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3 years ago

Thermal expansion is the increase in volume that results from an increase in temperature

Chemistry

1 answer:

sattari [20]3 years ago

4 0

Answer:

True

Explanation:

Thermal expansion is the increase of volume due to increase in tempature so this statement would in fact be correct. Have a fantastic day!

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7.Calculate the volume of nitrogen that reacts with 12dm3 of hydrogen with the volume of both gases measured at rtp:

marta [7]

Answer:

4.03dm³

Explanation:

The reaction expression is given as:

3H₂ + N₂ → 2NH₃

Volume of hydrogen = 12dm³

AT rtp:

1 mole of gas occupies volume of 22.4dm³

x mole of hydrogen will occupy a volume of 12dm³

Number of moles of hydrogen = $\frac{12}{22.4}$ = 0.54mole

From the balanced reaction equation:

3 mole of hydrogen gas combines with 1 mole of Nitrogen gas

0.54 mole of hydrogen as will therefore combine with $\frac{0.54}{3}$ = 0.18moles of nitrogen gas

Since ;

1 mole of gas occupies a volume of 22.4dm³

0.18moles of Nitrogen gas will occupy 0.18 x 22.4 = 4.03dm³

6 0

2 years ago

Why do you need many different indicators to span the entire ph spectrum?

marin [14]

The answer is each indicator has a narrow range. We need many different indicators to span the entire ph spectrum because each indicator has a narrow range.

4 0

2 years ago

A grating has 470 lines/mm. how many orders of the visible wavelength 538 nm can it produce in addition to the m = 0 order?

maria [59]

Three complete orders on each side of the m=0 order can be produced in addition to the m = 0 order.

The ruling separation is

d=1 / (470mm −1) = 2.1×10⁻³ mm

Diffraction lines occur at angles θ such that dsinθ=mλ, where λ is the wavelength and m is an integer.

Notice that for a given order, the line associated with a long wavelength is produced at a greater angle than the line associated with a shorter wavelength.

We take λ to be the longest wavelength in the visible spectrum (538nm) and find the greatest integer value of m such that θ is less than 90°.

That is, find the greatest integer value of m for which mλ<d.

since d / λ = 538×10⁻⁹m / 2.1×10 −6 m ≈ 3

that value is m=3.

There are three complete orders on each side of the m=0 order.

The second and third orders overlap.

Learn more about diffraction here : brainly.com/question/16749356

#SPJ4

7 0

1 year ago

When a particle is moving fast it has a lot of kinetic energy in it. ?

Sever21 [200]

Answer:

Random particle motion in liquids and gases is a difficult concept for in temperature, the particles move faster as they gain kinetic energy.

Explanation:

3 0

2 years ago

Which of the following (with specific heat capacity provided) would show the smallest temperature change upon gaining 200.0 J of

Brut [27]

Answer: The smallest temperature change is shown by water.

Explanation:

To calculate the heat absorbed or released, we use the equation:

$q=mC\times \Delta T$ ......(1)

where,

q = heat absorbed = 200.0 J

m = mass of the substance

C = specific heat of substance

$\Delta T$ = change in temperature

As, the same amount of heat is getting absorbed in all the cases. So, the change in temperature will depend on the product of mass and specific heat.

For the given options:

Option a: 50.0 g Fe, $C_{Fe}=0.449J/g^oC$

We are given:

$m=50.0g\\C_{Fe}=0.449J/g^oC$

Putting values in equation 1, we get:

$200.0J=50.0g\times 0.449J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 0.449}=8.99^oC$

Change in temperature = 8.99°C

Option b: 50.0 g water, $C_{water}=4.18J/g^oC$

We are given:

$m=50.0g\\C_{water}=4.18J/g^oC$

Putting values in equation 1, we get:

$200.0J=50.0g\times 4.18J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{50\times 4.18}=0.96^oC$

Change in temperature = 0.96°C

Option b: 25.0 g Pb, $C_{Pb}=0.128J/g^oC$

We are given:

$m=50.0g\\C_{Pb}=0.128J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.128J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.128}=62.5^oC$

Change in temperature = 62.5°C

Option d: 25.0 g Ag, $C_{Ag}=0.235J/g^oC$

We are given:

$m=25.0g\\C_{Ag}=0.235J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.235J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.235}=34.04^oC$

Change in temperature = 34.04°C

Option e: 25.0 g granite, $C_{granite}=0.79J/g^oC$

We are given:

$m=25.0g\\C_{Fe}=0.79J/g^oC$

Putting values in equation 1, we get:

$200.0J=25.0g\times 0.79J/g^oC\times \Delta T\\\\\Delta T=\frac{200}{25\times 0.79}=10.13^oC$

Change in temperature = 10.13°C

Hence, the smallest temperature change is shown by water.

5 0

3 years ago