17) How many grams are in 0.02 moles of beryllium iodide, Bel2?

Chemistry

2 answers:

Andreas93 [3]3 years ago

8 0

Answer: 5.26 g BeI2

Explanation: For 1 mole of BeI2 it is to its molar mass

Be mass is 9 and I mass is 127 x 2

The molar mass is 263 g BeI2

Solution:

0.02 moles BeI2 x 263 g BeI2/ 1 mole BeI2

= 5.26 g BeI2

Alinara [238K]3 years ago

5 0

Answer:

beryllium iodide has a molar mass of 262.821 g mol−1 , which means that 1 mole of beryllium iodide has a mass of 262.821 g . To find the mass of 0.02 moles of beryllium iodide, simply multiply the number of moles by the molar mass in conversion factor form.

Explanation:

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Neporo4naja [7]

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got it right on the test :)

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2 years ago

which of the following testable questions will provide evidence that elements in the same group have similar properties?

Usimov [2.4K]

The testable question which will provide evidence that elements in the same group have similar properties is valency

<h3>What is the valency?</h3>

In essence, how many electrons are present in their outermost shell.

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Complete question here:

which of the following testable questions will provide evidence that elements in the same group have similar properties A. Valency B. Orbit C. Group D. Period

Since, the reactive capacities of elements is dependent on the number of electrons on its outermost shell, we can conclude on this note that, Elements in the same group have similar properties.

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2 years ago

What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 500. mL of a buffer with the same pH as one made from 475 mL of 0.200 M

Fed [463]

36 ml of NaOh and 464 ml of HCOOH would be enough to form 500 ml of a buffer with the same pH as the buffer made with benzoic acid and NaOH.

What is benzoic acid found in?

Some natural sources of benzoic acid include: Fruits: Apricots, prunes, berries, cranberries, peaches, kiwi, bananas, watermelon, pineapple, oranges.
Spices: Cinnamon, cloves, allspice, cayenne pepper, mustard seeds, thyme, turmeric, coriander.

Calculate the amount of moles in NaOH and benzoic acid. This calculation is done by multiplying molarity by volume.

Amount of moles of NaOH -2 × 0.025 = 0.05 mol

Amount of moles of benzoic acid 2 × 0.475 = 0.095 mol

In this case, we can calculate the pH produced by the buffer of these two reagents, as follows

$pH = pKa + log\frac{base}{acid}$