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Daniel [21]
3 years ago
14

An example of an extensive property of matter is

Chemistry
2 answers:
timurjin [86]3 years ago
6 0
An example of an extensive property of matter is "mass"
N76 [4]3 years ago
3 0

Answer:

Examples of extensive properties are: temperature, mass, weight, volume, lenght, force, among others.

Explanation:

Extensive properties are properties that depends on the amount of mass (thus depends on the size) of an object.

For example:

It is not the same to have a package of 1 kg of sugar than a package of 5 kg of sugar. The amount of mass is different and also the weight and the volume (the 1 kg package take less place than the package of 5 kg).

Also exists  Intensive properties, these ones do not depends on the amount of mass, such as: density, organoleptic properties, boiling point, melting point, amog others.

In the previous example, the flavor of both packages of sugar will be the same independetly of the amount of sugar.

You might be interested in
What is the maximum number of moles of N-acetyl-p-toluidine can be prepared from 70. milliliters of 0.167 M p-toluidine hydrochl
Kitty [74]

Answer:

\large \boxed{\text{0.012 mol}}  

Explanation:

We will need a balanced equation with moles, so let's gather all the information in one place.

               CH₃C₆H₄NH₂·HCl + (CH₃CO)₂O ⟶ CH₃C₆H₄NHCOCH₃ + junk

V/mL:                    70.

c/mol·L⁻¹:             0.167

For simplicity in writing , let's call p-toluidine hydrochloride A and N-acetyl-<em>p</em>-toluidine B.

The equation is then

A + Ac₂O ⟶ B + junk

1. Moles of A

\text{Moles of A} = \text{70. mL A}\times \dfrac{\text{0.167 mmol A}}{\text{1 mL A}}= \text{12 mmol A}

2. Moles of B

The molar ratio is 1 mol B:1 mol A

Moles of B = moles of A = 12 mmol = 0.012 mol

\text{You can prepare $\large \boxed{\textbf{0.012 mol}}$ of N-acetyl-p-toluidine. }

3 0
3 years ago
A tank contains 90 kg of salt and 2000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain
sashaice [31]

Answer:

a) 90 kg

b) 68.4 kg

c) 0 kg/L

Explanation:

Mass balance:

-w=\frac{dm}{dt}

w is the mass flow

m is the mass of salt

-v*C=\frac{dm}{dt}

v is the volume flow

C is the concentration

C=\frac{m}{V+(6-3)*L/min*t}

-v*\frac{m}{V+(6-3)*L/min*t}=\frac{dm}{dt}

-3*L/min*\frac{m}{2000L+(3)*L/min*t}=\frac{dm}{dt}

-3*L/min*\frac{dt}{2000L+(3)*L/min*t}=\frac{dm}{m}

-3*L/min*\int_{0}^{t}\frac{dt}{2000L+(3)*L/min*t}=\int_{90kg}^{m}\frac{dm}{m}

-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)

-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)

m=90kg*[2000L/(2000L+3*L/min*t)]

a) Initially: t=0

m=90kg*[2000L/(2000L+3*L/min*0)]=90kg

b) t=210 min (3.5 hr)

m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg

c) If time trends to infinity the  division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.

6 0
3 years ago
An acid increases the hydroxide ion concentration of water.<br>True<br>or<br>False
ASHA 777 [7]
False is the answer!
3 0
3 years ago
Read 2 more answers
Elements that have complete valence electron shells are mostly found in the
nika2105 [10]
Noble gases have complete valence electron shells
6 0
3 years ago
Read 2 more answers
Able 1 Cell Type Operating Cell Potential for Commercial Batteries, E (V) Lithium-iodine Zinc-mercury +2.80 +1.35 Table 2 Standa
IrinaK [193]

Answer:

During the initial cell operation, each reaction is thermodynamically favorable, but the larger operating potential of the lithium-iodine cell indicates that its cell reaction is more thermodynamically favorable.  ( B )

During the initial cell operation, the oxidation of iodine is  thermodynamically favorable but the oxidation of mercury is not. ( C )

Explanation:

<u>The major Differences between The Zinc mercury cell and Lithium-iodine cell are :</u>

During the initial cell operation, each reaction is thermodynamically favorable, but the larger operating potential of the lithium-iodine cell indicates that its cell reaction is more thermodynamically favorable.   and

During the initial cell operation, the oxidation of iodine is thermodynamically favorable but the oxidation of mercury is not.

Given the relationship below,

Δ G = -nFE

E = emf of cell ,  G = free energy.

This relationship shows that if E is positive the reaction will be thermodynamically favorable also if E is large it will increase the negativity of free energy  also From the question we can see that with the reduction of mercury the value of E is more positive and this shows that Mercury is thermodynamically unfavorable

8 0
2 years ago
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