To find pH, use the following formula ---> pH= - log [H+]
so first we need to calculate the [H+] concentration using the OH concentration. to do this, we need to use this formula--> 1.0x10-14= [H+] X [OH-], so we solve for H+ and plug in
[H+]= 1.0X10-14/[OH-]---> 1.0 x 10-14/ 1.0 x 10-4= 1.0 x 10-10
now that we have the H+ concentration, we can solve of pH
pH= -log (1.0x10-10)= 10
answer is A
Non-metal atoms gain an electron, or electrons, from another atom to become >negatively charged ions.
True, both of the red and white phosphorus is a pure substance.
Both red and white phosphorus is made of phosphor element but they have a different structure which causes them to have a different color. This phenomenon called allotropes when a chemical element has two or more different form.They are not compounds as they are only made of phosphorus. A compound should be made by at least two different elements.
Answer:
1.18×10²³ atoms.
Explanation:
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ atoms.
From the above concept, 1 mole of sodium also contains 6.02×10²³ atoms.
1 mole of sodium = 23 g.
Thus,
23 g of sodium contains 6.02×10²³ atoms.
Therefore, 4.5 g of sodium will contain = (4.5 × 6.02×10²³)/23 = 1.18×10²³ atoms.
From the above calculation,
4.5 g of sodium contains 1.18×10²³ atoms.
<h3>
Answer:</h3>
1.2 × 10⁻⁸ mol Pb
<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>
- Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<u>Stoichiometry</u>
- Using Dimensional Analysis
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Given] 7.2 × 10¹⁵ atoms Pb
<u>Step 2: Identify Conversions</u>
Avogadro's Number
<u>Step 3: Convert</u>
- [DA] Set up:

- [DA] Multiply/Divide [Cancel out units]:

<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 2 sig figs.</em>
1.19562 × 10⁻⁸ mol Pb ≈ 1.2 × 10⁻⁸ mol Pb