What is the mass of 5.30×1021 platinum atoms?

1 answer:

3 0

<span>(5.08 * 10^21) / (6.022 * 10^23) = 0.0084357 moles * 195.084 g/mol = 1.65 g Atomic mass of platinum = 195.084 g/mol Avogadro's number = 6.022 * 10^23 so here you go</span>

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In addition, they are poor conductors of

MA_775_DIABLO [31]

Answer: electricity

Explanation:

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2 years ago

H2CO3(aq) + H200 H30 (aq) + HCO3 (aq).

timofeeve [1]

Answer:

$K_a=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}$

Explanation:

Several rules should be followed to write any equilibrium expression properly. In the context of this problem, we're dealing with an aqueous equilibrium:

an equilibrium constant is, first of all, a fraction;
in the numerator of the fraction, we have a product of the concentrations of our products (right-hand side of the equation);
in the denominator of the fraction, we have a product of the concentrations of our reactants (left-hand side o the equation);
each concentration should be raised to the power of the coefficient in the balanced chemical equation;
only aqueous species and gases are included in the equilibrium constant, solids and liquids are omitted.

Following the guidelines, we will omit liquid water and we will include all the other species in the constant. Each coefficient in the balanced equation is '1', so no powers required. Multiply the concentrations of the two products and divide by the concentration of carbonic acid:

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2 years ago

Calculate 4 Srxn for this equation. Round to the

masya89 [10]

Answer:

On edge nuity, it is 79.

Explanation:

Please read the explanation.

Since u have no coefficients, this problem is super easy.

You just add your products, 115.5+69.91 = 185.41

Then add your reactants: 49.8+56.5= 106.3

Then subtract your reactants from your product and round to the nearest whole number.

106.3-185.41=79

It's really easy. Once u understand it, you can do any problem like this. If there is coefficients, just multiply the coefficient by the amount given to you in the problem.

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$22.4269$ grams of sodium phosphate must be added to 1.4 L of this solution to completely eliminate the hard water ions