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zhuklara [117]
3 years ago
5

When an ionic compound such as sodium chloride (NaCl) is placed in water, the component atoms of the NaCl crystal dissociate int

o individual sodium ions (Na+) and chloride ions (Cl-). In contrast, the atoms of covalently bonded molecules (e.g., glucose, sucrose, glycerol) do not generally dissociate when placed in aqueous solution. Which of the following solutions would be expected to contain the greatest number of solute particles (molecules or ions)?
A) 1 L of 1.0 M NaCl
B) 1 L of 0.5 M NaCl
C) 1 L of 0.5 M glucose
D) 1 L of 1.0 M glucose
E) C and D will contain equal numbers of particles.
Chemistry
1 answer:
sesenic [268]3 years ago
8 0

Answer:

  • <em>The solution expected to contain the greatest number of solute particles is: </em><u>A) 1 L of 1.0 M NaCl</u>

Explanation:

The number of particles is calculated as:

a) <u>For Ionic compounds</u>:

  • molarity × volume in liters × number of ions per unit formula.

b) <u>For covalent compounds</u>:

  • molarity × volume in liters

The difference is a factor which is the number of particles resulting from the dissociation or ionization of one mole of the ionic compound.

So, calling M the molarity, you can write:

  • # of particles = M × liters × factor

This table show the calculations for the four solutions from the list of choices:

Compound    kind         Particles in solution  Molarity   # of particles

                                       (dissociation)              (M)          in 1 liter

A) NaCl          ionic            ions Na⁺ and Cl⁻        1.0            1.0 × 1 × 2 = 2

B) NaCl          ionic            ions Na⁺ anc Cl⁻        0.5           0.5 × 1 × 2 = 1

C) Glucose    covalent     molecules                   0.5          0.5 × 1 × 1 = 0.5

D) Glucose    covalent     molecules                   1.0           1.0 × 1  × 1 = 1

Therefore, the rank in increasing number of particles is for the list of solutions given is: C < B = D < A, which means that the solution expected to contain the greatest number of solute particles is the solution A) 1 L of 1.0 M NaCl.

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Answer:

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3 years ago
Calculate δs∘rxn for the reaction2no(g) o2(g)→2no2(g)express your answer to one decimal place and include the appropriate units
Anna35 [415]

The δs∘rxn  for the reaction 2NO(g) + O_{2} → 2NO_{2} (g) will be -146 J/K.

Entropy would be a measurable physical characteristic and a scientific notion that is frequently connected to a condition of disorder, unpredictability, or uncertainty.

Entropy would be a measurement of the system's unpredictability or disorder. The entropy increases as randomness do. It has broad properties as well as a state function. It has the unit JK^{-1} mol^{-1}.

Entropy of the reaction can be calculated by the reaction.

ΔS^{0} rxn = 2 mol × S^{0} (NO_{2} (g) - 2 mol × S^{0} NO (g) - 1 mol × S^{0} (O_{2} )

ΔS^{0} rxn  = 2 mol × 240 J/mol.K - 2 mol × 210 J/mol.K-1 mol  ×205.2 J/mol.K

ΔS^{0} rxn  = -146.8 J/K

Therefore, the δs∘rxn  for the reaction 2NO(g) + O_{2} → 2NO_{2} (g) will be -146 J/K.

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4 0
1 year ago
Consider the reaction below for which K = 78.2 atm-1. A(g) + B(g) ↔ C(g) Assume that 0.386 mol C(g) is placed in the cylinder re
borishaifa [10]

Answer:

1.65 L

Explanation:

The equation for the reaction is given as:

                        A            +            B           ⇄        C

where;

numbers of moles = 0.386 mol C  (g)

Volume =  7.29 L

Molar concentration of C = \frac{0.386}{7.29}

= 0.053 M

                        A            +            B           ⇄        C

Initial               0                           0                      0.530    

Change          +x                          +x                       - x

Equilibrium      x                           x                      (0.0530 - x)

K = \frac{[C]}{[A][B]}

where

K is given as ; 78.2 atm-1.

So, we have:

78.2=\frac{[0.0530-x]}{[x][x]}

78.2= \frac{(0.0530-x)}{(x^2)}

78.2x^2= 0.0530-x

78.2x^2+x-0.0530=0  

Using quadratic formula;

\frac{-b+/-\sqrt{b^2-4ac} }{2a}

where; a = 78.2 ; b = 1 ; c= - 0.0530

= \frac{-b+\sqrt{b^2-4ac} }{2a}   or \frac{-b-\sqrt{b^2-4ac} }{2a}

= \frac{-(1)+\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}  or \frac{-(1)-\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}

= 0.0204  or -0.0332

Going by the positive value; we have:

x = 0.0204

[A] = 0.0204

[B] = 0.0204

[C] = 0.0530 - x

     = 0.0530 - 0.0204

     = 0.0326

Total number of moles at equilibrium = 0.0204 +  0.0204 + 0.0326

= 0.0734

Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT

if we make V the subject of the formula; we have:

V = \frac{nRT}{P}

where;

P (pressure) = 1 atm

n (number of moles) = 0.0734 mole

R (rate constant) = 0.0821 L-atm/mol-K

T = 273.15 K  (fixed constant temperature )

V (volume) = ???

V=\frac{(0.0734*0.0821*273.15)}{(1.00)}

V = 1.64604

V ≅ 1.65 L

3 0
3 years ago
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aliya0001 [1]
The answer is the first one, Xe
5 0
3 years ago
22.5 g of silver nitrate reacts with excess magnesium bromide, determine the mass
Setler [38]

Answer:

9.82 g of Mg(NO₃)₂

Explanation:

Let's determine the reaction:

2AgNO₃  +  MgBr₂  → Mg(NO₃)₂  +  2AgBr

2 moles of nitrate silver reacts with MgBr₂ in order to produce 1 mol of magnesium nitrate and silver bromide.

We determine the moles of AgNO₃

22.5 g . 1mol / 169.87g = 0.132 moles

Ratio is 2:1.

2 moles of silver nitrate can produce 1 mol of magnesium nitrate

Then, our 0.132 moles may produce (0.132 . 1)/ 2 = 0.0662 moles

We convert moles to mass:

0.0662 mol . 148.3 g/ mol = 9.82 g

6 0
3 years ago
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