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3 years ago

Mei-Ling is given the task of studying the compound calcium bromide.

Chemistry

1 answer:

zhenek [66]3 years ago

7 0

Answer:

Meaning have no idea of what the compound given to her is and has to know about compound’s nature. But before beginning any kind of research over the compound the first thing. She can directly know about the compound by knowing its name is calcium.

Bromide is made out of two elements that is only two elements are involved in the making of this compound that is calcium and bromine. And that the compound formed is very different from its original elements hand it might not have the properties that the elements had before the reaction.

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Calculation of ph from hydrogen ion concentration what is the ph of a solution that has an h1 concentration of (a) 1.75 3 1025 m

faltersainse [42]

<h2>Solutions:</h2>

Case a: Finding pH for [H⁺] = 1.75 × 10⁻⁵ mol/L :

As we know pH is given as,

pH = -log [H⁺]

Putting value,

pH = -log [1.75 x 10⁻⁵]

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Case b: Finding pH for [H⁺] = 6.50 × 10⁻¹⁰ mol/L :

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pH = -log [H⁺]

Putting value,

pH = -log [6.50 × 10⁻¹⁰]

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Case c: Finding pH for [H⁺] = 1.0 × 10⁻⁴ mol/L :

As we know pH is given as,

pH = -log [H⁺]

Putting value,

pH = -log [1.0 × 10⁻⁴]

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Case d: Finding pH for [H⁺] = 1.50 × 10⁻⁵ mol/L :

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3 years ago

Account for the change when NO2Cl is added using the reaction quotient Qc. Match the words in the left column to the appropriate

ryzh [129]

Answer:

A. Disturbing the equilibrium by adding NO2Cl decreases Qc to a value less than Kc.

B. To reach a new state of equilibrium, Qc therefore increases which means that the denominator of the expression for Qc decreases.

C. To accomplish this, the concentration of reagents decreases, and the concentration of products increases.

Explanation:

Hello,

In this case, for the equilibrium reaction:

$NO_2Cl(g)+NO(g)\rightleftharpoons NOCl(g)+NO_2(g)$

Whose equilibrium expression is:

$Kc=\frac{[NO_2][NOCl]}{[NO_2Cl][NO]}$

The proper matching is:

A. Disturbing the equilibrium by adding NO2Cl decreases Qc to a value less than Kc, since the denominator becomes greater, therefore, Qc decreases.

B. To reach a new state of equilibrium, Qc therefore increases which means that the denominator of the expression for Qc decreases, since the lower the denominator, the higher Qc as it has the concentration of reactants.

C. To accomplish this, the concentration of reagents decreases, and the concentration of products increases, since the reactants must be consumed in order to reestablish equilibrium by shifting the reaction towards the products.

Best regards.

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4 years ago

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tekilochka [14]

Answer:

23.24%_23.17%

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