Calculate the number of moles of water molecules in 12 dm' of water
vapour at STP.

Answers

Answer 1

Answer:

[tex]22.4 \: {dm}^{3} \: are \: occupied \: by \: 1 \: mole \\ 12 \: {dm}^{3} \: will \: be \: occupied \: by \: ( \frac{12}{22.4} ) \: moles \\ = 0.536 \: moles[/tex]

Answer 2

Number of moles of water molecules in 12 dm^3 of water =667 moles

What is mole ?

A mole is defined as the amount of a substance that contain exactly 6.022 * 10 ^23 elementary entities of the given substance

1 dm ^3 = 1 L

12 dm^3 = 12 L

density of water = mass / volume

             mass = 1000kg/m^3 * 0.012 m^3

             mass = 12 kg = 12 * 10 ^ 3 g

moles = mass / molar mass

moles =  12 * 10 ^ (3 ) g / 18 g/mole

          = 0.667 *10 ^ (3 )

          = 667 moles

Number of moles of water molecules in 12 dm^3 of water =667 moles

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Related Questions

A rubber ball is dropped from a height of 1.5 m
(a) How long does it take to hit the ground?

Answers

Answer:

t =0.6

Explanation:

So this is just a basic kinematics where the only force acting on the ball is gravity (9.8 m/s2 [down] and air resistance is negligible.

∆d=vi∆t+1/2at2

Because you're dropping the ball, initial velocity is 0

1.5= 0*(∆t) + 1/2a*t2

After some rearranging, t= root ((2*∆d)/a) Therefore t= 0.5532833352 or root(15)/7 t=0.6s

in which reaction is there a transformation of mass into energy

Answers

Answer:

There is a transformation of mass into energy un nuclear fusion

40 points! Will give brainliest!

Which of the following describes an advantage of AC electricity over DC electricity?

A) AC is found in most low voltage operations.
B) AC is provided as strong, short bursts of electricity.
C) AC can be transported over long distances.
D) AC can be used is small electronic devices.

Answers

It’s C “ Ac can be transported over long distances.”

What is gravitational force? -,-​

Answers

Answer:

The force of attraction between all masses in the universe; especially the attraction of the earth's mass for bodies near its surface is called gravitational force.

An object of mass 25kg is falling from the height h=10 m. calculate
a. The total energy of an object at h=10m.
b. Potential energy of the object when it is at h= 4m
c. Kinetic energy of the object when it is at h= 4m
d. What will be the speed of the object when it hits the ground?​

Answers

Answer:

a=2500J,b=1000K,c=1000J,d=14.142m/s

Explanation:

V²=U²+2gh

V²=0 + 2×10×10=200m/s

a).kinetic energy=(1/2)mv²=(1/2)25×200=2500

potential energy=mgh

p.e=25×10×10=2500J

pe+ke=2500+2500=5KJ

b).mgh=25×10×4=1000J

c). V²=U²+2gh

V²=0+2×10×4

V²=80

kinetic energy=(1/2)mv²

=(1/2)25×80

=1KJ

d). From my first paragraph V²=200

V=√200

V=14.142m/s

What is the speed of the wave above of the frequency is 7.0 hertz

Answers

Answer:

Umm

Explanation:

Energy can be transferred from one _______ to another and one __________ to another. *

Answers

Answer:

Source, form

Explanation:

Answer:

Explanation:

Energy can be transferred from one _source_ to another and one __form____ to another.

How expensive is genetic counseling and reproductive technology?

Answers

Answer:

Genetic counseling without testing cost on average $213, whereas counseling, testing, and disclosure of results totaled $2057. A brief physician-based counseling instead of genetic counselor-based counseling would produce only small reductions in total costs.

hope it helps, please give me brainliest.

1. How would the forces from a header with such a light soccer ball cause a concussion? Draw
Two free body diagrams showing how the amount of peak force on the head would compare
to the amount of peak force on a soccer ball in a header that causes a collision.

Answers

Answer:

soccer when the ball hits an unprepared player in the head. He also gave examples of concussions occurring when players accidentally knock their heads into other players while attempting to head the ball, particularly if they are attempting to flick the ball backwards.

