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3 years ago

What is the enthalpy of fusion of water?

Chemistry

1 answer:

malfutka [58]3 years ago

5 0

Answer:

d- 334 kJ/g.

Explanation:

You can detect it from the units of the different choices.

a- has the unit J/g.°C that is the unit of the specific heat capacity (c).

b- has the unit Kelvin that is the unit of temperature.

c- has the unit g/mol which is the unit of the molar mass.

d- has the unit kJ/g which is the unit of the enthalpy divided by the no. of rams that is the specific entha;py of fusion.

<em>So, the right choice is: d- 334 kJ/g.</em>

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2. Classify the following solutions as acidic, basic, or neutral at 25OC.

PilotLPTM [1.2K]

Answer: a) pH = 13.00 : basic

b) $[H_3O^+]=1.0\times 10^{-12}$ : basic

c) pOH = 5.00 : basic

d) $[OH^-]=1.0\times 10^{-9}$ : acidic

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

$pH=-\log [H_3O^+]$

Acids have pH ranging from 1 to 6.9 and bases have pH ranging from 7.1 to 14.Neutral substances have pH of 7.

a) pH = 13.00

As pH is more than 7, the solution is basic.

b) $[H_3O^+]=1.0\times 10^{-12}$

Putting in the values:

$pH=-\log[1.0\times 10^{-12}]$

$pH=12$

As pH is more than 7, the solution is basic.

c) pOH = 5.00

$pH+pOH=14.0$

$pH=14.0-5.00=9.00$

As pH is more than 7, the solution is basic.

d) $[OH^-]=1.0\times 10^{-9}$

Putting in the values:

$pOH=-\log[1.0\times 10^{-9}]$

$pOH=9.00$

$pH+pOH=14.0$

$pH=14.0-9.00=5.00$

As pH is less than 7, the solution is acidic.

3 0

3 years ago

Identify the correct coefificients to balance it<br> -C3H8+O2 to -CO2 +-H2O

Iteru [2.4K]

Answer:

${\rm 1\; C_3H_8} + {\rm 5\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}$ .

Explanation:

${\rm ?\; C_3H_8} + {\rm ?\; O_2} \to {?\; \rm CO_2} + {?\; \rm H_2O}$ .

Among the four species in this reaction, $\rm C_3H_8$ is species with the largest number of atoms per molecule. Assume that the coefficient of this compound is $1$ .

${\rm 1\; C_3H_8} + {\rm ?\; O_2} \to {?\; \rm CO_2} + {?\; \rm H_2O}$ .

Number of atoms on the left-hand side of the reaction:

$\rm C$ : $1 \times 3 = 3$ .
$\rm H$ : $1 \times 8 = 8$ .
$\rm O$ : not found yet.

By the conservation of atoms, the number of atoms on the right-hand side of the reaction should match those on the left-hand side. In this reaction, $\rm CO_2$ is the only product with carbon atoms, whereas $\rm H_2O$ is the only product with hydrogen atoms. These $3$ carbon atoms and $8$ hydrogen atoms would correspond to:

$3 / 1 = 3$ $\rm CO_2$ molecules, and
$8 / 2 = 4$ $\rm H_2O$ molecules.

${\rm 1\; C_3H_8} + {\rm ?\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}$ .

Number of atoms on the right-hand side of the reaction:

$\rm C$ : $3 \times 1 = 3$ .
$\rm H$ : $4 \times 2 = 8$ .
$\rm O$ : $3 \times 2 + 4 \times 1 = 10$ .

The number of $\rm O$ atoms on the left-hand side should match those on the right-hand side. In this reaction, $\rm O_2$ is the only reactant with $\rm O\!$ atoms. These $10$ $\rm \! O$ atoms would correspond to: