Ask question

Ask question

All categories

3 years ago

8

What keeps an ionic bond held together?

2 answers:

andrew11 [14]3 years ago

6 0

Answer:

<u><em>D.</em></u><u><em> Attraction between oppositely- charged ions</em></u>

Explanation:

I took the test

Schach [20]3 years ago

5 0

D) Attraction between oppositely-charged ions.

In ionic bonds, one atom is negatively charged, and one atom is positively charged. The positively charged atom gives its electrons, and the negatively charged atom receives the electrons--thus, the opposite charge attracts them to one another and holds them together. That simple.

You might be interested in

Andrews [41]

it is at the bottom trust me i did the test

6 0

3 years ago

Read 2 more answers

Using the ideal gas law, PV=nRT, where R=0.0821 L atm/mol K, calculate the volume in liters of oxygen produced by the catalytic

AfilCa [17]

Answer:

V = 2.32 Liters

Explanation:

PV = nRT => V = nRT/P

n = 25.8g/122g/mole = 0.21 mole

R = 0.08206 L·atm/mol·K

T = 25.44°C + 273 = 298.44K

P = 2.22 atm (given in problem)

V = (0.21mol)(0.08206 L·atm/mol·K)(298.44K)/(2.22atm) = 2.32 Liters at 25.44°C & 2.22atm

8 0

3 years ago

Last question, I promise!!!!!! ... :>

VikaD [51]

Answer:

Which question?

Explanation:

3 0

4 years ago

Read 2 more answers

Qofudowqfowefjwobfwofbwfofnroenennefeoiqefopne FREEE PONITTSS

Thepotemich [5.8K]

Hey lol
explanation:)))

5 0

3 years ago

Read 2 more answers

A certain gas is present in a 10.0 LL cylinder at 4.0 atmatm pressure. If the pressure is increased to 8.0 atmatm the volume of

ICE Princess25 [194]

Answer:

The gas obeys Boyle’s law and the value of $k_i\&k_f$ both are equal to 40.0 atm L.

Explanation:

Initial volume of the gas = $V_1=10.0 L$

Initial pressure of the gas = $P_1=4.0 atm$

Final volume of the gas = $V_2=5.0 L$

Final pressure of the gas = $P_2=8.0 atm$

This law states that pressure is inversely proportional to the volume of the gas at constant temperature.

$PV=k$

The equation given by this law is:

$P_1V_1=P_2V_2$

$P_1\propto \frac{1}{V_1}$

$P_1V_1=k_i$

$k_i=4.0 atm\times 10.0 L = 40.0 atm L$

$P_2\propto \frac{1}{V_2}$

$P_V_2=k_f$

$k_f=8.0 atm\times 5.0 L = 40.0 atm L$

$k_i=k_f=40.0 atm L$

The gas in the cylinder is obeying Boyle's law.

The gas obeys Boyle’s law and the value of $k_i\&k_f$ both are equal to 40.0 atm L.

6 0

3 years ago