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Ivan
3 years ago
7

In liquid water, each molecule is hydrogen-bonded to approximately 3.4 molecules of water. What effect would freezing water have

on the number of hydrogen bonds? Heating water?
Chemistry
1 answer:
WINSTONCH [101]3 years ago
3 0

Answer:

1. Freezing water: The number of hydrogen bonds increase.

2. Heating water: Hydrogen bonds break

Explanation:

In the liquid state, hydrogen bonds are continuously formed and broken and the average amount is 3.4.  During freezing, molecule's energy decreases and there is less energy to break hydrogen bonds and for this reason, these bonds are more stable and average hydrogend bonds are close to 4 bonds.

When liquid water is heated, the kinetic energy raises and it causes the hydrogen bonds to break.

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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
from the balanced equation we see that h20 is 1 to 1 mole ratio the number of moles of acid and base. What is the number of mole
forsale [732]

Answer:

0.0498 mol

Explanation:

Number of moles = concentration in mol/L × volume in L

Concentration = 1 M = 1 mol/L

Volume = 49.8 mL = 49.8/1000 = 0.0498 L

Number of moles = 1×0.0498 = 0.0498 mol

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When liquid water evaporates to gaseous water
alina1380 [7]

Answer: liquid water 2

Explanation:

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hich quantity describes how spread out a set of data is? O A. Standard deviation O B. Percent error O C. Line of best fit D. Mea
juin [17]

The answer to this is B.

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4 years ago
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