ANALOGY: Evaporates are to chemical sedimentary as dead fish are to clastic sedimentary.

Helium has the lowest boiling point of any substance, at 4.2 k. part a what is this temperature in âc?

Annette [7]

Answer is: temperature of helium is -296,95°C.
The temperature T in degrees Celsius (°C) is equal to the temperature T in Kelvin (K) minus 273,15: T(°C) = T(K) - 273.15.
T(He) = 4,2 K.
T(He) = 4,2 K - 273,15.
T(He) = -268,95°C.
The Celsius scale was based on 0°C for the freezing point of water and 100°C for the boiling point of water at 1 atm pressure.

6 0

3 years ago

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What kind of energy can be transferred

AlekseyPX

The correct answer is:

Thermal energy.

$|Huntrw6|$

6 0

4 years ago

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A box of cleaning supplies weighs 15 N. If the box is lifted a distance of 0.60 m, how much work is done?

Nady [450]

Work Done = force x displacement. So in this case the 15N is the force (because weight is a force) and 0.60m is the displacement. Therefore 15 x 0.6 = 9 Joules of work done (btw, work done can also be referred to as energy transferred)

3 0

4 years ago

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A 10-gram aluminum cube absorbs 677 joules when its temperature is increased from 50°c to 125°c. what is the specific heat of al

faust18 [17]

What is the specific heat of aluminum?

The specific heat of aluminum is 0.90J/g°C.

Given:

Heat absorbs = 677 J

Mass of the substance = 10 g

Initial temperature = 50°C

Final temperature = 125°C

Formula used:

Q = m x c x ΔT

which can also be written as,

Q = m x c x ( $T_{final}$ - $T_{initial}$ )

where,

Q = heat absorbs

m = mass of a substance

c = heat capacity of aluminum

$T_{final}$ = Final temperature

$T_{initial}$ = Initial temperature

Now, put all the values in the formula given above to get the specific heat of aluminum,

677g = (10g) x c x (125 - 50)°C

c = 0.9026 J/g°C

c = 0.90 J/g°C

Learn more about the specific heat of aluminum here,

brainly.com/question/13696634

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7 0

2 years ago

A metal cylinder has a mass of 6.20 g. The density of the cylinder is 21.0 g/mL. What is the volume

Stolb23 [73]

Answer:

$\boxed {\boxed {\sf v \approx0.295 \ mL}}$

Explanation:

We are asked to find the volume of a metal cylinder. We are given the mass and the density.

We know that density is the mass per unit volume of a substance. It is found using the following formula.

$\rho= \frac {m}{v}$

We know the density is 21.0 grams per milliliter and the mass is 6.20 grams. We substitute these values into the formula.

ρ= 21.0 g/mL
m= 6.20 g

$21.0 \ g/mL = \frac{6.20 \ g}{v}$

We are solving for the volume, so we must isolate the variable v. First, cross multiply. Multiply the first numerator by the second denominator, then the first denominator and the second numerator.

$\frac {21.0 \ g/mL}{1} = \frac{6.20 \ g}{v}$

$21.0 \ g/ mL *v= 6.20 \ g * 1$

$21.0 \ g/ mL *v= 6.20 \ g$

v is being multiplied by 21.0 grams per milliliter. The inverse operation of multiplication is division. Divide both sides of the equation by 21.0 g/mL.

$\frac {21.0 \ g/ mL *v}{21.0 \ g/mL}= \frac{6.20 \ g }{21.0 \ g/mL}$

$v= \frac{6.20 \ g }{21.0 \ g/mL}$

The units of grams cancel.

$v= \frac{6.20 }{21.0 mL}$

$v=0.2952380952$

The original measurements of density and mass both have 3 significant figures, so our answer must have the same. For this number, 3 sig fig is the thousandth place. The 2 in the ten-thousandth place tells us to leave the 5.

$v \approx0.295 \ mL$

The volume of the metal cylinder is approximately 0.295 milliliters.

3 0

3 years ago