Answer:
317.6 mL
Explanation:
Step 1: Write the balanced neutralization equation
MgO + 2 HCl ⇒ MgCl₂ + H₂O
Step 2: Calculate the mass corresponding to 640.0 mg of MgO
The molar mass of MgO is 40.30 g/mol. The moles corresponding to 640.0 mg (0.6400 g) of MgO are:
0.6400 g × (1 mol/40.30 g) = 0.01588 mol
Step 3: Calculate the moles of HCl that react with 0.01588 moles of MgO
The molar ratio of MgO to HCl is 1:2. The moles of HCl are 2/1 × 0.01588 mol = 0.03176 mol
Step 4: Calculate the volume of 0.1000 M HCl that contains 0.03176 moles
0.03176 mol × (1 L/0.1000 mol) = 0.3176 L = 317.6 mL
<h3>
Answer:</h3>
0.024 kg CaO
<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>Aqueous Solutions</u>
- Molarity = moles of solute / liters of solution
<u>Atomic Structure</u>
- Reading a Periodic Tables
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
0.41 mol CaO
2.5 M Solution
<u>Step 2: Identify Conversions</u>
1000 g = 1 kg
Molar Mass of Ca - 40.08 g/mol
Molar Mass of O - 16.00 g/mol
Molar Mass of CaO - 40.08 + 16.00 = 56.08 g/mol
<u>Step 3: Convert</u>
- Set up:

- Multiply:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs as our lowest.</em>
0.024114 kg CaO ≈ 0.024 kg CaO
Explanation:
When we add a non-volatile solute in a solvent then due to the impurity added to the solution there will occur an increase in the boiling point of the solution.
This increase in boiling point will be known as elevation in boiling point.
As one beaker contains seawater (water having NaCl) will have some impurity in it. So, more temperature is required by seawater to escape into the atmosphere.
Whereas another beaker has only pure water so it is able to easily escape into the atmosphere since, it contains no impurity.
Thus, we can conclude that level of pure water will decrease more due to non-volatile solute present in it as compared to seawater.
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<u>Answer:</u> The amount of Iodine-131 remain after 39 days is 0.278 grams
<u>Explanation:</u>
The equation used to calculate rate constant from given half life for first order kinetics:

where,
= half life of the reaction = 8.04 days
Putting values in above equation, we get:

Rate law expression for first order kinetics is given by the equation:
![k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = 
t = time taken for decay process = 39 days
= initial amount of the sample = 8.0 grams
[A] = amount left after decay process = ?
Putting values in above equation, we get:
![0.0862=\frac{2.303}{39}\log\frac{8.0}{[A]}](https://tex.z-dn.net/?f=0.0862%3D%5Cfrac%7B2.303%7D%7B39%7D%5Clog%5Cfrac%7B8.0%7D%7B%5BA%5D%7D)
![[A]=0.278g](https://tex.z-dn.net/?f=%5BA%5D%3D0.278g)
Hence, the amount of Iodine-131 remain after 39 days is 0.278 grams