Answer:
CaS, CaBr₂, VBr₅, and V₂S₅.
Explanation:
- The ionic compound should be neutral; the overall charge of it is equal to zero.
- Binary ionic compound is composed of two different ions.
<u>Ca²⁺ can combined with either Br⁻ or S²⁻ to form binary ionic compounds.</u>
- CaS can be formed via combining Ca²⁺ with S²⁻ to form the neutral binary ionic compound CaS.
- CaBr₂ can be formed via combining 1 mole of Ca²⁺ with 2 moles of Br⁻ to form the neutral binary ionic compound CaBr₂.
<u>V⁵⁺ can combined with either Br⁻ or S²⁻ to form binary ionic compounds.</u>
- V₂S₅ can be formed via combining 2 moles of V⁵⁺ with 5 moles of S²⁻ to form the neutral binary ionic compound V₂S₅.
- VBr₅ can be formed via combining 1 mole of V⁵⁺ with 5 moles of Br⁻ to form the neutral binary ionic compound VBr₅.
<em>So, the empirical formula of four binary ionic compounds that could be formed is: CaS, CaBr₂, VBr₅, and V₂S₅.</em>
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Answer:
In a nuclear fusion reaction, the nuclei of two atoms combine to create a new atom. Most commonly, in the core of a star, two hydrogen atoms fuse to become a helium atom. Although nuclear fusion reactions require a lot of energy to get started, once they are going they produce enormous amounts of energy.
Explanation:
2.2 x 10^-2
0.055 / 250 = 0.00022 - This would be 2.2 x 10^-4, but the question is asking for percent, not proportion, so multiply by 100% to get the percentage.
0.00022 * 100% = 0.022% = 2.2 * 10^-2
Answer:
Exchange across cell membranes - diffusion. The higher the concentration gradient, the faster the rate of facilitated diffusion, up to a point. As equilibrium is reached the concentration gradient is much lower because the difference in concentrations is very small so the rate of facilitated diffusion will slow down and level off.
Explanation:
Explanation:
Reaction:
Cu + 2AgC₂H₃O₂ → Cu(C₂H₃O₂)₂ + 2Ag
The problem is to split the reaction into oxidation and reduction halves:
The oxidation half is the sub-reaction that undergoes oxidation
The reduction half is the one that undergoes reduction:
The ionic equation:
Cu + 2Ag⁺ + 2C₂H₃O₂⁻ → Cu²⁺ + 2C₂H₃O₂⁻ + 2Ag
Oxidation half:
Cu → Cu²⁺ + 2e⁻
Reduction half:
2Ag⁺ + 2e⁻ → 2Ag
C₂H₃O₂⁻ is neither oxidized nor reduced in the reaction.
learn more:
Oxidation state brainly.com/question/10017129
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