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3 years ago

if the Celsius temperature of a gas at constant pressure is increased from 10 Celsius to 20 Celsius the volume is

Chemistry

1 answer:

svet-max [94.6K]3 years ago

3 0

Answer:

The volume is increased.

Explanation:

According to Charles' Law, " at constant pressure the volume and temperature of the gas are directly proportional to each other". Mathematically this law is presented as;

V₁ / T₁ = V₂ / T₂ -----(1)

In statement the data given is,

T₁ = 10 °C = 283.15 K ∴ K = 273.15 + °C

T₂ = 20 °C = 293.15 K

So, it is clear that the temperature is being increased hence, we will find an increase in volume. Let us assume that the starting volume is 100 L, so,

V₁ = 100 L

V₂ = Unknown

Now, we will arrange equation 1 for V₂ as,

V₂ = V₁ × T₂ / T₁

Putting values,

V₂ = 100 L × 293.15 K / 283.15 K

V₂ = 103.52 L

Hence, it is proved that by increasing temperature from 10 °C to 20 °C resulted in the increase of Volume from 100 L to 103.52 L.

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3 years ago

explain the law of conservative of energy, give a specific example using kinetic and potential energy that shows how energy is c

KiRa [710]

Answer:

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Explanation:

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8 0

3 years ago

What is the mass of 4.5 x 10^22 molecules of hydrogen peroxide (H2O2)? Show your work in the space below.

valentinak56 [21]

Given :

Number of molecules of hydrogen peroxide, N = 4.5 × 10²².

To Find :

The mass of given molecules of hydrogen peroxide.

Solution :

We know, 1 mole of every compound contains Nₐ = 6.022 × 10²³ molecules.

So, number of moles of hydrogen peroxide is :

$n = \dfrac{N}{N_a}\\\\n = \dfrac{4.5\times 10^{22}}{6.022\times 10^{23}}\\\\n = 0.0747 \ moles$

Now, mass of hydrogen peroxide is given as :

m = n × M.M

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3 years ago

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3 years ago

How much energy is required to boil 65 grams of 100°C water And then heat the steam to 150°C?

Artyom0805 [142]

Answer:

13598 J

Explanation:

Q = m × c × ∆T

Where;

Q = amount of energy (J)

m = mass (grams)

c = specific heat capacity

∆T = change in temperature

m = 65g, specific heat capacity of water = 4.184J/g°C, initial temperature= 100°C, final temperature = 150°C

Q = 65 × 4.184 × (150 - 100)

Q = 271.96 × 50

Q = 13598 J

Hence, 13598 J of energy is required to boil 65 grams of 100°C water and then heat the steam to 150°C.

8 0

3 years ago