The reaction uses B) 9.0 g Br₂.
iron + bromine ⟶ product
2.0 g + <em>x</em> g ⟶ 11.0 g
According to the <em>Law of Conservation of Mass</em>, the total mass of the reactants must equal the total mass of the products.
∴2.0 + <em>x</em> = 11.0
<em>x</em> = 11.0 – 2.0 = 9.0
The reaction uses 9.0 g Br₂.
Answer:
Mg
Explanation:
The standard reduction potentials are
<u>E°/V
</u>
Au³⁺(aq ) + 3e⁻ ⟶ Au(s); 1.42
Hg²⁺(aq) + 2e⁻ ⟶ Hg(l); 0.85
Ag⁺(aq) + e⁻ ⟶ Ag(s); 0.80
Cu²⁺(aq) + 2e⁻ ⟶ Cu(s); 0.34
Mg2+(aq) + 2e- ⟶ Mg(s); -2.38
The more negative the standard reduction potential, the stronger the metal is as a reducing agent.
Mg is the only metal with a standard reduction potential lower than that of Cu, so
Only Mg will react spontaneously with Cu²⁺.
Answer: Solution A : ![[H_3O^+]=0.300\times 10^{-7}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D0.300%5Ctimes%2010%5E%7B-7%7DM)
Solution B : ![[OH^-]=0.107\times 10^{-5}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.107%5Ctimes%2010%5E%7B-5%7DM)
Solution C : ![[OH^-]=0.177\times 10^{-10}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.177%5Ctimes%2010%5E%7B-10%7DM)
Explanation:
pH or pOH is the measure of acidity or alkalinity of a solution.
pH is calculated by taking negative logarithm of hydrogen ion concentration and pOH is calculated by taking negative logarithm of hydroxide ion concentration.

![[H_3O^+][OH^-]=10^{-14}](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%5BOH%5E-%5D%3D10%5E%7B-14%7D)
a. Solution A: ![[OH^-]=3.33\times 10^{-7}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D3.33%5Ctimes%2010%5E%7B-7%7DM)
![[H_3O^+]=\frac{10^{-14}}{3.33\times 10^{-7}}=0.300\times 10^{-7}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D%5Cfrac%7B10%5E%7B-14%7D%7D%7B3.33%5Ctimes%2010%5E%7B-7%7D%7D%3D0.300%5Ctimes%2010%5E%7B-7%7DM)
b. Solution B : ![[H_3O^+]=9.33\times 10^{-9}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D9.33%5Ctimes%2010%5E%7B-9%7DM)
![[OH^-]=\frac{10^{-14}}{9.33\times 10^{-9}}=0.107\times 10^{-5}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5Cfrac%7B10%5E%7B-14%7D%7D%7B9.33%5Ctimes%2010%5E%7B-9%7D%7D%3D0.107%5Ctimes%2010%5E%7B-5%7DM)
c. Solution C : ![[H_3O^+]=5.65\times 10^{-4}M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%3D5.65%5Ctimes%2010%5E%7B-4%7DM)
![[OH^-]=\frac{10^{-14}}{5.65\times 10^{-4}}=0.177\times 10^{-10}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5Cfrac%7B10%5E%7B-14%7D%7D%7B5.65%5Ctimes%2010%5E%7B-4%7D%7D%3D0.177%5Ctimes%2010%5E%7B-10%7DM)
Answer:
Because warm air is less dense
Explanation: