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Anarel [89]
3 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]3 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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How do I solve this problem?
JulijaS [17]
First question. Applying ideal gas equation PV=nRT, P= 101.3 x 10³Pa = 1atm. therefore, 1 x 260 x 10^-3 = n x 0.082 x 294.( Temperature in kelvin=273+21). n = 0.01 moles. Volume of gas at STP= n x 22.4 = 0.01x22.4 = 0.224L. Hope this helps
5 0
3 years ago
A 4.36-g sample of an unknown alkali metal hydroxide is dissolved in 100.0 mL of water. An acid-base indicator is added, and the
BARSIC [14]

Answer:

Rb+

Explanation:

Since they are telling us that the equivalence point was reached after 17.0 mL of   2.5 M HCl were added , we can calculate the number of moles of HCl which neutralized our unknown hydroxide.

Now all the choices for the metal cation are monovalent, therefore the general formula for our unknown is XOH and  we know the reaction is 1 equivalent acid to 1 equivalent base. Thus we have the number of moles, n,  of XOH and from the relation n = M/MW we can calculate the molecular weight of XOH.

Thus our calculations are:

V = 17.0 mL x 1 L / 1000 mL = 0.017 L

2.5 M HCl x 0.017 L = 2.5 mol/ L x 0.017 L = 0.0425 mol

0.0425 mol = 4.36 g/ MW XOH

MW of XOH = (atomic weight of X + 16 + 1)

so solving the above equation we get:

0.0425 = 4.36 / (X + 17 )

0.7225 +0.0425X = 4.36

0.0425X = 4.36 -0.7225 = 3.6375

X = 3.6375/0.0425 = 85.59

The unknown alkali is Rb which has an atomic weight of 85.47 g/mol

6 0
3 years ago
A sample of table sugar (sucrose, C12H22O11) has a mass of 1.202 g.
Archy [21]

Answer:

a) 0.003512 moles

b) Moles C= 0.04214 moles carbon

Moles H = 0.07726 moles hydrogen

Moles O = 0.03863 moles of oxygen

c) C atoms = 2.54 *10^22 carbon atoms

H atoms = 4.65 *10^22 hydrogen atoms

O atoms = 2.33 *10^22 oxygen atom

Explanation:

Step 1: Data given

Mass of sucrose = 1.202 grams

Molar mass of sucrose = 342.3 g/mol

Step 2: Calculate moles of sucrose

Moles sucrose = Mass sucrose / molar mass sucrose

Moles sucrose = 1.202 grams / 342.3 g/mol

Moles sucrose = 0.003512 moles

Step 3: Calculate moles of each element

For 1 mol of C12H22O11 we have 12 moles of carbon, 22 moles of hydrogen and 11 moles of oxygen

Moles C: 12*0.003512 = 0.04214 moles carbon

Moles H: 22* 0.003512 = 0.07726 moles hydrogen

Moles O: 11* 0.003512 = 0.03863 moles of oxygen

Step 4: Calculate the number of atoms

C atoms = 6.022 *10^23 / mol * 0.04214 moles = 2.54 *10^22 atoms carbon

H atoms = 6.022 * 10^23 / mol * 0.07726 moles = 4.65 *10^22 atoms H

O atoms = 6.022 * 10^23 / mol * 0.03863 moles = 2.33 *10^22 atoms O

0.003512 moles of sucrose containse 6.022 *10^23 * 0.003512 = 2.11 * 10^21 sucrose molecules

7 0
3 years ago
What is an atomic weight?
S_A_V [24]
The average weight of an atom of an element, formerly based on the weight of one hydrogen atom taken as a unit or on 1/16 (0.0625) the weight of an oxygen atom, but after 1961 based on 1/12 the weight of the carbon-12 atom.
3 0
3 years ago
Which would be considered a substance? Element, Compound, Mixture.
Romashka [77]

Answer:

MIXTURE , ELEMENT AND SUBSTANCE

3 0
3 years ago
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