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Lapatulllka [165]
3 years ago
14

Which of the following pairs of compounds would you expect to form homogeneous solutions when combined? a)CH3CH2CH2CH2CH3 AND CH

3CH2CH2CH2CH2CH3 b) CBr4 and H2O c)LiNO3 and H2O d)CH3OH and CH3CH2CH2CH2CH3
Chemistry
1 answer:
Stolb23 [73]3 years ago
8 0
<span>Which of the following pairs of compounds would you expect to form homogeneous solutions when combined?
</span><span>a) CH3CH2CH2CH2CH3 AND CH3CH2CH2CH2CH2CH3
and
</span><span>c) LiNO3 and H2O</span>
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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
How many miles of C are in 32.6 g of C2H6
sammy [17]
32.6 grams divided by the molar mass of C2H6, which is 18.0584g/mol = 1.8 moles of C2H6.

As there are two carbon atoms per C2H6, we must multiply the number of moles of C2H6 by 2 to get the number of moles of Carbon which is 3.6 moles.
The answer is 3.6 moles.

Hope this helps.

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5 0
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Which law states that the volume and absolute temperature of a fixed quantity of gas are directly proportional under constant pr
seropon [69]

Charles’ Law........

7 0
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"A sphere of radius 0.50 m, temperature 27oC, and emissivity 0.85 is located in an environment of temperature 77oC. What is the
Aleksandr-060686 [28]

Explanation:

It is known that formula for area of a sphere is as follows.

                     A = 4 \pi r^{2}

                        = 4 \times 3.14 \times (0.50 m)^{2}

                        = 3.14 m^{2}

    T_{a} = (27 + 273.15) K = 300.15 K

          T = (77 + 273.15) K = 350.15 K

Formula to calculate the net charge is as follows.

             Q = esA(T^{4} - T^{4}_{a})

where,    e = emissivity = 0.85

               s = stefan-boltzmann constant = 5.6703 \times 10^{-8} Wm^{-2} K^{-4}

                A = surface area

Hence, putting the given values into the above formula as follows.

                 Q = esA(T^{4} - T^{4}_{a})

                     = 0.85 \times 5.6703 \times 10^{-8} Wm^{-2} K^{-4} \times 3.14 \times ((350.15)^{4} - (300.15)^{4})

                     = 1046.63 W

Therefore, we can conclude that the net flow of energy transferred to the environment in 1 second is 1046.63 W.

8 0
3 years ago
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