The empirical formula for a compound composed of 0.0683 mol of carbon ( C ), 0.0341 mol of hydrogen ( H ), and 0.1024 mol of nitrogen ( N ) is 
.
<h3>What is the empirical formula?</h3>
An empirical formula tells us the relative ratios of different atoms in a compound.
Given data:
Moles of carbon = 0.0683 mol
Moles of hydrogen = 0.0341 mol
Moles of nitrogen = 0.1024 mol
Dividing each mole using the smallest number that is divided by 0.0341 moles.
We get:
Carbon= 2
Hydrogen=1
Nitrogen=3
The empirical formula for a compound is.
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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:
1.59x10⁻¹⁰m
Explanation:
To solve this question we must know that the length of the cubic cell, X, is equal to:
X = √8 * R
<em>Where R is the atomic radius of germanium</em>
And that in 1 unit cell there are 4 atoms of germanium.
To solve this question we must find the mass in 1 unit cell, with this mass we can find the volume of the cube and the length. With the length we can know the atomic radius:
<em>Mass in 1 unit cell -Molar mass Ge = 72.64g/mol:</em>
4 atoms Ge * (1mol / 6.022x10²³ atoms) = 6.64x10⁻²⁴ moles Ge
6.64x10⁻²⁴ moles Ge * (72.64g / mol) = 4.825x10⁻²²g Ge
<em>Volume unit cell:</em>
4.825x10⁻²²g Ge * (1cm³ / 5.32g) = 9.07x10⁻²³cm³
<em>Length unit cell:</em>
∛9.07x10⁻²³cm³ = 4.49x10⁻⁸cm * (1m / 100cm) = 4.49x10⁻¹⁰m
<em>Atomic radius Ge:</em>
4.49x10⁻¹⁰m / √8 =
<h3>1.59x10⁻¹⁰m</h3>
The answer should be A, they turn litmus paper blue!
Given information:
Mass of pure silver = 195 mg
To determine :
Mass of silver chloride required to plate 195 mg of pure silver
Calculation:
The silver chloride used for silver plating usually contains 75.27 % silver
i.e.
every 100 mg of silver chloride will contain 75.27 mg silver
therefore, amount of silver chloride corresponding to 195 mg of silver would be = 195 mg * 100 mg/75.27 mg
= 259.1 mg
