Hydrogen reacts with oxygen based on the following equation:
2 H2<span> + O</span>2<span> → 2 H</span>2<span>O
</span>
From the periodic table:
molar mass of hydrogen = 1 gram
molar mass of oxygen = 16 grams
From the balanced equation above, we can find that:
4 grams of hydrogen react with 32 grams of oxygen to produce 36 grams of water.
This means that: 0.73 grams of hydrogen require (0.73x32) / 4 = 5.84 grams of oxygen to react with.
Since only 3.28 grams of oxygen are reacting, this means that oxygen is our limiting reagent and that the reaction would stop once the amount of oxygen is consumed.
So, we will base our calculations on oxygen.
mass of water produced from 3.28 grams of oxygen can be calculated as follows:
mass of water = (3.28 x 36) / 32 = 3.69 grams
Answer:
1
Explanation:
outershell atoms of an element are also known as valency of that element
so the valency and number of elctron in the outershell of a sodium atom is +1.
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Answer:
Si, Sn, Cs
Explanation:
Atomic no. Atomic radius
Silicon Si_______14___________0.117
Tin Sn______50___________0.140
cesium Cs______55___________0.262
Silicon is the smallest among these
<span>C2H5
First, you need to figure out the relative ratios of moles of carbon and hydrogen. You do this by first looking up the atomic weight of carbon, hydrogen, and oxygen. Then you use those atomic weights to calculate the molar masses of H2O and CO2.
Carbon = 12.0107
Hydrogen = 1.00794
Oxygen = 15.999
Molar mass of H2O = 2 * 1.00794 + 15.999 = 18.01488
Molar mass of CO2 = 12.0107 + 2 * 15.999 = 44.0087
Now using the calculated molar masses, determine how many moles of each product was generated. You do this by dividing the given mass by the molar mass.
moles H2O = 11.5 g / 18.01488 g/mole = 0.638361 moles
moles CO2 = 22.4 g / 44.0087 g/mole = 0.50899 moles
The number of moles of carbon is the same as the number of moles of CO2 since there's just 1 carbon atom per CO2 molecule.
Since there's 2 hydrogen atoms per molecule of H2O, you need to multiply the number of moles of H2O by 2 to get the number of moles of hydrogen.
moles C = 0.50899
moles H = 0.638361 * 2 = 1.276722
We can double check our math by multiplying the calculated number of moles of carbon and hydrogen by their respective atomic weights and see if we get the original mass of the hydrocarbon.
total mass = 0.50899 * 12.0107 + 1.276722 * 1.00794 = 7.400185
7.400185 is more than close enough to 7.40 given rounding errors, so the double check worked.
Now to find the empirical formula we need to find a ratio of small integers that comes close to the ratio of moles of carbon and hydrogen.
0.50899 / 1.276722 = 0.398669
0.398669 is extremely close to 4/10, so let's reduce that ratio by dividing both top and bottom by 2 giving 2/5.
Since the number of moles of carbon was on top, that ratio implies that the empirical formula for this unknown hydrocarbon is
C2H5</span>
<h3>
Answer:</h3>
P₂ = 0.67 atm
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality<u>
</u>
<u>Chemistry</u>
<u>Gas Laws</u>
Boyle's Law: P₁V₁ = P₂V₂
- P₁ is pressure 1
- V₁ is volume 1
- P₂ is pressure 2
- V₂ is volume 2
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] P₁ = 2.02 atm
[Given] V₁ = 4.0 L
[Given] V₂ = 12.0 L
[Solve] P₂
<u>Step 2: Solve</u>
- Substitute in variables [Boyle's Law]: (2.02 atm)(4.0 L) = P₂(12.0 L)
- [Pressure] Multiply: 8.08 atm · L = P₂(12.0 L)
- [Pressure] [Division Property of Equality] Isolate unknown: 0.673333 atm = P₂
- [Pressure] Rewrite: P₂ = 0.673333 atm
<u>Step 3: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs as our smallest.</em>
0.673333 atm ≈ 0.67 atm