Explanation:

Heading in soccer can increase your risk of concussions. Over time, repeated subconcussive injuries can also accumulate and cause brain damage.

what is the focalength of the combination of two thin renses of power -5D and -2D placeci in contact with each other?​

Answers

Answer:

f = - 0.143 m = - 14.3 cm

Explanation:

First, we will calculate the power of the combination of lenses:

P = P₁ + P₂

where,

P = Power of Combination = ?

P₁ = Power of first lens = - 5D

P₂ = Power of second lens = - 2D

Therefore,

P = - 5D - 2D

P = - 7D

Now, the focal length can be given as:

[tex]f = \frac{1}{P} \\\\f = \frac{1}{-\ 7\ D} \\\\[/tex]

f = - 0.143 m = - 14.3 cm

Negative focal length indicates that combination will act as diverging lens.

Q: Winding roads and sharp curves are inherently dangerous
because -

A- You can't see far ahead

B- The speed limit is so slow

C- There may be other vehicles traveling in the
opposite direction

D- You may struggle to stay at the same speed

Answers

Answer:

A) you can't see far ahead

D. You may struggle to stay at the same speed.

Winding roads and sharp curves are dangerous because of centrifugal forces while taking curves, whose classic formula is described below:

[tex]F = m\cdot \frac{v^{2}}{r}[/tex] (1)

Where:

[tex]F[/tex] - Centrifugal force, in newtons.[tex]v[/tex] - Speed, in meters per second.[tex]r[/tex] - Radius of curvature, in meters per second.

Sharps curves are characterized by low radii of curvature. Given that centrifugal force is inversely proportional to the square radius of curvature and it is opposed to friction force between tires and pavement, if centrifugal force is higher than maximum static friction force, then a car accident could if driver does not decelerate.

Therefore, correct choice is D.

We kindly invite to see this question on centrifugal forces: https://brainly.com/question/545816

For variables control, a circuit voltage will be measured using a sample of five circuits. The past average voltage for samples of 5 units has been 3.4 volts, and the range has been 1.3 volts.

Required:
What would the upper and lower control limits be for the resulting control charts (average and range)?

Answers

Answer:

Average :

UCL = 4.15

LCL = 2.65

Range :

UCL = 2.75

LCL = 0

Explanation:

Given :

Sample size, n = 5

Average, X = 3.4

Range, R = 1.3

A2 for n = 5 ; equals 0.577 ( X chart table)

For the average :

Upper Control Limit (UCL) :

X + A2*R

3.4 + 0.577(1.3) = 4.1501

Lower Control Limit (LCL) :

X - A2*R

3.4 - 0.577(1.3) = 2.6499

FOR the range :

Upper Control Limit (UCL) :

UCL = D4*R

D4 for n = 5 ; equals = 2.114

UCL = 2.114*1.3 = 2.7482

Lower Control Limit (LCL) :

LCL = D3*R

D3 for n = 5 ; equals = 0

LCL = 0 * 1.3 = 0

What is the period of a wave, with a frequency of 0.75 Hertz?

Answers

Answer:

wavelength  4 cm

Explanation:

PLEASE HELP ME ASAP!!!!

Answers

Both mic and speaker

Magnesium hydroxide is a common ___?

Answers

Answer:

Hello There!!

Explanation:

The answer is=> It is a common component of antacids.

hope this helps,have a great day!!

~Pinky~

Explain how newtons first law applies to the image to the left.

Answers

can i get the images plz

A 60-W light bulb radiates electromagnetic waves uniformly in all directions. At a distance of 1.0 mm from the bulb, the light intensity is Io, the average energy density of the waves is u0, and the rms electric and magnetic field values are Eo and Bo, respectively.

Required:
a. What is the light intensity?
b. What is the average energy density of the waves?
c. What is the rms magnetic field value?

