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Lady_Fox [76]
3 years ago
15

Which of the possible compounds has a mass of 163 grams when

Chemistry
1 answer:
Simora [160]3 years ago
6 0

Answer:

CH4

Explanation:

In solving this problem, we must remember that one mole of a compound contains Avogadro's number of elementary entities. These elementary entities include atoms, molecules, ions etc. Recall that one mole of a substance is the amount of substance that contains the same number of elementary entities as 12g of carbon-12. The Avogadro's number is 6.02 × 10^23.

Hence we can now say;

If 163 g of the compound contains 6.13 ×10^24 molecules

x g will contain 6.02 × 10^23 molecules

x= 163 × 6.02 × 10^23 / 6.13 × 10^24

x= 981.26 × 10^23/ 6.13 ×10^24

x= 160.1 × 10^-1 g

x= 16.01 g

x= 16 g(approximately)

16 g is the molecular mass of methane hence x must be methane (CH4)

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<h2>Answer:</h2>

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Low energy waves have (3 points)
Sergeeva-Olga [200]

Answer:

Low energy waves have <u>a long wavelength.</u>

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3 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

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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

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T_f=22.7^oC

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

4 0
3 years ago
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