Answer:
Idk if this is right but i hope it helps... sorry if it's wrong
Explanation:
<h3>
Answer:</h3>
2.51 mol Cu
<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>
- Using Dimensional Analysis
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
1.51 × 10²⁴ atoms Cu
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- Set up:
![\displaystyle 1.51 \cdot 10^{24} \ atoms \ Cu(\frac{1 \ mol \ Cu}{6.022 \cdot 10^{23} \ atoms \ Cu})](https://tex.z-dn.net/?f=%5Cdisplaystyle%201.51%20%5Ccdot%2010%5E%7B24%7D%20%5C%20atoms%20%5C%20Cu%28%5Cfrac%7B1%20%5C%20mol%20%5C%20Cu%7D%7B6.022%20%5Ccdot%2010%5E%7B23%7D%20%5C%20atoms%20%5C%20Cu%7D%29)
- Multiply/Divide:
![\displaystyle 2.50747 \ mol \ Cu](https://tex.z-dn.net/?f=%5Cdisplaystyle%202.50747%20%5C%20mol%20%5C%20Cu)
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
2.50747 mol Cu ≈ 2.51 mol Cu
Answer:
Explanation:The final homogenous solution, after cooling it to 40°C, will contain 47 g of potassium sulfate disolved in 150 g of water, so you can calculate the amount disolved per 100 g of water in this way:
[47 g of solute / 150 g of water] * 100 g of g of water = 31.33 grams of solute in 100 g of water.
So, when you compare with the solutiblity, 15 g of solute / 100 g of water, you realize that the solution has more solute dissolved with means that it is supersaturated.
To make a saturated solution, 15 grams of potassium sulfate would dissolve in 100 g of water.
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Neutralization reactions are the reactions type which form salts.
Explanation:
Salts are formed by ionic bonds when the oxidation states of anions and cations are equal and have opposite signs. So one should be highly electronegative in nature and another should be highly electropositive in nature. So the electropositive element will be ready to give electrons and the electronegative element will be ready to accept all the electrons given by the electropositive element. As a whole the compound will be neutrally charged by adding of equal number of positively charged and negatively charged ions.
The reduction or addition of electrons will be occurring in cations and the oxidation or removal of electrons will be occurring in anions.
So the salt formation is based on neutralization reactions.