<h3>
Answer:</h3>
1.39 × 10²³ particles CuCr₂O₇
<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>
- Reading a Periodic Table
- Moles
- 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] 64.5 g CuCr₂O₇
[Solve] particles CuCr₂O₇
<u>Step 2: Identify Conversions</u>
Avogadro's Number
[PT] Molar Mass of Cu - 63.55 g/mol
[PT] Molar Mass of Cr - 52.00 g/mol
[PT] Molar Mass of O - 16.00 g/mol
Molar Mass of CuCr₂O₇ - 63.55 + 2(52.00) + 7(16.00) = 279.55 g/mol
<u>Step 3: Convert</u>
- [DA] Set up:
- [DA] Divide/Multiply [Cancel out units]:
<u>Step 4: Check</u>
<em>Follow sig fig rules and round. We are given 3 sig figs.</em>
1.38944 × 10²³ particles CuCr₂O₇ ≈ 1.39 × 10²³ particles CuCr₂O₇
Answer:
45.2%
Explanation:
To calculate the percent of an element in a compound we divide the molar mass of the element by the compound and multiply that by 100
First lets find the molar mass of Fluoride
Looking at the periodic table Fluoride has a molecular mass of 18.998 g
Now we need to find the molecular mass of NaF
Looking at a periodic table, Sodium (Na) has a molecular mass of 22.990g and Fluoride has a molecular mass of 18.998 so NaF has a molecular mass of 22.990(1) + 18.998(1) = 41.988g
Now we divide the mass of fluoride by the mass of sodium fluoride and multiply that by 100 to find the percentage of fluoride that is present in NaF
Mass of Fluoride = 18.998g
Mass of Sodium Fluoride = 41.988g
Percentage of fluoride present in NaF = (18.998g / 41.988g) * 100 = 45.2%
Answer:
Explanation:
Mining:
Mining is when certain substances are removed from the Earth, usually in large quantities. Mining can occur below the surface of the Earth via tunnels and caves. It can also occur at surface level through methods such as strip mining, which removes huge amounts of soil in the process
Answer:
See whole explanation to understand
Explanation:
the reason why there is such a large jump from 2nd to 3rd ionization energy for calcium is because to remove the third electron, a larger amount of energy is required, since the shell is closer to the nucleus, and higher attraction exists between them. This is why the second ionization energy is 1125.4 and then the third IE is 4912.4 which is a very big difference. It's all about the elections and energy!!
Answer:
Option D. 30 g
Explanation:
The balanced equation for the reaction is given below:
2Na + S —> Na₂S
Next, we shall determine the masses of Na and S that reacted from the balanced equation. This is can be obtained as:
Molar mass of Na = 23 g/mol
Mass of Na from the balanced equation = 2 × 23 = 46 g
Molar mass of S = 32 g/mol
Mass of S from the balanced equation = 1 × 32 = 32 g
SUMMARY:
From the balanced equation above,
46 g of Na reacted with 32 g of S.
Finally, we shall determine the mass sulphur, S needed to react with 43 g of sodium, Na. This can be obtained as follow:
From the balanced equation above,
46 g of Na reacted with 32 g of S.
Therefore, 43 g of Na will react with = (43 × 32)/46 = 30 g of S.
Thus, 30 g of S is needed for the reaction.
