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Zielflug [23.3K]
3 years ago
9

How much nano3 is needed to prepare 225 ml of a 1.55 m solution of nano3?

Chemistry
1 answer:
andrezito [222]3 years ago
5 0

Answer:

= 29.64 g  NaNO3

Explanation:

Molarity is given by the formula;

Molarity = Moles/Volume in liters

Therefore;

Number of moles = Molarity × Volume in liters

                             = 1.55 M × 0.225 L

                             = 0.34875 moles NaNO3

Thus; 0.34875 moles of NaNO3 is needed equivalent to;

   = 0.34875 moles × 84.99 g/mol

   = 29.64 g

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Explanation:

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If a frog initially contained 2 grams of carbon-14 and the half-life of carbon-14 is 5,730 years, how much carbon-14 remains in
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3 years ago
A dehydration reaction starting with 3.8 g cyclohexanol produces 2.6 g cyclohexene. Calculate the theoretical yield for this rea
diamong [38]

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Theoretical yield of C6H10 = 3.2 g.

Explanation:

Defining Theoretical yield as the quantity of product obtained from the complete conversion of the limiting reactant in a chemical reaction. It can be expressed as grams or moles.

Equation of the reaction

C6H11OH --> C6H10 + H2O

Moles of C6H11OH:

Molar mass of C6H110H = (12*6) + (1*12) + 16

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4 0
4 years ago
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