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2 years ago

7

PLEASE HELP 100 POINTS

1 answer:

OleMash [197]2 years ago

3 0

Answer:

Hexenes + Dioxygen = Carbon Dioxide + Water

The reaction type is combustion.

Its reactants are Hexenes - C6H12 and Dioxygen - O2

its products are Carbon Dioxide - CO2 and Water - H2O

Explanation: This was my yesterdays class

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A balloon contains 1.00 mol helium gas at STP. What will the volume of the balloon be if the pressure increases to 2.00 atm? 11.

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A chemist adds 0.10 mol argon gas to the syringe. The pressure and temperature remain constant. What will be the volume on the syringe after the argon is added? Acetylene gas (C2H2) reacts with oxygen gas (O2) to produce carbon dioxide (CO2) and water vapor (H2O).

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3 years ago

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Triple covalent bond can be formed between:

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A triple covalent bond can be formed between d) C & O

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3 years ago

A concentrated salt solution has a mass of 5.29 g for a 5.00 mL sample. What is the specific gravity of this solution

TiliK225 [7]

Answer:

$\rho_s=1.07g/cc^3$

Explanation:

From the question we are told that:

Mass $M=5.29g$

Volume $V=5.00mL=>5.0cc^3$

Generally the equation for Specific Gravity momentum is mathematically given by

$Specific\ Gravity\ g= Density\ of\ Salt\ Solution\ \rho_s / Density\ of\ Water\ \rho_w$

$g=\frac{ \rho_s} { \rho_w}$

$\rho_s=\frac{ g} { \rho_w}$

$\rho_s=\frac{ 5.29} { 5}$

$\rho_s=1.07g/cc^3$

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3 years ago

An isotope undergoes radioactive decay by emitting radiation that has a –1 charge. What other characteristic does the radiation

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4 years ago

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Given the two reactions PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq), K3 = 1.85×10−10, and AgCl(aq)⇌Ag+(aq)+Cl−(aq), K4 = 1.17×10−4, what is the

kramer

Answer:

The equilibrium constant of the given reaction is 0.01351.

Explanation:

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

The equilibrium constant of the reaction = $K_3=1.85\times 10^{-10}$

$K_3=\frac{[Pb^{2+}][Cl^-]^2}{[PbCl_2]}$ ...[1]

$AgCl(aq)\rightleftharpoons Ag^{+}(aq)+Cl^-(aq)$

The equilibrium constant of the reaction = $K_4=1.17\times 10^{-4}$

$K_4=\frac{[Ag^+][Cl^-]}{[AgCl]}$ ..[2]

$[Cl^-]=\frac{K_4\times [AgCl]}{[Ag^+]}$

$PbCl_2(aq)+2Ag^+(aq)\rightleftharpoons 2AgCl(aq)+Pb^{2+}(aq)$

The expression of equilibrium constant of the creation is ;

$K=\frac{[AgCl]^2[Pb^{2}]}{[PbCl_2]][Ag^+]^2}$

Dividing [1] by [2]

$\frac{K_3}{K_4}=\frac{\frac{[Pb^{2+}][Cl^-]^2}{[PbCl_2]}}{\frac{[Ag^+][Cl^-]}{[AgCl]}}$

$\frac{K_3}{K_4}=\frac{[Pb^{2+}][Cl^-][AgCl]}{[PbCl_2][Ag^+]}$

Substituting the value of $[Cl^-]$ from [2] :

$\frac{K_3}{K_4}=\frac{[Pb^{2+}][AgCl]}{[PbCl_2][Ag^+]}\times \frac{K_4\times [AgCl]}{[Ag^+]}$

$\frac{K_3}{K_4}=K_4\times K$

$K=\frac{K_3}{(K_4)^2}=\frac{1.85\times 10^{-10}}{(1.17\times 10^{-4})^2}$

$K=0.01351$

The equilibrium constant of the given reaction is 0.01351.

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3 years ago