Answers

Answer:

The appropriate solution is:

(a) [tex]\frac{1}{4}(I_o)[/tex]

(b) [tex]\frac{1}{4} (u_o)[/tex]

(c) [tex]\frac{1}{2}B_o[/tex]

Explanation:

According to the question, the value is:

Power of bulb,

= 60 W

Distance,

= 1.0 mm

Now,

(a)

⇒  [tex]\frac{I}{I_o} =\frac{r_o_2}{r_2}[/tex]

On applying cross-multiplication, we get

⇒  [tex]I=I_o\times \frac{1_2}{2^2}[/tex]

⇒     [tex]=I_o\times \frac{1}{4}[/tex]

⇒     [tex]=\frac{1}{4} (I_o)[/tex]

(b)

As we know,

⇒ [tex]\frac{u}{u_o} =\frac{I}{I_o}[/tex]

By putting the values, we get

⇒ [tex]u=\frac{1}{4}(u_o)[/tex]

(c)

⇒ [tex]\frac{B^2}{B_o^2} =\frac{u}{u_o}[/tex]

         [tex]=\frac{I}{I_o}[/tex]

⇒ [tex]B=B_o\times \sqrt{\frac{1}{4} }[/tex]

⇒     [tex]=\frac{1}{2}(B_o)[/tex]

One way to overcome gravity is

Answers

Answer:

the ultimate way to overcome the Earth's gravity is to leave the Earth's gravitational influence altogether. The force of gravity reduces in inverse proportion to the square of the distance from the centre of the Earth.

Explanation:

A block with mass M is placed on an inclined plane with slope angle q and is connected to a second hanging block with mass m by a cord passing over a small, frictionless pully. The coefficient of static friction μs and the coefficient of kinetic friction is μk. For what range values of of m will the blocks remain at rest if they are released from rest.

Answers

Answer:

The mass of the block m is:

[tex]m=M(sin(\theta)+\mu_{s}cos(\theta))[/tex]

Explanation:

Let's analyze the block by parts

For the block M

[tex]T-W_{x}-f_{f}=0[/tex] (1)

Where:

T is the tensionW(x) is the component of the weight in the x-directionF(f) is the friction force

[tex]T-Mgsin(\theta)-\mu_{s}N=0[/tex]  

[tex]T-Mgsin(\theta)-\mu_{s}Mgcos(\theta)=0[/tex]  

For the block m

[tex]T-W=0[/tex]

[tex]T=mg[/tex] (2)

Now, let's combines equation (1) and (2):

[tex]mg-Mgsin(\theta)-\mu_{s}Mgcos(\theta)=0[/tex]  

Finally, let's solve it for block m.

[tex]mg-Mg(sin(\theta)+\mu_{s}cos(\theta))=0[/tex]

[tex]m=M(sin(\theta)+\mu_{s}cos(\theta))[/tex]

I hope it helps you!

Explain how Law 1 applies to the image to the left.

Answers

Answer:

12

Explanation:

Answer:

12?

Explanation:

The 5 kg red book is about to fall from a 2.1 m high bookshelf.

What is its PE?

Use g = 9.8 m/s2 for the acceleration due to gravity.

Answers

Answer:

w = d×f = 2.1 × ( 5×9.8)= 102.9J

Explanation:

because the question is asking for the energy PE

Answer:

[tex]\boxed {\boxed {\sf 102.9 \ J}}[/tex]

Explanation:

Potential energy is the energy an object possesses due to position. The formula is:

[tex]PE= m \times g \times h[/tex]

where m is the mass, g is the acceleration due to gravity, and h is the height.

We know the book is 5 kilograms and is on a 2.1 meter high bookshelf. Since this is on Earth, the acceleration due to gravity is 9.8 meters per square second.

m= 5 kg g= 9.8 m/s²h= 2.1 m

Substitute the values into the formula.

[tex]PE= 5 \ kg \times 9.8 \ m/s^2 \times 2.1 \ m[/tex]

Multiply the numbers together.

[tex]PE= 49 kg *m/s^2 *2.1 \ m[/tex]

[tex]PE= 102.9 \ kg*m^2/s^2[/tex]

1 kilogram square meter per square second ( 1 kg*m²/s²) is equal to 1 Joule (J). So, our answ er of 102.9 kg*m²/s² is equal to 102.9 J

[tex]PE= 102.9 \ J[/tex]

The book has 102.9 Joules of gravitational potential energy.

what are conductors and insulators​

Answers

Conductors are materials that permit electrons to flow freely from particle to particle. ... Conductors allow for charge transfer through the free movement of electrons. In contrast to conductors, insulators are materials that impede the free flow of electrons from atom to atom and molecule to molecule.

