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3 years ago

The minimum amount of 100-degree-c steam needed to melt 1 gram of 0-degree-celsius ice is

1 answer:

7 0

The mass of ice to be melted is
1 g = 10⁻³ kg.

Let m = the mass of steam required to melt the ice.

For water,
The latent heat of vaporization is 2260 kJ/kg
The latent heat of fusion is 334 kJ/kg

The steam should give up its latent heat of vaporization to melt the ice, which gives up its latent heat of fusion.
Therefore
(m kg)*(2260 kJ/kg) = (10⁻³ kg)*(334 kJ/kg)
m = 1.478 x 10⁻⁴ kg = 0.1478 g

Answer: 0.148 g

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PLEASE HELP CHEMISTRY

eduard

When sodium carbonate is dissolved in water, the equation is $Na_2CO_3 (s) --- > 2Na^+ (aq) + CO_3^{2-} (aq)$ .

When carbon dioxide is placed in water, aqueous carbon dioxide is formed: $CO_2 (g) --- > CO_2(aq)$

<h3>Dissolution of compounds in water</h3>

Some compounds are water-soluble, some are just partially soluble, while others are insoluble in water. Some soluble or partially soluble substances dissociate in water into their component ions. These substances are said to be ionic.

Sodium carbonate, like every other sodium salt, is soluble in water. It dissolves in water to form an aqueous solution of sodium carbonate.

While in solution, sodium carbonate dissociates into its component ions according to the following equation:

$Na_2CO_3 (s) --- > 2Na^+ (aq) + CO_3^{2-} (aq)$

Carbon dioxide, on the other hand, does not dissociate in water. Instead, it dissolves in water where most of it remains as aqueous carbon dioxide in equilibrium with a small amount of hydronium ion and hydrogen carbonate ion.

Since the hydronium and hydrogen carbonate ions formed are so minute, the equation of the reaction can be written as: $CO_2 (g) --- > CO_2(aq)$

More on the dissolution of substances can be found here: brainly.com/question/28580758

#SPJ1

7 0

1 year ago

Three 6−l flasks, fixed with pressure gauges and small valves, each contain 6 g of gas at 276 k. flask a contains ch4, flask b c

Margaret [11]

First, please check the missing part in your question in the attachment.

a) So first, the Rank of pressure:

according to this formula PV = nRT and when n = m/Mw

PV = m/Mw * R*T

when we have the same mass m and the same V volume so P will proportional with the mole weight M as when the M is smaller the pressure will be greater

when Mw of H2(A) = 2 g / Mw of He (B) = 4 g and Mw of CH4(C) = 16 g

∴ Pressure :

(A) > (B) > (c)

B) The rank of average molecular kinetic energy:

when K = 3/2 KB T

when K is the average kinetic energy per molecule of gas

and KB is Boltzmann's constant

and T is the temperature (K)

So from this equation, we can know that K only depends on T value, and when we have the T constant here for A, B, and C So the rank of K will be like the following:

∴ A = B = C

C) the rank of diffusion rate after the valve is opened:

according to this formula:

R2/R1 = √M1/M2

from this equation, we can see that diffusion is proportional to the reciprocal of the molecular mass M so,

when Mw H2 (A) = 2 g & Mw He(B) = 4 g & CH4 (C) = 16 g

∴ the rank of diffusion:

A > B > C

D) The rank of the Total kinetic energy of the molecules:

when we have the Mw different so it will make the no.of molecules differs as when the Mw is low the no.of molecules will be hight, and when the average molecular kinetic energy equals. so the total kinetic energy will depend on no. of molecules

∵ Mw A < Mw B < Mw C

∴no .of molecules of A > B >C

∴ the rank of total kinetic energy is:

A > B > C

e) the rank of density:

when ρ = m/ v

and m is the mass & v is the volume and we have both is the same for A, B, and C

so the density also will be the same, ∴ the rank of the density is:

A = B = C

F) the rank of the collision frequency:

as the no.of molecules increase the collision frequency increase and depend also on the velocity and it's here the same.

∴ Collision frequency will only depend on the no.of molecules

we have no.of molecules of A > B > C as Mw A < B < C

∴the rank of the collision frequency is:

A > B > C

6 0

3 years ago

The volume of a gas is reduced from 4 L to 0.5 L while the temperature is held constant. How does the gas pressure change?

liberstina [14]

Answer:

It increases by a factor of eight

Explanation:

When temperature is held constant, gas pressure changes according the volume, in undirectly proportion.

Volume increases → Pressure decreases

Volume decreases → Pressure increases

As volume gas, was reducted from 4L to 0.5L, it was reduced by 1/8, so the pressure gas was increased by a factor of eight.

5 0

3 years ago

Is a faded picture a chemical or physical change?

nalin [4]

It is a chemical change

5 0

3 years ago

Read 2 more answers

NEED HELP ASAP WITH THESE QUESTIONS GIVING FAIR AMOUNT OF POINTS IF HELPED WITH ALL QUESTIONS Violet light has a wavelength of 4

scoray [572]

Answer:

a) 7.14e19 Hz

b) 2.298e-27 J

c) 2.793e-19 J; 7.117e9 nm

d) 7.5e14 Hz; 4.96e-19 J

e) 6.2947e-18 J; 31.6 nm

f) 2.21e-22 J

g) 7.1e-19 J; 1.1e15 Hz

h) 3.422e-19 J; 581 nm

i) 4.2e14 Hz

j) 1.92e8 m

k) 7.14e16 Hz; Ultraviolet

Explanation:

Frequency: ν Wavelength: λ Energy: E Speed of light: C (3.00e8) Planck's Constant: h (6.626e-34)

ν -> λ λ = C/ν

λ -> ν ν = C/λ

For either of these equations, wavelength must be converted to meters or nanometers, depending on the equation.

For ν -> λ, after doing the equation, convert the wavelength into nanometers by dividing by 1e-9.

For converting λ -> ν, convert the wavelength into meters by multiplying by 1e-9.

For energy: E = hν = hc/λ

Now that the setup is out of the way:

a) Violet light has a wavelength of 4.20 x 10-12 m. What is the frequency?

λ -> ν ν = C/λ

$\frac{3.00e8}{4.20e-12}$ = 7.14e19 Hz

b) A photon has a frequency (n) of 3.468 x 106 Hz. Calculate its energy

E = hν = hc/λ

(6.626e-34) (3.468e6) = 2.298e-27 J

c) Calculate the energy (E) and wavelength (l) of a photon of light with a frequency of 4.215 x 1014 Hz.

E = hν = hc/λ

(6.626e-34) (4.215e14) = 2.793e-19 J

ν -> λ λ = C/ν

$\frac{3.00e8}{4.215e14}$ = 7.117 m

$\frac{7.117m}{1}*\frac{1nm}{1e-9m}$ = 7.117e9 nm

d) Calculate the frequency and the energy of blue light that has a wavelength of 400 nm (h = 6.62 x 10-34 J-s).

λ -> ν ν = C/λ

$\frac{400 nm}{1} *\frac{1e-9m}{1nm}$ = 4e-7 m

$\frac{3.00e8}{4e-7}$ = 7.5e14 Hz

E = hν = hc/λ

(6.626e-34) (7.5e14) = 4.96e-19 J

e) Calculate the wavelength and energy of light that has a frequency of 9.5 x 1015 Hz.

ν -> λ λ = C/ν

$\frac{3.00e8}{9.5e15}$ = 3.16e-8 m

$\frac{3.16e-8m}{1}*\frac{1nm}{1e-9m}$ = 31.6 nm

E = hν = hc/λ

(6.626e-34) (9.5e15) = 6.2947e-18 J

f) A photon of light has a wavelength of 0.090 cm. Calculate its energy.

E = hν = hc/λ

Convert the wavelength from cm to meters:

$\frac{0.090cm}{1} *\frac{1m}{100cm}$ = 9e-4 m

$\frac{(6.626e-34)(3.00e8)}{9e-4}$ = 2.21e-22 J

g) Calculate the energy and frequency of red light having a wavelength of 2.80 x 10-5 cm.

E = hν = hc/λ

Convert the wavelength from cm to meters:

$\frac{2.80e-5cm}{1} *\frac{1m}{100cm}$ = 2.8e-7 m

$\frac{(6.626e-34)(3.00e8)}{2.8e-7}$ = 7.1e-19 J

λ -> ν ν = C/λ

Convert the wavelength from cm to meters:

$\frac{2.80e-5cm}{1} *\frac{1m}{100cm}$ = 2.8e-7 m

$\frac{3.00e8}{2.8e-7}$ = 1.1e15 Hz

h) Calculate the energy (E) and wavelength (l) of a photon of light with a frequency of 5.165 x 1014 Hz.

E = hν = hc/λ

(6.626e-34) (5.165e14) = 3.422e-19 J

ν -> λ λ = C/ν

$\frac{3.00e8}{5.165e14}$ = 5.81e-7 m

$\frac{5.81e-7m}{1}*\frac{1nm}{1e-9m}$ = 581 nm

i) The wavelength of green light from a traffic signal is centered at 7.20 x 10-5 cm. Calculate the frequency.

λ -> ν ν = C/λ

Convert the wavelength from cm to meters:

$\frac{7.20e-5 cm}{1} *\frac{1m}{100cm}$ = 7.2e-7 m

$\frac{3.00e8}{7.2e-7}$ = 4.2e14 Hz

j) If it takes 1.56 seconds for radio waves (which travel at the speed of light) to reach the moon from Earth, how far away is the moon?

All we want to do here is to convert frequency (speed) to wavelength (distance). This problem requires a bit of thought, but it isn't bad once you realize that frquency is speed and wavelength is distance. It becomes just like the other problems after that. Also, I'll leave this distance in meters, but I think you can figure out how to convert it if it wants it in another unit.

One second is equal to 1 Hertz, so our frequency is 1.56 Hz.

ν -> λ λ = C/ν

$\frac{3.00e8}{1.56}$ = 1.92e8 m

The actual distance from the earth to the moon via google is 3.84e7, but sometimes problems like this will mess with the numbers to make sure that you didn't just look up the answer. I'm still pretty sure that this is right, however.

k) Calculate the frequency of light that has a wavelength of 4.20 x 10-9m. Identify the type of electromagnetic radiation.

First, we convert wavelength to frequency, as normal:

λ -> ν ν = C/λ

$\frac{3.00e8}{4.20e-9}$ = 7.14e16 Hz

Then we identify the electromagnetic wave type. You can look up a conversion chart for these on google, but since our frequency is in the e15 - e17 range, this light is considered ultraviolet.

5 0

3 years ago