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Alona [7]
3 years ago
8

A 0.35 L balloon found at 19 C is heated in an oven to 250 C. What is the new volume of the balloon?

Chemistry
2 answers:
Ymorist [56]3 years ago
7 0

Answer:

0.63L

Explanation:

V1 = 0.35L

T1 = 19°C = (19 + 273.15)K = 292.15K

V2 = ?

T2 = 250°C = (250 + 273.15)K = 523.15k

Applying Charle's law, the volume of a given mass of gas is directly proportional to its pressure provided temperature remains constant

V1 / T1 = V2 / T2

V2 = (V1 * T2) / T1

V2 = (0.35 * 523.15) / 292.15

V2 = 0.626L

V2 = 0.63L

Svet_ta [14]3 years ago
5 0

Answer:

V = 0.63 L

Explanation:

To solve this problem, we need to use the Charle's law which is a law that involves temperature and volume, assuming we have a constant pressure. The problem do not state that the pressure is being altered, so we can safely assume that the pressure is constant (Maybe 1 atm).

Now, as the pressure is constant, the Charle's law is the following:

V₁ / T₁ = V₂ / T₂   (1)   V is volume in Liter, and T is temperature in Kelvin.

Using this law with the given data, we solve for V₂:

V₂ = V₁T₂ / T₁

Before we use this expression, let's convert the temperatures to Kelvin:

T₁ = 19 + 273 = 292 K

T₂ = 250 + 273 = 523 K

Now, let's calculate the volume of the balloon:

V₂ = 0.35 * 523 / 292

<h2>V₂ = 0.63 L</h2>
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dsp73

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180.166889

Explanation:

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Damm [24]

1) For example, sodium oxide (Na₂O).

One one molecule of sodium oxide has one sodium atome (Na) and two oxygen atoms (O).

The subscript after element shows the number of elements in a molecule.

2) Yes, it possible for two different compounds to be made from the exact same two elements.

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3) There are large number of compounds in this world, around 40 millions.

7 0
3 years ago
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How much work (in JJ) is required to expand the volume of a pump from 0.0 LL to 2.5 LL against an external pressure of 1.1 atmat
stira [4]

Answer:

- 278.85 J  

Explanation:

Given that:

Pressure = 1.1 atm

The initial volume V₁ = 0.0 L

The final volume V₂ = 2.5 L

The work that takes place in a reaction at constant pressure can be expressed by using the equation:

W = P(V₂ - V₁ )

Since the volume of the gas is expanded from 0 to 2.5 L when 1.1 atm pressure is applied. Then, the work can be given by the expression:

W = - P(V₂ - V₁ )

W = -1.1 atm ( 2.5 - 0.0) L

W = -1.1 atm (2.5 L)

W = -2.75 atm L

Recall that:

1 atm L = 101.4 J

Therefore;

-2.75 atm L = ( -2.75 × 101.4 )J

= -278.85 J  

Thus, the work required at the chemical reaction when the pressure applied is 1.1 atm  = - 278.85 J  

8 0
3 years ago
Help Pls!
Salsk061 [2.6K]

Answer:

d= 50.23 g/cm³

Explanation:

Given data:

radius = 137.9 pm

mass is = 5.5 × 10−22 g

density = ?

Solution:

volume of sphere= 4/3π r³

First of all we calculate the volume:

v= 4/3π r3

v= 1.33× 3.14× (137.9)³

v= 1.33 × 3.14 × 2622362.939 pm³

v= 1.095 × 10∧7 pm³

v= 1.095 × 10∧-23 cm³

Formula:

Density:

d=m/v

d= 5.5 × 10−22 g/ 1.095 × 10∧-23 cm³

d= 5.023 × 10∧+1 g/cm³

d= 50.23 g/cm³

8 0
3 years ago
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