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2 years ago

What is the hydrogen ion concentration, [H+], of a solution with a pH of 5.43?

Chemistry

1 answer:

grigory [225]2 years ago

6 0

Answer: 3.7 x10−6 Mole per dm^3

Explanation:

pH is the negative logarithm of hydrogen ion concentration in a solution.

So, pH = - log(H+)

Since the solution has a pH of 5.43

5.43 = -log(H+)

To get hydrogen ion concentration, find the Antilog of 5.43

(H+) = Antilog (-5.43)

(H+) = 0.000003715

Then, 0.000003715 in standard form becomes 3.7 x10−6 M

Thus, the concentration of hydrogen ion in the solution is 3.7 x10−6 Mole per dm^3

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Answer:

CERN uses a series of particle accelerators to accelerate protons to their end speeds in the LHC. The first accelerator is the linear accelerator Linac 2. At one end of Linac 2, hydrogen from a bottle is passed through an electric field to strip off the electrons, leaving positively charged protons.

Explanation:

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2 years ago

The charge on the most stable ion of zinc

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3 years ago

What is the conjugate acid in the following equation:

yanalaym [24]

Answer:

HNO₂

Explanation:

An acid is a proton donor; a base is a proton acceptor.

Thus, NO₂⁻ is the base, because it accepts a proton from the water.

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NO₂⁻/HNO₂ make one conjugate acid/base pair, and H₂O/OH⁻ are the other conjugate acid/base pair.

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base acid conj. conj.

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5 0

2 years ago

Chemists can use moles to calculate: A. How much of the products are needed and how much reactant will be made. B. How much prod

damaskus [11]

Answer:

c.- How much of the reactants are needed and how much product will made.

Explanation:

The moles is the matter unit used in chemistry to simplify some calculations, instead of using grams. Also the moles are very useful because the chemical reaction can be balanced.

When a Chemical reaction is balanced, then it can be easily to calculate how many moles are necessary to add in a process to obtain a quantity of grams of a product.

5 0

3 years ago

Part III.The two reactions involved in quantitatively determining the amount of iodate in solution are: IO3-(aq) 5 I-(aq) 6 H (a

slega [8]

Answer:

$\large \boxed{\math{\dfrac{\text{6 mol thiosulfate}}{\text{1 mol iodate}}}}$

Explanation:

The I₂ is the common substance in the two equations.

(1) IO₃⁻ + 5I⁻ + 6H⁺ ⟶ 3I₂ + 3H₂O

{2) I₂ + 2S₂O₃²⁻ ⟶ 2I⁻ + S₄O₆²⁻

From Equation (1), the molar ratio of iodate to iodine is

$\dfrac{\text{I}_{2}}{\text{IO}_{3}^{-}} = \dfrac{3}{1}$

From Equation (2), the molar ratio of iodine to thiosulfate is

$\dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{I}_{2}} = \dfrac{2}{1}$

Combining the two ratios, we get

$\text{Stoichiometric factor} = \dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{IO}_{3}^{-}} = \dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{I}_{2}} \times \dfrac{\text{I}_{2}}{\text{IO}_{3}^{-}} = \dfrac{2}{1} \times \dfrac{3}{1} = \mathbf{\dfrac{6}{1}}\\\\\text{The stoichiometric factor is $\large \boxed{\mathbf{\dfrac{\text{6 mol thiosulfate}}{\text{1 mol iodate}}}}$}$

7 0

2 years ago