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3 years ago

Chemical property

Chemistry

1 answer:

Tasya [4]3 years ago

4 0

Answer:

Characteristic of a substance that Cannot be observed without altering the identity of the substance.

Explanation:

A physical property is a characteristic of a substance that can be observed or measured without changing the identity of the substance. Physical properties include color, density, hardness, and melting and boiling points.

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How much energy will a sample of copper alloy ( 63.0 g) lose if it is cooled from 48.0° to 23.0°C ?

Ymorist [56]

Answer:

Heat lost would be 606.375J

Explanation:

Mass = 63.0g

T1 = 48°C

T2 = 23°C

Specific heat capacity of copper (c) = 0.385J/g°C

Q = Mc∇T

Q = Mc(T₂ - T₁)

Q = 63 * 0.385 * (23 - 48)

Q = 24.255 * -25

Q = -606.375J

The heat lost would be 606.375J

4 0

3 years ago

Read 2 more answers

What is the structure of epithelial cells?

EleoNora [17]

Your answer is C for sure

5 0

3 years ago

A chemist wants to extract a solute from 100 mL of water using only 300 mL of ether. The partition coefficient between ether and

devlian [24]

Answer:

a. X = 0,909

b. X = 0,965

c X = 0,997

Explanation:

The partition coefficient (k) is defined as:

k = Solute in ether / solute in water

a. $3,34 = \frac{\frac{X}{300mL} }{\frac{1-X}{100mL} }$

Where X is the fraction of solute extracted

3,34 = X / 3-3X

10,02-10,02X = X

10,02 = 11,02X

X = 0,909

b. First extraction:

$3,34 = \frac{\frac{X}{100mL} }{\frac{1-X}{100mL} }$

3,34 = X / 1-X

3,34 - 3,34X = X

3,34 = 4,34X

X = 0,770

That means solute in water will be: 1-0,770 = 0,23

Second extraction:

The second extraction will extract the same fraction of solute, as now you have 0,230 of solute in water you will extract:

0,230×0,770 = 0,177

Third extraction:

In the same way, the third extraction will extract:

(0,230-0,177)×0,770 = 0,018

Fraction in water×Fraction extracted

That means total solute extracted is:

0,770 + 0,177 + 0,018 = 0,965

c. Extracting with 50mL of ether:

First extraction

$3,34 = \frac{\frac{X}{50mL} }{\frac{1-X}{100mL} }$

3,34 = 2X / 1-X

3,34 - 3,34X = 2X

3,34 = 5,34X

X = 0,625

Second extraction:

(1-0,625)×0,625= 0,234

Third Extraction:

(1-0,625-0,234)×0,625= 0,088

Fourth extraction:

(1-0,625-0,234-0,088)×0,625= 0,033

Fifth extraction:

(1-0,625-0,234-0,088-0,033)×0,625= 0,013

Sixth extraction:

(1-0,625-0,234-0,088-0,033-0,013)×0,625= 0,004

Total extractions gives:

0,625+0,234+0,088+0,033+0,013+0,004 = 0,997

I hope it helps!

8 0

4 years ago

Determine the nuclear composition (number of protons and neutrons) of the following isotopes. (a) chromium-52 protons neutrons (

madreJ [45]

Answer: a) chromium-52 : protons = 24, neutrons = 28

selenium - 80 : protons = 34, neutrons = 46

molybdenum-98: protons = 42, neutrons = 56

xenon-132: protons = 54, neutrons = 78

ytterbium-174 : protons = 70, neutrons = 104

Explanation:

Atomic number : It is defined as the number of electrons or number of protons present in a neutral atom.

Thus, number of protons = atomic number

Mass number is the number of the entities present in the nucleus which is the equal to the sum of the number of protons and electrons.

Mass number = Number of protons + Number of neutrons

a) chromium-52 :

Atomic number of Chromium is 24 hence number of protons = 24

52= 24+ Number of neutrons

Number of neutrons = 28

b) selenium - 80 :

Atomic number of selenium is 34 hence number of protons = 34

80 = 34 + Number of neutrons

Number of neutrons = 46

c) molybdenum - 98 :

Atomic number of molybdenum is 42 is hence number of protons = 42

98 = 42 + Number of neutrons

Number of neutrons = 56

d) xenon-132:

Atomic number of xenon is 54 hence number of protons = 54

132 = 54 + Number of neutrons

Number of neutrons = 78

e) ytterbium-174:

Atomic number of ytterbium is 70 and hence number of protons = 70

174 = 70 + Number of neutrons

Number of neutrons = 104

6 0

3 years ago

What is the hybridization around the O atom?

baherus [9]

Your answer is......... with little bit explains....and read correctly.......The nitrogen is sp3 hybridized which means that it has four sp3 hybrid orbitals. Two of the sp3 hybridized orbitals overlap with s orbitals from hydrogens to form the two N-H sigma bonds. One of the sp3 hybridized orbitals overlap with an sp3 hybridized orbital from carbon to form the C-N sigma bond. The lone pair electrons on the nitrogen are contained in the last sp3 hybridized orbital. Due to the sp3 hybridization the nitrogen has a tetrahedral geometry. However, the H-N-H and H-N-C bonds angles are less than the typical 109.5o due to compression by the lone pair electrons.

7 0

3 years ago