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CaHeK987 [17]
1 year ago
12

The density of a gas sample in a balloon is 1.50 g/l at 75°c. what is the density of this gas when the temperature is changed t

o 25°c at constant pressure?
Physics
1 answer:
maksim [4K]1 year ago
8 0

At 25°C, a gas sample's density in the balloon is 4.500 g/L.

When density decreases, temperature increases. When more temperature increases, density reduces. When the temperature decrease, density increases.

When a substance is heated, the molecules move faster and slightly farther apart, taking up more space and causing the density to drop. When something is cooled, the molecules slow down and get a little closer together, taking up less space and becoming denser.

The density of a gas sample in a balloon = 1.50 g/L

Temperature = 75°C

Temperature increased = 25°C

The density of a gas sample in the balloon at 25°C = (1.50 × 75)/ 25

                                                                                      = 4.500 g/L

Therefore, the density of a gas is 4.500 g/L.

Learn more about density here:

brainly.com/question/6838128

#SPJ4

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Answer:

Minimum height of metal = 5 inches

Explanation:

Volume of the cylindrical metal = πR²H = 125π

cancelling out π on both sides

R²H = 125

Hence it can be deduced that R² = 25 and H = 5

Hence minimum height of metal = 5 inches

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3 years ago
A note of frequency 200Hz has a velocity of 400m/s. what is the wavelength of the note​
Xelga [282]

Answer:

\huge\boxed{\sf \lambda = 2 m}

Explanation:

<h3><u>Given data:</u></h3>

Frequency = f = 200 Hz

Velocity = v = 400 m/s

<h3><u>Required:</u></h3>

Wavelength = λ = ?

<h3><u>Formula:</u></h3>

v = fλ

<h3><u>Solution:</u></h3>

Put the givens in the formula

400 = (200)λ

Divide 200 to both sides

400/200 = λ

2 m = λ

λ = 2 m

\rule[225]{225}{2}

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Find the final temperature of 375 grams of tea (c = 4.184 J/g°C) if its initial temperature is 95°C just before it is placed in
Minchanka [31]

Answer:

the final temperature of the tea is 7.39⁰C.

Explanation:

Given;

mass of the tea, m = 375 g

specific heat capacity of the tea, C = 4.184 JJ/g°C

initial temperature of the tea, t₁ = 95°C

the final temperature of the tea, t₂ = ?

Energy lost by the refrigerator, Q = 137,460 J

The energy lost by the refrigerator is given by the following formula;

-Q = mc(t₂ - t₁)

-137,460 =375 x 4.184(t₂ - 95°C)

-137,460 = 1569(t₂ - 95°C)

t_2-95 = \frac{-137,460}{1569} \\\\t_2-95 = -87.61\\\\t_2 = -87.61 + 95\\\\t_2 = 7.39 \ ^0 C

Therefore, the final temperature of the tea is 7.39⁰C.

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3 years ago
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