soaps and detergents, and as a drain cleaner. Worldwide production in 2004 was approximately 60 million tons, while demand was 51 million tons.[13]
Answer is: <span>the molarity of the diluted solution 0,454 M.
</span>V₁(NaOH) = 100 mL ÷ 1000 mL/L = 0,1 L.
c₁(NaOH) = 0,75 M = 0,75 mol/L.
n₁(NaOH) = c₁(NaOH) · V₁(NaOH).
n₁(NaOH) = 0,75 mol/L · 0,1 L.
n₁(NaOH) = 0,075 mol
n₂(NaOH) = n₁(NaOH) = 0,075 mol.
V₂(NaOH) = 165 mL ÷ 1000 mL/L = 0,165 L.
c₂(NaOH) = n₂(NaOH) ÷ V₂(NaOH).
c₂(NaOH) = 0,075 mol ÷ 0,165 L.
c₂(NaOH) = 0,454 mol/L.
