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2 years ago

If 14.5 kJ of heat were added to 485 g of liquid water, How much would its temperature increase?

Chemistry

1 answer:

olga55 [171]2 years ago

3 0

Answer: The increase in temperature is 7.14°C

Explanation:

To calculate the change in temperature, we use the equation:

$q=mc\Delta T$

where,

q = heat absorbed = 14.5 kJ = 14500 J (Conversion factor: 1 kJ = 1000 J)

m = mass of water = 485 g g

c = specific heat capacity of water = 4.184 J/g.°C

$\Delta T$ = change in temperature = ?

Putting values in above equation, we get:

$14500J=485g\times 4.184J/g.^oC\times \Delta T\\\\\Delta T=7.14^oC$

Hence, the increase in temperature is 7.14°C

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The acid-dissociation constant, ka, for gallic acid is 4.57 ⋅ 10-3. what is the base-dissociation constant, kb, for the gallate

LiRa [457]

Hello!

To determine the Kb of gallic acid is actually very simple.

The dissociation reaction of Gallic Acid (HGal) is the following:

HGal+H₂O ⇄ H₃O⁺ + Gal⁻

The equation for converting from Ka to Kb is the following:

$Ka*Kb=Kw \\ \\ Kb= \frac{Kw}{Ka}= \frac{1*10^{-14} }{4,57*10^{-3} }=2,19 * 10^{-12}$

So, the Kb is 2,19*10⁻¹²

Have a nice day!

3 0

2 years ago

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Wet steam at 1100 kPa expands at constant enthalpy (as in a throttling process) to 101.33 kPa, where its temperature is 105°C. W

nikklg [1K]

Answer:

There are 3 steps of this problem.

Explanation:

Step 1.

Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.

Step 2.

Enthalpy of saturated liquid Haq = 781.124 J/g

Enthalpy of saturated vapour Hvap = 2779.7 J/g

Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g

Step 3.

In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy

So, H1=H2

H2= (1-x)Haq+XHvap.........1

Putting the values in 1

2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}

= 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)

1904.976 (J/g) = x1998.576 (J/g)

x = 1904.976 (J/g)/1998.576 (J/g)

x = 0.953

So, the quality of the wet steam is 0.953

7 0

2 years ago

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Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation N2O4(g) 2N

Amanda [17]

Answer : The correct option is, (a) 0.44

Explanation :

First we have to calculate the concentration of $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

Now we have to calculate the dissociated concentration of $N_2O_4$ .

The balanced equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(aq)$

Initial conc. 1.0 M 0

At eqm. conc. (1.0-x) M (2x) M

As we are given,

The percent of dissociation of $N_2O_4$ = $\alpha$ = 28.0 %

So, the dissociate concentration of $N_2O_4$ = $C\alpha=1.0M\times \frac{28.0}{100}=0.28M$

The value of x = $C\alpha$ = 0.28 M

Now we have to calculate the concentration of $N_2O_4\text{ and }NO_2$ at equilibrium.

Concentration of $N_2O_4$ = 1.0 - x = 1.0 - 0.28 = 0.72 M

Concentration of $NO_2$ = 2x = 2 × 0.28 = 0.56 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.56)^2}{0.72}=0.44$

Therefore, the equilibrium constant $K_c$ for the reaction is, 0.44

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2 years ago

Draw a structure containing only carbon and hydrogen that is a stable alkyne of five carbons containing a ring.

stepladder [879]

Answer:

Ethynylcyclopropane is the stable isomer for given alkyne.

Explanation:

In order to solve this problem we will first calculate the number of Hydrogen atoms. The general formula for alkynes is as,

CₙH₂ₙ₋₂

Putting value on n = 5,

C₅H₂.₅₋₂

C₅H₈

Also, the statement states that the compound contains one ring therefore, we will subtract 2 hydrogen atoms from the above formula i.e.

C₅H₈ ------------(-2 H) ----------> C₅H₆

Hence, the molecular formula for given compound is C₅H₆

Below, 4 different isomers with molecular formula C₅H₆ are attached.

The first compound i.e. ethynylcyclopropane is stable. As we know that alkynes are sp hybridized. The angle between C-C-H in alkynes is 180°. Hence, in this structure it can be seen that the alkyne part is linear and also the cyclopropane part is a well known moiety.

Compounds 3-ethylcycloprop-1-yne, cyclopentyne and 3-methylcyclobut-1-yne are highly unstable. The main reason for the instability is the presence of triple bond in three, five and four membered ring. As the alkynes are linear but the C-C-H bond in these compound is less than 180° which will make them highly unstable.

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2 years ago

When balanced, which equation would have the coefficient 3 in front of any of the reactants? Zn + HCl ZnCl2 + H2 H2SO4 + B(OH)3

Gwar [14]

The correct answer is B. H2SO4 + B(OH)3 B2(SO4)3 + H2O

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2 years ago

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