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In-s [12.5K]
2 years ago
10

Because of friction?

Chemistry
2 answers:
Zanzabum2 years ago
5 0

Answer:

what do you mean?

Explanation:

Pepsi [2]2 years ago
5 0

Answer:

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Explanation:

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Explain why different liquids boil at the different temperatures
Gennadij [26K]

The boiling point of the fluid depends on the intermolecular forces between the fluid atoms and molécules, as these forces must be disrupted to switch from a fluid to a gas. The stronger the intermolecular forces, the greater the point of boiling.

4 0
2 years ago
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The amount of matter in an object is its
andrezito [222]

The amount of matter in an object is its Mass...

7 0
2 years ago
If 45.0 mL of ethanol (density=0.789 g/mL) initially at 9.0 C is mixed with 45.0 mL of water (density=1.0 g/mL) initially at 28.
Klio2033 [76]

Answer : The final temperature of the mixture is 22.7^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

And as we know that,

Mass = Density × Volume

Thus, the formula becomes,

(\rho_1\times V_1)\times c_1\times (T_f-T_1)=-(\rho_2\times V_2)\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of ethanol = 2.3J/g^oC

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of ethanol

m_2 = mass of water

\rho_1 = density of ethanol = 0.789 g/mL

\rho_2 = density of water = 1.0 g/mL

V_1 = volume of ethanol = 45.0 mL

V_2 = volume of water = 45.0 mL

T_f = final temperature of mixture = ?

T_1 = initial temperature of ethanol = 9.0^oC

T_2 = initial temperature of water = 28.6^oC

Now put all the given values in the above formula, we get

(0.789g/mL\times 45.0mL)\times (2.3J/g^oC)\times (T_f-9.0)^oC=-(1.0g/mL\times 45.0mL)\times 4.18J/g^oC\times (T_f-28.6)^oC

T_f=22.7^oC

Therefore, the final temperature of the mixture is 22.7^oC

4 0
3 years ago
What vitamin is produced by a mammalian skin?
Contact [7]
It is Vitamin D, hope that helps
7 0
2 years ago
What volume of a 1.0 M HCl is required to completely neutralize 25.0 ml of a 1.0 M KOH?
kirza4 [7]
The question above can be solved by using this equation: 
CAVA =CBVB
Where:
CA =Concentration of acid = 1.0 M
VA = Volume of acid = ?
CB = Concentration of base = 1.0 M
VB = Volume of base = 25 ml
VA = CBVB / CA
VA = [1 * 25] / 1 = 25 / 1 = 25
VA = 25 ml
Therefore, the volume of acid that is required to completely neutralize the base is 25 ml.<span />
8 0
3 years ago
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