3.54g H2 means you have 3.54 / 2 = 1.77 moles of H2, which means you have 1.77*2 = 3.54 moles of H atom.
Since 1 mole of NH3 has 3 moles of H, so you can produce 3.54 / 3 moles of NH3, which is (3.54/3) * (12+3) = 17.7 grams
<span>Theory of Matter state that matter is composed of?
</span><span>
Answer : Large number of small particles—individual atoms or molecules.</span>
Answer:
Explanation:
Columns and rows used in the periodic classification of elements shows trends in the properties of elements.
The periodic columns are called groups and they denote vertical arrangement of elements.
The horizontal rows are the periods.
- Groups and Periods provides a very simple way to classify elements.
- They reflect trends among different elements that are found in nature. Following these groups and periods, it is even possible to predict the nature of an undiscovered element.
- Certain parts of the table have some unique elements on them with some distinctive properties that distinguish them from others.
- The atomic number is the basis of this classification. By this, it is possible to delineate that elements with similar properties are located on the same group.
Answer:
Na⁺ (aq) + OH⁺ (aq) + H⁺ (aq) + Cl⁻ (aq) → Na⁺ (aq) + Cl⁻ (aq) + H₂O (l)
General Formulas and Concepts:
<u>Atomic Structure</u>
<u>Aqueous Solutions</u>
- Solubility Rules
- States of matter
<u>Stoichiometry</u>
- Reaction RxN Prediction
- Balancing Reactions RxN
Explanation:
<u>Step 1: Define</u>
NaOH reacting w/ HCl
NaOH is soluble
HCl is soluble
[RxN] NaOH (aq) + HCl (aq) → NaCl (aq) + H₂O (l)
<u>Step 2: Total Ionic Equation</u>
<em>Break up soluble compounds into ionic form.</em>
[T.I.E] Na⁺ (aq) + OH⁺ (aq) + H⁺ (aq) + Cl⁻ (aq) → Na⁺ (aq) + Cl⁻ (aq) + H₂O (l)
Answer:
Molar mass of the unknown gas: 146 g /mol
Explanation:
At STP, any "ideal" gas is contained in a volume of 22.4L. This situation, for 1 mol of gas. In this case we have:
0.5 g / 4 g/mol = 0.125 moles of He
These amount of He occupies 0.125 mol . 22.4L / 1 mol = 2.8 L so the double of volume will be occupied by the unkown gas.
22.4 L / 1 mol = (2.8 L . 2) / x moles
22.4L / 1 mol = 5.6 L / x mol
x mol = 5.6L / 22.4L → 0.25 moles
To determine the molar mass of the unknown gas → 36.5 g / 0.25 mol = 146 g /mol