A student pushes a laptop cart down the hallway by applying 20 Newtons force. The student pushes it 10 meters and
then stops at the water fountain with the cart and haves a sip of water. While he drinks from the water fountain, what
type of work is being done to the laptop cart.

Answers

Answer:

None.

Explanation:

By definition, work is a process that it happens when an applied force

        causes an object to change its position, i.e. to have a displacement.

        Since the laptop cart is at rest while he stops at the water fountain, no  

       net work done is on the laptop cart.

the official world land speed record is1228.0km/h are on October 15,1997,by Andy green in the jet engine car then express the speed in the meters per second​

Answers

Answer:

v = 341.11 m/s

Explanation:

Given that,

The official world land speed was recorded as 1228.0km/h on October 15,1997.

We need to find the speed in m/s.

We know that,

1 km = 1000 m

1 h = 3600 s

So,

[tex]v=1228\ km/h\\\\=1228\times \dfrac{1000\ m}{3600\ s}\\\\v=341.11\ m/s[/tex]

So, the required speed is 341.11 m/s.

A scientist is asking you what would happen to Radon-226 (represented by Ra on the graph and 226 is the atomic mass) if instead of a purple arrow after Radon-226 there was a single green arrow. What would the result of this reaction on Radon be?

Answers

The sky legit green today

Smoke detectors fall into two major classes. Ionization detectors, the most common units, contain two parallel electrodes that are typically separated by 3 cm with a 5-V potential difference across them. The air molecules between the electrodes are ionized by collisions with helium nuclei that are produced by a radioactive source. Most units are initially fueled with 60 million nuclei of radioactive americium 241 (half-life 430 years). The now-ionized air molecules drift toward one of the electrodes with an average speed of 0.1 m/s and thus support a small current between the two electrodes. Smoke particles that enter and combine with the ions reduce the current and initiate an alarm.
Photoelectric detectors, by contrast, contain a light-emitting diode that sends a beam of unpolarized light across a small chamber. The light beam usually has a wavelength of 6.0 × 10–7m and has an intensity of 1.0 × 10–3 W. When smoke particles enter the chamber, the light scatters in all directions. A photocell then senses either the increase in the scattered light or the reduced intensity of the light beam and sets off the alarm. The speed of light in air is 3.0 × 108 m/s.
Ionization detectors respond faster to the large smoke particles of flaming fires; photoelectric detectors sense the small particles of smoldering fires more quickly. Modern units have both types of detectors.
When fewer than 3.75 × 106 americium nuclei remain, the ionization smoke detector will not operate due to insufficient ionization. How much time will pass before there are this many nuclei remaining?
a. 1720 years
b. 2150 years
c. 4300 years
d. 6880 years

Answers

Answer:

1720 years is the amount of time that will pass.

Option a) 1720 years is the correct answer

Explanation:

Given the data in the question;

 Number of nuclei initially N₀  = 60 million = 60,000,000

After time t, Number of nuclei remaining N[tex]_{rem[/tex] = 3.75 × 10⁶

Also given that; half-life of radioactive americium [tex]t_{1/2[/tex] = 430 years.

so;

λ = ln2 / [tex]t_{1/2[/tex]

we substitute

λ = ln2 / 430 years

N[tex]_{rem[/tex] = N₀e^(-λt)

solve for t

t = 1/λ × ln( N₀/N[tex]_{rem[/tex]  )

so we substitute

t = 1 / (ln2 / 430 years) × ln( 60,000,000 / ( 3.75 × 10⁶ ) )

t = ( 430 years / ln2 ) × ln( 60,000,000 / ( 3.75 × 10⁶ ) )

t = ( 430 years / 0.693147 ) × ln( 16 )

t = 620.359 years × 2.7725887

t =  1720.0003 ≈ 1720 years

Therefore, 1720 years is the amount of time that will pass.

Option a) 1720 years is the correct answer

A box has base dimensions of 30 cm x 30 cm and a mass of 3 kg. Calculate to what
depth it sinks when placed in a tank of water (assume the sides are sufficiently high,
so that it floats). Density of water = 1000 kgm-3. Assume g=10 ms-2 for this problem.​

Answers

Answer:

60

Explanation:

A box has base dimensions of 30 cm x 30 cm and a mass of 3 kg. Calculate to what

depth it sinks when placed in a tank of water (assume the sides are sufficiently high,

so that it floats). Density of water = 1000 kgm-3. Assume g=10 ms-2 for this problem.​

Describe the formation of galaxies.

