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Chemistry

1 answer:

Aleksandr [31]3 years ago

5 0

Answer:

H2O= +2

CCl4 = +4

Na+= 0

CuCl2= -2

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Classify HBr as a strong acid or a weak acid.<br><br><br> Strong Acid or Weak Acid?

IceJOKER [234]

HBr is a Strong acid

4 0

2 years ago

Name two metalloids that are semiconductors

guajiro [1.7K]

The Metalloids are:

Boron.

Silicon.

5 0

3 years ago

The maximum allowable concentration of pb2+ ions in drinking water is 0.05 ppm (i.e., 0.05 g of pb2+ in 1 million grams of water

trasher [3.6K]

PbSO₄ partially dissociates in water. the balanced equation is;

PbSO₄(s) ⇄ Pb²⁺(aq) + SO₄²⁻(aq)
Initial - -
Change -X +X +X
Equilibrium X X

Ksp = [Pb²⁺(aq)] [SO₄²⁻(aq)]
1.6 x 10⁻⁸ = X * X
1.6 x 10⁻⁸ = X²
X = 1.3 x 10⁻⁴ M

Hence the Pb²⁺ concentration in underground water is 1.3 x 10⁻⁴ M.
[Pb²⁺] = 1.3 x 10⁻⁴ M.
= 1.3 x 10⁻⁴ mol / L x 207 g / mol
= 26.91 ppm

8 0

3 years ago

The question is in the picture below

Rus_ich [418]

Answer:

$\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3$

Explanation:

Hess's Law of Constant Heat Summation states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation is equal to the sum of the enthalpy changes of the other chemical equations. Thus, the reaction that involves the conversion of reactant A to B, for example, has the same enthalpy change even if you convert A to C, before converting it to B. Regardless of how many steps it takes for the reactant to be converted to the product, the enthalpy change of the overall reaction is constant.

With Hess's Law in mind, let's see how A can be converted to 2C +E.

$\bf{\text{A} \rightarrow 2\text{B}}$ (Δ $\text{H}_1$ ) -----(1)

Since we have 2B, multiply the whole of II. by 2:

$\bf{2\text{B} \rightarrow 2\text{C} +2\text{D}}$ (2Δ $\text{H}_2$ ) -----(2)

This step converts all the B intermediates to 2C +2D. This means that the overall reaction at this stage is $\text{A} \rightarrow 2\text{C} +2\text{D}$ .

Reversing III. gives us a negative enthalpy change as such:

$\bf{2\text{D} \rightarrow \text{E}}$ (-Δ $\text{H}_3$ ) -----(3)

This step converts all the D intermediates formed from step (2) to E. This results in the overall equation of $\text{A} \rightarrow 2\text{C} +\text{E}$ , which is also the equation of interest.