Sulfur trioxide (SO3) is a chemical compound that is a significant pollutant in gaseous form as it is involved in the production of acid rain.
Industrially, sulfur trioxide is an important precursor to sulfuric acid and is formed from the reaction between sulfur dioxide (SO2) and oxygen gas (O2) as shown in the chemical equation below.
Answer:
1.4 × 10² mL
Explanation:
There is some info missing. I looked at the question online.
<em>The air in a cylinder with a piston has a volume of 215 mL and a pressure of 625 mmHg. If the pressure inside the cylinder increases to 1.3 atm, what is the final volume, in milliliters, of the cylinder?</em>
Step 1: Given data
- Initial volume (V₁): 215 mL
- Initial pressure (P₁): 625 mmHg
- Final pressure (P₂): 1.3 atm
Step 2: Convert 625 mmHg to atm
We will use the conversion factor 1 atm = 760 mmHg.
625 mmHg × 1 atm/760 mmHg = 0.822 atm
Step 3: Calculate the final volume of the air
Assuming constant temperature and ideal behavior, we can calculate the final volume of the air using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁ / P₂
V₂ = 0.822 atm × 215 mL / 1.3 atm = 1.4 × 10² mL
Answer:
CH2O
Explanation:
Firstly, we need to convert the masses of the elements to percentage compositions. This can be done by placing the mass of each element over the total mass multiplied by 100% . We can start with carbon.
C = 5.692/14.229 * 100 = 40%
O = 7.582/14.229 * 100 = 53.29%
H = 0.955/14.229 * 100 = 6.71%
We then proceed to divide each percentage composition by their atomic mass of 12, 16 and 1 respectively.
C = 40/12 = 3.333
O = 53.29/16 = 3.33
H = 6.71/2 = 6.71
Dividing by the smaller value which is 3.33
C = 3.33/3.33 = 1
O = 3.33/3.33= 1
H = 6.71/3.33 = 2
The empirical formula of the compound ribose is CH2O
Molarity is given as,
Molarity = Moles / Volume of Solution ----- (1)
Also, Moles is given as,
Moles = Mass / M.mass
Substituting value of moles in eq. 1,
Molarity = Mass / M.mass × Volume
Solving for Mass,
Mass = Molarity × M.mass × Volume ---- (2)
Data Given;
Molarity = 2.8 mol.L⁻¹
M.mass = 101.5 g.mol⁻¹
Volume = 1 L (I have assumed it because it is not given)
Putting values in eq. 2,
Mass = 2.8 mol.L⁻¹ × 101.5 g.mol⁻¹ × 1 L
Mass = 284.2 g of CuF₂