<h3>
Answer:</h3>
0.6 g NaCl
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] Na₂CO₃ (aq) + CaCl₂ (aq) → CaCO₃ (s) + 2NaCl (aq)
[Given] 0.5 g Na₂CO₃ reacted with excess CaCl₂
<u>Step 2: Identify Conversions</u>
[RxN] Na₂CO₃ → 2NaCl
Molar Mass of Na - 22.99 g/mol
Molar Mass of C - 12.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of Cl - 35.45 g/mol
Molar Mass of Na₂CO₃ - 2(22.99) + 12.01 + 3(16.00) = 105.99 g/mol
Molar Mass of NaCl - 22.99 + 35.45 = 58.44 g/mol
<u>Step 3: Stoichiometry</u>
- Set up:

- Multiply/Divide:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 1 sig fig.</em>
0.551373 g NaCl ≈ 0.6 g NaCl
Answer: im thinking its gonna be d.C2H6 and also
the explanation is on the research i had did before i had answered this question so i really hope this help :)
Explanation:
Ar = van de waals forces or london forces
C
H
4
= van de waals forces or london forces
HCl=permanent dipole-dipole interactions
CO = permanent dipole-dipole interactions
HF = hydrogen bonding
N
a
N
O
3
= permanent dipole-dipole interactions
C
a
C
l
2
= van de waals forces or london forces
They have low boiling points
Answer:
Only Reaction 1
Explanation:
In reaction 1, there is a change in state from solid to liquid. Hence, there is an increase in number of ways particles and their energies could be arranged. As a result, entropy increases.
In reaction 2, there is a decrease in amount of gas particles (4 mol to 2 mol). Hence there is a decrease in the number of ways particles and their energies could be arranged. As a result entropy decreases
Answer:
Eu(ClO3)3
Explanation:
The chlorate ion is written as follows, ClO⁻ ₃. We can see from this that the ion is univalent.
From the formula, Eu203, it is easy to see that the europium ion is trivalent.
Hence, when a compound is formed between the europium ion and chlorate ion, the compound will be written as Eu(ClO3)3.
This is so because, when ionic compounds are formed, there is an exchange of valence between the ions in the compound. This gives the final formula of the ionic substance.