Answer:
Explanation:
There is a formula for this:
M = DRT/P where M = molar mass. This just derived from PV = nRT where you say n = grams/molar mass. However, just with this formula, we can get D which is density at STP (1 atm and 273K). We find that D = 6.52g/L.
Answer:
The answer to your question is letter B. 9
Explanation:
Unbalanced reaction
Al₂(SO₄)₃ + Ca(OH)₂ ⇒ Al(OH)₃ + CaSO₄
Reactants Elements Products
2 Al 1
3 S 1
14 O 7
1 Ca 1
2 H 3
Balanced reaction
Al₂(SO₄)₃ + 3Ca(OH)₂ ⇒ 2Al(OH)₃ + 3CaSO₄
Reactants Elements Products
2 Al 2
3 S 3
18 O 18
3 Ca 3
6 H 6
The sum of the coefficients is 1 + 3+ 2+ 3 = 9
Answer:
The specific rotation of D is 11.60° mL/g dm
Explanation:
Given that:
The path length (l) = 1 dm
Observed rotation (∝) = + 0.27°
Molarity = 0.175 M
Molar mass = 133.0 g/mol
Concentration in (g/mL) = 0.175 mol/L × 133.0 g/mol
Concentration in (g/mL) = 23.275 g/L
Since 1 L = 1000 mL
Concentration in (g/mL) = 0.023275 g/mL
The specific rotation [∝] = ∝/(1×c)
= 0.27°/( 1 dm × 0.023275 g/mL
)
= 11.60° mL/g dm
Thus, the specific rotation of D is 11.60° mL/g dm
From 5 L to moles, just divide 5 by 22.4. I got 0.22 moles of H2.
From 5 moles to liters, just multiply 5 by 22.4. I got 112 L of H2.
You must add 7.5 pt of the 30 % sugar to the 5 % sugar to get a 20 % solution.
You can use a modified dilution formula to calculate the volume of 30 % sugar.
<em>V</em>_1×<em>C</em>_1 + <em>V</em>_2×<em>C</em>_2 = <em>V</em>_3×<em>C</em>_3
Let the volume of 30 % sugar = <em>x</em> pt. Then the volume of the final 20 % sugar = (5 + <em>x</em> ) pt
(<em>x</em> pt×30 % sugar) + (5 pt×5 % sugar) = (<em>x</em> + 5) pt × 20 % sugar
30<em>x</em> + 25 = 20x + 100
10<em>x</em> = 75
<em>x</em> = 75/10 = 7.5
