All categories

2 years ago

Which of the following pairs of elements would form a covalent compound?

Chemistry

1 answer:

qwelly [4]2 years ago

8 0

Answer:

Nitrogen and Oxygen

Explanation:

they are both nonmetals

You might be interested in

According to Table I, which equation represents a change resulting in the greatest quantity of energy released?

love history [14]

The answer is 3. The releasing of energy means exothermic reaction. So the ΔH should be negative. And the greatest quantity of energy released means that the greatest number. So according to the table I, the answer is 3.

8 0

3 years ago

Read 2 more answers

What happens to these physical properties as the strength of intermolecular forces increases?Increase or decrease?a) melting poi

docker41 [41]

Boiling Point, Melting Point, Viscosity, Surface Tension. Decrease: Vapor Pressure.

6 0

3 years ago

Number 7 plz tysm guys

morpeh [17]

It is letter C: Continental crust is thicker than oceanic crust.

6 0

3 years ago

Read 2 more answers

If 10 staples have a mass of 1.33 g, what will be the mass of 225 staples?

Leona [35]

Answer:

A) 29.9g

Explanation:

first find the weight of 1 staple.

then multiply with 225

5 0

3 years ago

Read 2 more answers

If you have a solution that contains 46.85 g of codeine, C18H21NO3, in 125.5 g of ethanol, C2H5OH, what is the mole fraction of

irakobra [83]

Answer:

0.22

Explanation:

Given, Mass of $C_{18}H_{21}NO_3$ = 46.85 g

Molar mass of $C_{18}H_{21}NO_3$ = 299.4 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{46.85\ g}{299.4\ g/mol}$

$Moles\ of\ C_{18}H_{21}NO_3= 0.1565\ mol$

Given, Mass of $C_{2}H_{5}OH$ = 125.5 g

Molar mass of $C_{2}H_{5}OH$ = 46.07 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{125.5\ g}{46.07\ g/mol}$

$Moles\ of\ C_{2}H_{5}OH= 0.5535\ mol$

So, according to definition of mole fraction:

$Mole\ fraction\ of\ codeine=\frac {n_{codeine}}{n_{codeine}+n_{ethanol}}$

$Mole\ fraction\ of\ codeine=\frac{0.1565}{0.1565+0.5535}=0.22$

4 0

3 years ago