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pshichka [43]
3 years ago
14

Which substance is a gas at room temperature and atmospheric pressure?

Chemistry
1 answer:
natulia [17]3 years ago
8 0
Ar is a substance that is a gas at room temperature and atmospheric pressure.
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What is a percent volume of ethanol c2h60 in the final solution when 85 ml of it is diluted to a volume?
Ilya [14]
34% I believe

Hope this helps!
STSN
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3 years ago
A metal entrance door swings freely in the winter, but when
ivann1987 [24]

Answer:

This phenomenon occurs because the door, being metal and leading to changes in temperature, undergoes proportional and morphological changes, metals face expansion and expansion in the presence of heat, called thermal expansion.

On the other hand, against the cold, thermal contraction is suffered, that is why its volume decreases, and it contracts.

Explanation:

The expansion phenomenon of the door is not linear, since it increases its volume in width and height, therefore simultaneously on the entire surface.

When an area or surface expands, it does so by increasing its dimensions in the same proportion. For example, a metal sheet increases its length and width, which means an increase in area. Area dilation differs from linear dilation in that it involves an increase in area.

The area expansion coefficient is the increase in area that a body of a certain substance experiences, with an area equal to unity, as its temperature rises one degree centigrade. This coefficient is represented by the Greek letter gamma.

Regarding shrinkage, a clear example of this is when a metal foundry or a weld shrinks, sometimes it is difficult to understand with examples like these (doors) because it is little noticeable by our eyes and the dimensional changes for our perspective. it is infima.

3 0
3 years ago
(*Anohter one* For my friends!!! ^^)
Sliva [168]

Answer:

Thx Have a Fantastic day :)

Explanation:

7 0
3 years ago
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Based on the information in the table, which of the following arranges the bonds in order of decreasing
tatuchka [14]

Answer:

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3 0
2 years ago
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Calculate the change in energy when 75.0 grams of water drops from<br> 31.0C to 21.6.
zysi [14]

Answer: Step 1: Calculate qsur (the surrounding is

usually the water)

qsur = ? J

m = 75.0 g water

c = 4.184 J/g

oC

ΔT = (Tfinal- Tinitial)= (21.6 – 31.0) = -9.4 oC

qsur = m · c · (ΔT)

qsur = (75.0g) (4.184 J/g

oC) (-9.4 oC)

qsur = - 2949.72 J

First, using the information we know that we

must solve for qsur, which is the water. We know

the mass for water, 75.0g, the specific heat of

the water, 4.184 j/g

o

c, and the change in

temperature, 21.6-31.0 = -9.4 oC. Plugging it

into the equation, we solve for qsur.

Step 2: Calculate qsys qsys = - (qsur)

qsys = - (- 2949.72 J)

qsys = + 2949.72

In this case, the qsur is negative, which means

that the water lost energy. Where did it go? It

went to the system. Thus, the energy of the

system is negative, opposite, the energy of the

surrounding.

Step 3: Calculate moles of the substance

that is the system

Given: 12.8 g KCl

Mol system = (g system given)

(molar mass of system)

Mol system = (12.8 g KCl)

(39.10g + 35.45g)

Mol system = 12.8 g KCl

74.55 g

Mol system = 0.172

Here, we solve for the mol in the system by

using the molar mass of the material in the

system.

Step 4: Calculate ΔH ΔH = q sys .

Mol system

ΔH= + 2949.72 J

0.172 mol

ΔH= +17179.81 J/mol or +1.72 x 104

J/mol

i hope this helps

5 0
3 years ago
Read 2 more answers
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