<h3>
Answer:</h3>
16.7 g H₂O
<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>Stoichiometry</u>
- Reading a Periodic Table
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[RxN - Balanced] 2NaOH (s) + CO₂ (g) → Na₂CO₃ (s) + H₂O (l)
[Given] 1.85 mol NaOH
<u>Step 2: Identify Conversions</u>
[RxN] 2 mol NaOH → 1 mol H₂O
Molar Mass of H - 1.01 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 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 3 sig figs.</em>
16.6685 g H₂O ≈ 16.7 g H₂O
Answer:
This question is incomplete but the correct option is B
Explanation:
This question is incomplete because of the absence of the "Reference Table S", however the question can still be answered in the absence of the table. The energy described in the question is the ionization energy (energy required to remove the most loosely bound electron in an atom). This question seeks to know the atom (from the options provided) with the least ionization energy.
Ionization energy increases from left to right across the period because it's easier to remove a single electron (valence electron) from the outermost shell than to remove two electrons from the same shell; thus the more the valence electrons (in a shell), the higher the ionization energy. Thus, bromine (Br) and tin (Sn) have high ionization energies because they have more number of electrons in there outermost shell.
<u>Berylium (Be) and strontium (Sr) are both in the group 2 of the periodic table because they both have 2 electrons in there outermost shell. Ionization energy decreases down a group. This is because the farther an electron is from the nucleus, the weaker the force of attraction between the nucleus and the electron. Thus, strontium (Sr) would have a lesser ionization energy between the two and would indeed have the least ionization among the options provided</u>. Hence, the correct option is B
Answer:This is what's known as a metal displacement reaction: the lead substitutes for the copper and ends up precipitating out of solution as insoluble lead (II) sulfate. ... The weight of copper deposited was 15.86gm.
Explanation:
Answer:
CasH52(1) + 38 O2(g) → 25 CO2(g) + 26 H2O(g)
Explanation:
