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Anarel [89]
2 years ago
5

When NaCl dissolves in water, aqueous Na+ and Cl- ions result. The force of attraction that exists between Na+ and H2O is called

a(n) __________ interaction. Select one: a. dipole-dipole b. ion-ion c. hydrogen bonding d. ion-dipole e. London dispersion force
Chemistry
1 answer:
kiruha [24]2 years ago
4 0

Answer:

b. ion-ion

Explanation:

When NaCl dissolves in water , it dissociates completely into the corresponding ions , i.e. , Na⁺ and Cl⁻ ,

Since , the charges have complete positive and negative charges , hence , they interact with each other via ion - ion interaction , i.e. the interaction between the positive and negative charges , is referred to as ion - ion interaction .

Hence , the correct option is b. ion-ion interaction .

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The beta rays emitted from atomic nuclei are _____.<br><br> an electron<br> a neutron<br> a proton
Sliva [168]

The answer I believe is An Electron


6 0
2 years ago
Read 2 more answers
2. Matt went to his friend’s party. He ate a big meal and drank a keg of beer. He felt heartburn after the meal and took Tums to
evablogger [386]

Answer:

390.85mL

Explanation:

Step 1:

Data obtained from the question.

Initial pressure (P1) = 780 torr

Initial volume (V1) = 400mL

Initial temperature (T1) = 40°C = 40°C + 273 = 313K

Final temperature (T2) = 25°C = 25°C + 273 = 298K

Final pressure (P2) = 1 atm = 760torr

Final volume (V2) =?

Step 2:

Determination of the final volume i.e the volume of the gas outside Matt's body.

The volume of the gas outside Matt's body can be obtained by using the general gas equation as shown below:

P1V1/T1 = P2V2/T2

780 x 400/313 = 760 x V2 /298

Cross multiply to express in linear form

313 x 760 x V2 = 780 x 400 x 298

Divide both side by 313 x 760

V2 = (780 x 400 x 298) /(313 x 760)

V2 = 390.85mL

Therefore, the volume of the gas outside Matt's body is 390.85mL

3 0
2 years ago
Thsof<br> is which acid H2S04
Crank

Explanation:

H2So4=sulphuric acid , strong acid

6 0
10 months ago
Is 6.022x 10^23 carbon atoms a mol
Kamila [148]

Answer:

Yes

Explanation:

1 mol of any substance is equal to the avagadro constant

8 0
2 years ago
molecules of i2 are produced by the reaction of 0.4321 mol of cucl2 according to the following equation. 2 cucl2 4 ki → 2 cui 4
Juli2301 [7.4K]

\: 1.237 \times 10^{23}   molecules \: of \:I _{2} \: are

are\: produced \: and \: 53.74 \: g \: of \: I _{2} \: are

produced from the reaction.

The overall balanced equation for the reaction is,

2CuCl_{2} +   KI→2CuI + 4KCl  + I_{2}

Number  \: of  \: moles  \: ofCuCl_{2}=0.4235 \: mol

2 \: moles \: of \: CuCl_{2} \: produce \: \: 1 \: mole \:

of \:  I_{2}.

2 \: mole \: of \: CuCl_{2}  = 1 \: mole \: of \:  I_{2}

1 \: mole \: of \: I_{2} \:  = 6.022 \times 10 ^{23}  \: molecules \: of  I_{2}

So, the number of molecules produced in the reaction of

I_{2} \: are

=  \frac{0.4235 \times 6.022 \times 10 ^{23} }{2 }

= 1.237 \times 10^{23}  \: molecules \:

1.237 \times 10^{23}  \: molecules \: of \: I_{2} \: are \:

produced from the reaction.

Mass  \: of \:   I _{2} \: produced \: are \: is ,

=  \frac{0.4235 \times 253.8}{2}

= 53.74 \: g

53.74 \: g \: of \:  I _{2} \: produced \: in \: the \:

reaction.

Therefore , \: 1.237 \times 10^{23}   molecules

of  \: I_{2} \: are\: produced \: and \: 53.74 \: g \: of

are produced in the reaction.

To know more about moles, refer to the below link:

brainly.com/question/26416088

#SPJ4

7 0
2 years ago
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