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2 years ago

How much energy is required to melt 13.00 kg of ice at 273 K? The specific latent heat of fusion of ice is 3.36 x 10^5 J kg-1.

Chemistry

1 answer:

Lostsunrise [7]2 years ago

7 0

Hello!

The energy required to melt 13 kg of ice at 273k is 5040000 J

Explanation:

For solving this exercise, we are going to use the equation that relates the energy in Joules (E) with the Specific Latent Heat of Fusion (l) and the mass of ice (m):

$E= m * l=15 kg*3,36*10^{5} J/kg=5040000J=5040kJ$

Have a nice day!

You might be interested in

How much maleic anhydride would you need to react 178 mg of anthracene? Assume 1:1 ratio from maleic anhydride to anthracene.

Licemer1 [7]

Answer:

(1) 0.10 (2) 17.8 g

Explanation:

Since the reaction ratio is 1:1 what we need is to convert the given masses to moles and you will have the answer:

MW anthracene = 178.23 g/mol

MW maleic anhydride = 98.06 g/mol

a) mass anthracene = 178 mg x 1 g/ 1000 mg = 0.178 g anthracene

Moles anthracene = 0.178 g anthracene/ 178.23 g/mol

= 0.001 mol anthracene

0.001 mol anthracene x 1 mol maleic acid/mol anthracene

= 0.001 mol maleic anhydride

mass maleic anhydride = 0.001 mol x 98.06 g/mol = 0.10 g

b) moles maleic anhydride = 9.8 g/ 98.06 g/mol = 0.099 moles

0.099 moles maleic anhydride x 1 mol anthracene/mol maleic anhydride =

0.099 mol anthracene

g anthracene = 0.10mol x 178 g/mol = 17.8 g

8 0

2 years ago

A column is filled with four different liquids of different densities. A red liquid, a blue liquid, a green liquid, and a purple

Svet_ta [14]

Answer:

The order of liquid from top to bottom is as follow

Purple, green, red, blue

Explanation:

Chart of densities:

Red = 1.2 g/cm∧3
Blue = 1.6 g/cm∧3
green = 0.8 g/cm∧3
Purple = 0.1 g/cm∧3

Density:

Density is the ratio of mass and volume as follow

d = m/v

Summary:

Greater the density, it will be at bottom and vice versa.
Blue liquid has greater density so it will be at bottom
Purple liquid has low density, so it will be at top.

5 0

2 years ago

Read 2 more answers

Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll

makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature $T_1$ = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

The truncated viral equation is expressed as:

$\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}$

where; B = - $152.5 \ cm^3 /mol$ C = -5800 $cm^6/mol^2$

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

$\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

$4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

Multiplying through with V² ; we have

$4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0$

$4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0$

V = 2250.06 cm³ mol⁻¹

Z = $\frac{PV}{RT}$

Z = $\frac{1800*2250.06}{8.314*10^3*523.15}$

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

$T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345$

$T__{\gamma}} = \frac{T}{T_c}$

$T__{\gamma}} = \frac{523.15}{647.1}$

$T__{\gamma}} = 0.808$

$P__{\gamma}} = \frac{P}{P_c}$

$P__{\gamma}} = \frac{1800}{22055}$

$P__{\gamma}} = 0.0816$

$B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}$

$B_o = 0.083 - \frac{0.422}{0.808^{1.6}}$

$B_o = 0.51$

$B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}$

$B_1 = -0.282$

The compressibility is calculated as:

$Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}$

$Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}$

Z = 0.9386

$V= \frac{ZRT}{P}$

$V= \frac{0.9386*8.314*10^3*523.15}{1800}$

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At $T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa$

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V = 0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = $\frac{PV}{RT}$

Z = $\frac{1800*10^3 *0.1249}{729.77*523.15}$

Z = 0.588

3 0

2 years ago

The highest principal enegy level of period 2 elements is 2. Period 3 elements all have six 3p electrons. Period 4 elements have

Dmitrij [34]

Explanation:

The highest principal energy level of period 2 elements is 2. (True)

The highest principle energy level of elements of any period is equal to period number. So the given statement is true.

Period 3 elements all have six 3p electrons – (False)

For example, sodium and sulfur are period 3 elements but do not contain six 3p electrons. Only Argon of period possesses six p-electrons.

Period 4 elements have an inner electron configuration of [Ar]- (True)

Atomic number of elements belonging to period 4 have atomic number more than 18 (atomic number of Ar). So they must have an inner electron configuration of [Ar]

The valence electrons of group 5A elements are in the 6s subshell – [FALSE]

Elements of group 5A are nitrogen, arsenic, antimony and bismuth. They are p-block elements. Therefore, valence electrons are present in p-subshell.

Group 8A elements have full outer principal s and p subshells –(True)

Group 8A elements are also known as noble gas element. Octets of noble gases are complete and hence, have full outer principal s and p subshells.

The valence electrons of group 2A elements are in an s subshell – (True)

Elements of group 2A are barium, magnesium, calcium, etc. They belong to s-block elements. Therefore, their valence electrons will be in s-subshell.

The highest principal energy level of period 3 elements is 4 – (False)

The highest principle energy level of elements of any period is equal to period number.

For example, sodium is present in 3rd period, so its principle quantum number will be 3.

Period 5 elements have an inner electron configuration of [Xe] – (False)

Period 5 elements have an inner electron configuration of [Kr]. The first member of this period is rubidium. 37 and that of Kr is 36.

6 0

2 years ago

Which of the following shows an accurate combustion reaction?

Jet001 [13]

Answer:

Cu + O₂ → CuO₂

Explanation:

A combustion reaction is the reaction with oxygen along with the release of energy in form of heat or light.

Organic compounds (like CH₄) undergo combustion forming water and CO₂.

The combustion reaction of CH₄ is:

CH₄ + 2O₂ → CO₂ + 2H₂O

Hence, the first equation from the choices is not showing the combustion reaction of CH₄.

Not only organic compounds can undergo combustion. Metals and no metals can undergo combustion, i.e. metals and no metals can react with oxygen releasing light or heat.

The reaction of copper and oxygen (second choice) is a combustion reaction:

Cu + O₂ → CuO₂

The formation of water (2H₂ + O₂ → 2H₂O) is other example of a combustion reaction where no organic compounds are involved.

On the other hand, the other two equations from the choice list are not reactions with oxygen, so they do not show combustion reactions.

5 0

2 years ago