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3 years ago

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.

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Consider the following reaction. SO2Cl2 → SO2 + Cl2. After collecting experimental data you found that plotting ln[SO2Cl2] vs. t

Nikitich [7]

Answer:

$[SO_2Cl_2]_{600}= 0.0842 M$

Explanation:

Some theoretical knowledge is required here. We should understand that whenever we plot the natural logarithm, ln, of a concentration vs. time and obtain a straight line, this indicates a first-order reaction. That said, since this is the case here, we have a first-order reaction with respect to $SO_2Cl_2$ .

The linear equation has the following terms:

$y = -0.000290t - 2.30$

It is a linear form of the integrated first-order law equation:

$ln[SO_2Cl_2]_t = -kt + ln[SO_2Cl_2]_o$

Therefore, the rate constant, k, is:

$k = 0.000290 s^{-1}$

The natural logarithm of initial molarity is:

$ln[SO_2Cl_2]_o = -2.30$

Using the equation, we may substitute for t = 600 s and obtain the natural logarithm of the concentration at that time:

$ln[SO_2Cl_2]_{600} = -0.000290 s^{-1}\cdot 600 s - 2.30 = -2.474$

Take the antilog of both sides to find the actual molarity:

$[SO_2Cl_2]_{600}=e^{-2.474} = 0.0842 M$

4 0

2 years ago

. Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:

barxatty [35]

A. HCl:
pH= -log [H3O+]
PH=-log (0.200)
= 0.699

poH= 14-0.699
= 13.301

b. NaOH:
PoH= -log [OH-]
= -log (0.0143)
= 1.845
pH= 14-poH
= 14- 1.845
= 12.16

c. HNO3:
PH= -log[H3O+]
=-log(3.0)
= -0.4771
poH= 14-pH
= 14-9-0.4771
= 14.4771

pH= -0.4771, poH= 14.4771

d. [Ca(OH)2] = 0.0031M
[OH-]= 2X0.0031
[OH-] = 0.0062M

PoH= - log[OH-]
=-log(0.0062)
=-log(6.2x10-3)
=-(-2.21)
= 2.21
PH=14-poH
=14-2.21
=11.79
POH=2.21, PH= 11.79

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Mamont248 [21]

Atomic number
atomic mass
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andreev551 [17]

Answer:

Explanation:

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