Answer: 24.13 g Cu
Explanation:
<u>Given for this question:</u>
M of CuO = 30 g
m of CuO = 79.5 g/mol
Number of moles of CuO = (given mass ÷ molar mass) = (30 ÷ 79.5) mol
= 0.38 mol
The max number of CuO (s) that can be produced by the reaction of excess methane can be solved with this reaction:
CuO(s) + CH4(l) ------> H2O(l) + Cu(s) + CO2(g)
The balanced equation can be obtained by placing coefficients as needed and making sure the number of atoms of each element on the reactant side is equal to the number of atoms of each element on the product side
4CuO(s) + CH4(l) ----> 2H2O(l) + 4Cu(s) + CO2(g)
From the stoichiometry of the balanced equation:
4 moles of CuO gives 4 moles of Cu
1 mole of CuO gives 1 mol of Cu
0.38 mol of CuO gives 0.38 mol of Cu
Therefore, the grams of Cu that can be produced = 0.38 × molar mass of Cu
= 0.38 × 63.5 g
= 24.13 grams
Therefore, 24.13 grams of copper could be produced by the reaction of 30.0 of copper oxide with excess methane
Answer:
The answer is option D, that is, carbon and carbon monoxide
C) Weathering can be chemical or physical; deposition
is only chemical
Answer:- 47.62 mL
Solution:- It is a dilution problem where we are asked to calculate the volume of 15.75 M perchloric acid solution required to make 500.0 mL of 1.500 M solution.
For solving this type of problems we use the dilution equation:
Where, 
is the concentration of the concentrated solution and 
is it's volume.
is the concentration of the diluted solution and 
is it's volume. Let's plug in the values in the equation and solve it for .
On rearranging this for :
So, 47.62 mL of 15.75 M perchloric acid are required to make 500.0mL of 1.500 M solution.
Answer:
Biggest Radii V²⁺ > V³⁺ > V⁴⁺ > V⁵⁺ Smallest Radii
General Formulas and Concepts:
- Periodic Trends: Atomic/Ionic Radii
- Coulomb's Law
Explanation:
The Periodic Trend for Atomic Radii is down and to the left. Therefore, the element with the largest radius would be in the bottom left corner of the Periodic Table.
Anions will always have a bigger radii than the parent radii. When we add e⁻ to the element, we are increasing the e⁻/e⁻ repulsions. This will cause e⁻ to repel themselves more and thus create more space, increasing the radii size.
Cations will always have smaller radii than the parent radii. When we remove e⁻ from the element, we are decreasing e⁻/e⁻ repulsions. Since there are less e⁻, there is no need for more space and thus decreases the radii size.
Since Cations are smaller than the parent radii, the more e⁻ we remove, the smaller it will become.
Therefore, the least removed e⁻ Vanadium would be the largest and the most removed e⁻ Vanadium would be the smallest.
