If the total amount of the sample is 230g and they give each component of the sample, we add the given amount of grams of each element and subtract the given from 230g.
That means the rest of oxygen is 35.2g
This is Percent Composition.
The formula to find the percent composition of an element inside a compound is
Now we implement the variables into the formula to find the percent composition oxygen.
The answer is <span>
15.30% = O</span>
Answer:
300 K
General Formulas and Concepts:
<u>Atomic Structure</u>
<u>Gas Laws</u>
Ideal Gas Law: PV = nRT
- <em>P</em> is pressure
- <em>V</em> is volume
- <em>n</em> is moles
- <em>R</em> is gas constant
- <em>T</em> is temperature
Explanation:
<u>Step 1: Define</u>
<em>Identify variables</em>
[Given] <em>n</em> = 0.5 mol N₂
[Given] <em>V</em> = 6 L
[Given] <em>P</em> = 2 atm
[Given] <em>R</em> = 0.0821 L · atm · mol⁻¹ · K⁻¹
[Solve] <em>T</em>
<em />
<u>Step 2: Solve for </u><em><u>T</u></em>
- Substitute in variables [Ideal Gas Law]: (2 atm)(6 L) = (0.5 mol)(0.0821 L · atm · mol⁻¹ · K⁻¹)T
- Multiply [Cancel out units]: 12 atm · L = (0.04105 L · atm · K⁻¹)T
- Isolate <em>T</em> [Cancel out units]: 292.326 K = T
- Rewrite: T = 292.326 K
<u>Step 3: Check</u>
<em>Follow sig fig rules and round. We are given 1 sig fig as our lowest.</em>
292.326 K ≈ 300 K
Kc = [CS2] / [S2][C] but, since C is solid, it doesn't fit here
<span>Kc = [CS2] / [S2] </span>
<span>S2(g) + C(s) ---> CS2(g) </span>
<span>13.8mole S2 / 6.05L = 2.28M </span>
<span>Kc = x / 2.28-x </span>
<span>9.4 x (2.28-x) = x </span>
<span>21.43 - 9.4x = x </span>
<span>21.42 = 10.4x </span>
<span>x = 2.05M = [CS2] </span>
<span>2.05M x 6.05L = 12.46moles </span>
<span>12.46moles x 76gmole = 946.96g CS2</span>
Answer:
Mg+2HCL-Magnesium Chloride +Hygrogen
Explanation:
N.O of moles=Mass\Molar Mass
192÷24=8
1:1
8×2=16
