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Iteru [2.4K]
3 years ago
11

Define a physical change and give one example

Chemistry
1 answer:
ladessa [460]3 years ago
7 0

Answer:

The definition of a physical change is: A physical change is a type of change in which the form of matter is altered but one substance is not transformed into another. The size or shape of matter may be changed but no chemical reactions occur.

An example of physical change is: crumbling a piece of paper or freezing water into ice.

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How many atoms are in 1.75 moles of CHCl?
jok3333 [9.3K]
1.75 moles ChCl3 x (6.02 x 10 ^-23) / 1 mole = 1.0535 x 10^-22 atoms.

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How many minutes are in 17 years?
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Explanation:

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2 years ago
1. In what ways does an organism’s mode of reproduction impact the diversity of its offspring? (Asexual vs. Sexual)
Katen [24]

Answer:

<u>Asexual reproduction</u> = An exact copy of the reproducer

<u>Sexual reproduction</u> = Genetic variation into the organism if a random mutation in the organism's DNA is transmitted to offspring.

Explanation:

Example of Asexual Reproduction diversity - A red apple has apple seeds. You plant the red apple seeds, and it grows up to be no different than the apple before it.

Sexual reproduction Example of diversity -Two parents are of different ethnicities. The female gets pregnant, but because of the different genetics from both of the parents, the child will inherit their genetics + some random mutation in the DNA.  

This impacts the diversity of the offspring because it can be an exact copy of the reproducer (Asexual), or significantly different with some similarities if the mode of reproduction was sexual.

I really hope this helps you! Please tell me if it did or not. Good luck with your assignment/exam/quiz!

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7 0
2 years ago
Read 2 more answers
10-kg of R-134a at 300 kPa fills a rigid container whose volume is 14 L. Determine the temperature and total enthalpy in the con
Mariulka [41]

Answer:

Temperature = 0.605°C

Total enthalpy at 300kpa = 545.2 kJ

Total enthalpy at 600kpa = 846.45 kJ.

Explanation:

Checking the table for 134a pressure table. It is given that the specific volume of saturated liquid and the specific volume of the saturated vapor of 280kpa is 0.0007699 m^3/kg and 0.072352 m^3/kg respectively.

Also, the specific volume of saturated liquid and the specific volume of the saturated vapor of 320kpa is 0.0007772 m^3/kg and 0.063604 m^3/kg respectively.

The first thing to do is to determine the value for the specific volume of saturated liquid.

At 300 kpa, the specific volume of saturated liquid,n is given below as;

300 - 280/ 320 - 280 = (n - 0.0007699)/ 0.0007772 - 0.0007699.

Therefore, n = specific volume of saturated liquid = 0.0007735 m^3/kg.

300 - 280/ 320 - 280 = n - 0.072352/ (0.063604 - 0.072352).

n = 0.0679 m^3/kg.

The second thing to do is to determine the value of the specific volume.

Specific volume = 14 × 10^-3/ 10 = 0.0014 m^3/kg.

Determine the enthalpy of the mixture,b(I). This is given below as;

300 - 280/ 320 - 280 = b(I) - 199.54/ (196.7 - 199.54).

b(I) = 198.125 kJ/Kg.

Hence, b = [ 300 - 280/ 320 - 280 = j - 50.18 / 55.16 - 50.18] + [ ( 0.0014 - 0.00077735) / 0.067978 - 0.00077735] × 198.125.

b = 54.517 KJ/Kg.

Total enthalpy = 10 × 54.517 = 545.17 kJ.

Temperature can be Determine as below;

300 - 280/ 320 - 280 = T + 1.25 / 2.46 - 1.25.

Temperature = 0.605°C.

Hence, at 600kpa, the total enthalpy = [81.51 + ( 0.0014 - 0.0008199/ 0.034295 - 0.0008199) × 180.90] × 10

Total enthalpy at 600kpa = 846.45 kJ.

3 0
2 years ago
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