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Dimas [21]
3 years ago
7

Consider the following: you mix 10.0 mL of CHCl3 (d = 1.492 g/mL) and 5.0 mL of CHBr3 (d = 2.890 g/mL), giving 15.0 mL of soluti

on. What is the density of this mixture?
Chemistry
2 answers:
shepuryov [24]3 years ago
7 0

Answer:

1.958 g/mL

Explanation:

the density of mixture will be calculated by taking ratio of total mass of the mixture obtained by mixing the two liquids and the total volume of the mixture

the total volume = volume of CHCl₃ + volume of  CHBr₃ = 10.0+5.0 =15.0mL

the mass of CHCl₃=density of CHCl₃ X volume of CHCl₃

                               = 1.492 X 10.0  = 14.92 g

the mass of CHBr₃=density of CHBr₃ X volume of CHBrl₃

                               = 2.890 X 5.0  = 14.45 g

Thus total mass = 14.92+14.45 = 29.37

Density of mixture = total mass / total volume = 29.37 / 15.0 = 1.958g/mL

Allushta [10]3 years ago
4 0
d_{c}=\frac{m_{1}+m_{2}}{V_{1}+V_{2}}=\frac{10mL*1,492\frac{g}{mL}+5mL*2,89\frac{g}{mL}}{15mL}=\\ \\ =\frac{14,92g+14,45g}{15mL}=\frac{29,37g}{15mL}=1,958\frac{g}{mL}
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<u>Answer:</u> The final equation has hydroxide ions which indicate that the reaction has occurred in a basic medium.

<u>Explanation:</u>

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Reduction half-reaction: MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-

  • Multiply each half-reaction by the correct number in order to balance charges for the two half-reactions

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Reduction half-reaction: MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-             ( × 2)

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Reduction half-reaction: 2MnO_4^-+4H_2O+3e^-\rightarrow 2MnO_2+8OH^-

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