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STatiana [176]
3 years ago
8

What two categorical ways of thinking about matter are discussed in the passage?

Chemistry
1 answer:
Mamont248 [21]3 years ago
4 0

Answer:

There's no passage but I can say that matter are that ''thing'' which extend through all Universe, matter is everything. It can be seen or felt as liquid, gas or solid. This changes of its state depend on the amount of energy that is involved.

So, here is used categorical thinking because is a abstract concept that is hard to understand sometimes. It's abstract when we say that matter is all, not  just what we see, but even what we don't see like ''Dark Matter'', which is the majority in the Universe.

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An electron charge e mass m and a positron charge e mass m revolve around their common center of mass under the influence of the
Katen [24]

Answer: v = 2π2 Kme2 Z / nh

Explanation:

The formula for velocity of an electron in the nth orbit is given as,

v = 2π2 Kme2 Z / nh

v = velocity

K = 1/(4πε0)

m= mass of an electron

e = Charge on an electron

Z= atomic number

h= Planck’s constant

n is a positive integer.

3 0
3 years ago
A gas occupies a volume of 1.00 L at 25.0°C. What volume will the gas occupy at 1.00 x10^2 °C?
Leno4ka [110]

Answer : The volume of gas occupy at 1.00\times 10^2^oC is, 1.25 L

Explanation :

Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=1.00L\\T_1=25.0^oC=(25.0+273)K=298K\\V_2=?\\T_2=1.00\times 10^2^oC=((1.00\times 10^2)+273)K=373K

Putting values in above equation, we get:

\frac{1.00L}{298K}=\frac{V_2}{373K}\\\\V_2=1.25L

Therefore, the volume of gas occupy at 1.00\times 10^2^oC is, 1.25 L

3 0
3 years ago
Which of the following instrument would you use to see a plant cell?
Gemiola [76]
B . microscope
this is the answer .
3 0
2 years ago
Read 2 more answers
How do you decide whether an observed property of matter is a physical or a chemical property?
Otrada [13]
By the object's composition and the acting agent to which factors the change of the substance or entity.
<span>The effect of physical property of matter on the object or substance can be better identified when the object doesn't change in composition or in nature. Unlike chemical property which has changed in its composition and atomic structure that was caused by chemical change or reaction due to an agent. Physical property is identified thru physical reactions or changes that has never changed the object in an atomic level, like cutting paper into smaller sizes. The aforementioned example illustrates physical proerty but being cut into smaller buts without changing the object from paper to any other substance, thus, its structure remain and its still called paper regardless of size, mass and texture. <span>
</span></span>
8 0
3 years ago
Read 2 more answers
A 1.800-g sample of solid phenol (C6H5OH(s)) was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/?C. The temp
vichka [17]

Answer:

The balanced chemical equation:

C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)

Heat of combustion per gram of phenol is 32.454 kJ/g

Heat of combustion per gram of phenol is 3,050 kJ/mol

Explanation:

C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)

Heat capacity of calorimeter = C = 11.66 kJ/°C

Initial temperature of the calorimeter = T_1= 21.36^oC

Final temperature of the calorimeter = T_2= 26.37^oC

Heat absorbed by calorimeter = Q

Q=C\times \Delta T

Heat released during reaction = Q'

Q' = -Q ( law of conservation of energy)

Energy released on combustion of 1.800 grams of phenol = Q' = -(58.4166 kJ)

Heat of combustion per gram of phenol:

\frac{Q'}{1.800 g}=\frac{-58.4166 kJ}{1.800 g}=32.454 kJ/g

Molar mass of phenol = 94 g/mol

Heat of combustion per gram of phenol:

\frac{Q'}{\frac{1.800 g}{94 g/mol}}=\frac{-58.4166 kJ\times 94 g/mol}{1.800 g}=3,050 kJ/mol

3 0
3 years ago
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