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IRISSAK [1]
2 years ago
10

A 4.0 L balloon has a pressure of 406 kPa. When the pressure increases to 2,030 kPa, what is the volume? *

Chemistry
1 answer:
elena-14-01-66 [18.8K]2 years ago
3 0

The new volume when pressure increases to 2,030 kPa is 0.8L

BOYLE'S LAW:

The new volume of a gas can be calculated using Boyle's law equation:

P1V1 = P2V2

Where;

  1. P1 = initial pressure (kPa)
  2. P2 = final pressure (kPa)
  3. V1 = initial volume (L)
  4. V2 = final volume (L)

According to this question, a 4.0 L balloon has a pressure of 406 kPa. When the pressure increases to 2,030 kPa, the volume is calculated as:

406 × 4 = 2030 × V2

1624 = 2030V2

V2 = 1624 ÷ 2030

V2 = 0.8L

Therefore, the new volume when pressure increases to 2,030 kPa is 0.8L.

Learn more about Boyle's law calculations at: brainly.com/question/1437490?referrer=searchResults

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6H2O represents 6 molecules of water

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1) HNO3/H2SO4, 2) CH3CH2CH2Cl/AlCl3

Explanation:

Benzene is a stable aromatic compound hence it undergoes substitution rather than addition reaction.

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If I want to synthesize m-nitropropylbenzene, I will first carry out the nitration of benzene using HNO3/H2SO4 since the -nitro group is a meta director. This is now followed by Friedel Craft's alkykation using CH3CH2CH2Cl/AlCl3.

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A 21.0 kg child is running at a velocity of 2.50 m/s. The child is carrying a 2.25 kg gallon of milk. What is the momentum of th
KIM [24]

Answer: The momentum of the child and milk together is 58.125 kg.m/s

Explanation:

Momentum is defined as the product of object's mass and velocity.

Mathematically,

p=mv

where, p = momentum

m = mass of the object

v = velocity of the object

In the given question, we are given that a child of mass 21.0 kg is carrying a gallon of milk having mass 2.25 kg and running with a velocity of 2.5 m/s. Hence, the momentum by both milk and child will be:

p=(m_{child}+m_{milk})v     ....(1)

Given:

m_{child}=21.0kg\\m_{milk}=2.25kg\\v=2.5m/s\\p=?kg.m/s

Putting values in equation 1, we get:

p=(21+2.25)kg\times 2.5m/s=58.125kg.m/s

Hence, the momentum of the child and milk together is 58.125 kg.m/s

7 0
3 years ago
How many calories is required to change the temperature of 2.18g of water from 15.3°C to 69.5°C. The specific heat of liquid wat
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The number  of calories that are  required  to change the temperature  of 2.18 g of water from 15.3 c to 69.5 c is  <u>118.16 cal</u>


    <u><em> calculation</em></u>

  •    Heat in calories  = MCΔ T where,
  • M(mass)= 2.18 g
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 heat is therefore= 2.18 g x 1.00 cal/g/c  x 54.2 c=118.16  cal

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