Answers

Today's cosmologists assume that matter was not uniformly distributed in the universe after the Big Bang. Dense places attract more matter than the surrounding area according to their gravitational forces. Over the course of billions of years, these gas agglomerations eventually led to the formation of the galaxies we see today.

A bat emits a sound at a frequency of 30.0 kHz as it approaches a wall. The bat detects beats such that the frequency of the echo is 900 Hz higher than the frequency the bat is emitting. The speed of sound in air is 340 m/s at emits a sound at a frequency of 30.0 kHz
(a) What is the speed of the bat?
(b) What is the wavelength of the sound that the bat hears15?

Answers

Answer:

a) the speed of the bat is 5.02 m/s

b) the wavelength of the sound that the bat hears is 0.011 m

Explanation:

Given the data in the question;

Frequency of sound emitted by a bat f = 30.0 kHz = 30000 Hz

detected frequency by the bat δf = 900 Hz

speed of sound in air c = 340 m/s

Let speed of sound and speed of bat be c and [tex]v_s[/tex] respectively;

Now, frequency of the sound that is coming from the bat towards the wall due to DROPPLER EFFET will be;

f₁ = ( c / ( c - [tex]v_s[/tex] ) )f ----- let this be equ 1

Also, frequency does not change after deflection. The bat becomes an observer as the dropper is shifted because the reflected sound wave is coming towards it;

Hence, Doppler shifted frequency will be;

f₂ = ( (c + [tex]v_s[/tex] ) / c )f₁

from equ 1, f₁ = ( c / ( c - [tex]v_s[/tex] ) )f, so we substitute

f₂ = ( (c + [tex]v_s[/tex] ) / c ) × ( c / ( c - [tex]v_s[/tex] ) )f

f₂ = ( (c + [tex]v_s[/tex] ) / ( c - [tex]v_s[/tex] ) )f

∴ beat frequency will be;

δf = f₂ - f = ( (c + [tex]v_s[/tex] ) / ( c - [tex]v_s[/tex] ) )f - f

δf = ( 2[tex]v_s[/tex] / c - [tex]v_s[/tex] )f

δf = ( 2[tex]v_s[/tex] / c - [tex]v_s[/tex] )f

2f/δf = c - [tex]v_s[/tex] / [tex]v_s[/tex]

2f/δf = c/[tex]v_s[/tex] - [tex]v_s[/tex] / [tex]v_s[/tex]

2f/δf = c/[tex]v_s[/tex] - 1

c/[tex]v_s[/tex] = 2f/δf + 1

[tex]v_s[/tex] =  c / (2f/δf + 1)

now, we substitute in our values;

[tex]v_s[/tex] =  340 / ((2×30000 / 900 ) + 1)

[tex]v_s[/tex] =  340 / (66.6666 + 1)

[tex]v_s[/tex] =  340 / 67.6666

[tex]v_s[/tex] =  5.02 m/s

Therefore, the speed of the bat is 5.02 m/s

b) the wavelength of the sound that the bat hears

frequency of reflected wave is;

f₂ = f + δf = 30000 + 900 = 30900 Hz

λ₂ = c / f₂

we substitute

λ₂ = 340 / 30900

λ₂ = 0.011 m

Therefore, the wavelength of the sound that the bat hears is 0.011 m

how hot can the desert get​

Answers

Answer:

134 f

Explanation:

The hottest temperature ever reliably measured in a desert was 134 degrees F, in Death Valley of the Mojave Desert in 1913.

not as hot as me.jk.

Temperature. During the day, desert temperatures rise to an average of 38°C (a little over 100°F). At night, desert temperatures fall to an average of -3.9°C (about 25°F). At night, desert temperatures fall to an average of -3.9 degrees celsius (about 25 degrees fahrenheit).

Desert surfaces receive a little more than twice the solar radiation received by humid regions and lose almost twice as much heat at night. Many mean annual temperatures range from 20-25 degrees Celsius. The extreme maximum ranges from 43.5-49 degrees Celsius. Minimum temperatures sometimes drop to -18 degrees Celsius.

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