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3 years ago

Beryllium oxide, Beo, is an electrical insulator. How

Chemistry

1 answer:

egoroff_w [7]3 years ago

3 0

Answer:

There are 10.0 moles of beryllium oxide in a 250 grams sample of the compound.

Explanation:

We can calculate the number of moles (η) of BeO as follows:

$\eta = \frac{m}{M}$

Where:

m: is the mass = 250 g

M: is the molar mass = 25.0116 g/mol

Hence, the number of moles is:

$\eta = \frac{250 g}{25.0116 g/mol} = 10.0 moles$

Therefore, there are 10.0 moles of beryllium oxide in a 250 grams sample of the compound.

I hope it helps you!

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What is the difference in energy between an electron in 1s and an electron in 4s.

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An electron in 4s is farther away from nucleus and it has higher energy when compared to electron from 1s.

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3 years ago

Which is a living part of an ecosystem? A.soil B.bacteria C.animal bones D.minerals

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3 years ago

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2 years ago

An aqueous solution containing 9.82 g9.82 g of lead(II) nitrate is added to an aqueous solution containing 5.76 g5.76 g of potas

n200080 [17]

Answer:

The limiting reactant is lead(II) nitrate.

7.20 g of precipitate are formed.

1.9 g of the excess reactant remain.

Explanation:

The reaction that takes place is:

Pb(NO₃)₂(aq) + 2KCl(aq) → PbCl₂(s) + 2KNO₃(aq)

With a percent yield of 87.5%.

To determine the limiting reactant, first we convert the masses of each reactant to moles, using their molar mass:

9.82 g Pb(NO₃)₂ ÷ 331.2 g/mol = 0.0296 mol Pb(NO₃)₂

5.76 g KCl ÷ 74.55 g/mol = 0.0773 mol KCl

Looking at the stoichiometric coefficients, we see that 1 mol of Pb(NO₃)₂ would react completely with 2 moles of KCl. Following that logic, 0.0296 mol Pb(NO₃)₂ would react completely with (2x0.0296) 0.0592 mol of KCl. We have more than that amount of KCl, this means KCl is the reactant in excess and Pb(NO₃)₂ is the limiting reactant.

To calculate the mass of precipitate (PbCl₂) formed, we use the moles of the limiting reactant:

0.0296 mol Pb(NO₃)₂ $\frac{1molPbCl_{2}}{1molPb(NO_{3})_{2}}$ * $\frac{278.1g}{1molPbCl_{2}}$ * 87.5/100 = 7.20 g PbCl₂

- Keeping in mind the reaction yield, the moles of Pb(NO₃)₂ that would react are:

0.0296 mol Pb(NO₃)₂ * 87.5/100 = 0.0259 mol Pb(NO₃)₂

Now we convert that amount to moles of KCl and finally into grams of KCl:

0.0259 mol Pb(NO₃)₂ $\frac{2molKCl}{1molPb(NO_{3})_2}$ * $\frac{74.55g}{1molKCl}$ = 3.86 g KCl

3.86 g of KCl would react, so the amount remaining would be:

5.76 - 3.86 = 1.9 g KCl

8 0

3 years ago

If your front lawn is 18.0 feet wide and 20.0 feet long. And each square foot of lawn accumulates 1450 new snow flakes every min

creativ13 [48]

To solve this problem, we begin by first calculating the area of the front lawn. The length and width of the lawn was given and the area of a rectangle is given by the formula: Area = length x width. Thus, the area of the front lawn can be obtained by multiplying 18 ft by 20 ft, wherein we get 360 ft^2 as the area.

Second, the problem indicated that each square foot of lawn accumulates 1450 new snow flakes per minute. This can be translated into the expression 1450 snow flakes/ (minute·ft^2). In this way, we can convert it to units of mass (kg). Afterwards, we simply need to multiply it to the area of the lawn and convert minute to hour. The following expression is then used:

1450 snow flakes/ (minute·ft^2) x 1.90 mg/snow flake x 1 g/1000 mg x 1kg/1000 g x 360 ft^2 x 60 minutes/hour = 59.508 kg snow flake/hour

It is then calculated that 59.508 kg of snow flake accumulates in the lawn every hour.

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3 years ago