Answer:
The answer is
<h2>59.6 g </h2>
Explanation:
The mass of a substance when given the density and volume can be found by using the formula
<h3>mass = Density × volume</h3>
From the question
Density of aluminum = 2.00 g/mL
volume = 29.8 mL
The mass is
mass = 2 × 29.8
We have the final answer as
<h3>59.6 g</h3>
Hope this helps you
<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:
![\displaystyle 1.85 \ mol \ NaOH(\frac{1 \ mol \ H_2O}{2 \ mol \ NaOH})(\frac{18.02 \ g \ H_2O}{1 \ mol \ H_2O})](https://tex.z-dn.net/?f=%5Cdisplaystyle%201.85%20%5C%20mol%20%5C%20NaOH%28%5Cfrac%7B1%20%5C%20mol%20%5C%20H_2O%7D%7B2%20%5C%20mol%20%5C%20NaOH%7D%29%28%5Cfrac%7B18.02%20%5C%20g%20%5C%20H_2O%7D%7B1%20%5C%20mol%20%5C%20H_2O%7D%29)
- Multiply/Divide:
![\displaystyle 16.6685 \ g \ H_2O](https://tex.z-dn.net/?f=%5Cdisplaystyle%2016.6685%20%5C%20g%20%5C%20H_2O)
<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:
<h2>━☆゚.*・。゚Fine, but not as good</h2>
<u>Answer:</u> The hydroxide ion concentration and pOH of the solution is
and 2.88 respectively
<u>Explanation:</u>
We are given:
Concentration of barium hydroxide = 0.00066 M
The chemical equation for the dissociation of barium hydroxide follows:
![Ba(OH)_2\rightarrow Ba^{2+}+2OH^-](https://tex.z-dn.net/?f=Ba%28OH%29_2%5Crightarrow%20Ba%5E%7B2%2B%7D%2B2OH%5E-)
1 mole of barium hydroxide produces 1 mole of barium ions and 2 moles of hydroxide ions
pOH is defined as the negative logarithm of hydroxide ion concentration present in the solution
To calculate pOH of the solution, we use the equation:
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
We are given:
![[OH^-]=(2\times 0.00066)=1.32\times 10^{-3}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%282%5Ctimes%200.00066%29%3D1.32%5Ctimes%2010%5E%7B-3%7DM)
Putting values in above equation, we get:
![pOH=-\log(1.32\times 10^{-3})\\\\pOH=2.88](https://tex.z-dn.net/?f=pOH%3D-%5Clog%281.32%5Ctimes%2010%5E%7B-3%7D%29%5C%5C%5C%5CpOH%3D2.88)
Hence, the hydroxide ion concentration and pOH of the solution is
and 2.88 respectively