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Firdavs [7]
3 years ago
15

3.9 X 1022 molecules of CO2 = _ moles of CO2

Chemistry
1 answer:
agasfer [191]3 years ago
7 0

Answer:

moles CO2 = 0.065 moles

Explanation:

  • mole = 6.022 E23 molecules.....Avogadro

∴ molecules CO2 = 3.9 E22 molecules

⇒ moles CO2 = (3.9 E22 molecules CO2)×( moles CO2/ 6.022 E23 molecules CO2)

⇒ moles CO2 = 0.065 moles

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Mechanical advantage = load/_______​
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Mechanical advantage = load/<u>effort</u>

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How do you convert 134kj to Calories?
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1 kilo joule = 0.239006 calories
134 kilo joule = 134 x 0.239006
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When light bends as it passes from one transparent object to another, the light is .
Aleks [24]
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Assume that you did the following dilutions
MrMuchimi
Below is the solution. I hope it helps. 

CfVf = CiVi 
<span>Cf = (CiVi)/ Vf </span>

<span>i. Cf = [ (10^-6 mol / L) (1 mL) (1L / 1000 mL) ] / [ (1kL) (1000L / 1 kL) ] = 1x10^-12 M → use as Ci in next dilution </span>
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3 years ago
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Calculate ℰ° values for the galvanic cells described below. (a) cr3+(aq) + cl2(g) equilibrium reaction arrow cr2o72-(aq) + cl -(
Ad libitum [116K]

Answer:

\boxed{\text{(a) 0.00 V; (b) 0.424 V}}

Explanation:

We must look up the standard reduction potentials for the half-reactions.

                                                                  <u>   ℰ°     </u>

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ ⇌ 2Cr³⁺ + 7H₂O     1.36

Cl₂ + 2e⁻ ⇌ 2Cl⁻                                       1.35827

2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O             1.195

Fe³⁺ + e⁻ ⇌ Fe²⁺                                      0.771

(a) Cr³⁺/Cl₂

We reverse the sign of ℰ° for the oxidation half-reaction. Then we add the two half-reactions and their ℰ° values.

                                                                               <u>   ℰ°/V     </u>

2Cr³⁺ + 7H₂O ⇌ Cr₂O₇²⁻ + 14H⁺ + 6e⁻                  -1.36

<u>Cl₂ + 2e⁻ ⇌ 2Cl⁻                                                </u>     <u>1.358 27 </u>

2Cr³⁺ + 3Cl₂ + 7H₂O ⇌ Cr₂O₇²⁻  + 6Cl⁻ + 14H⁺     0.00

(b) Fe²⁺/IO₃⁻

                                                                           <u> ℰ°/V </u>

Fe²⁺ ⇌ Fe³⁺ + e⁻                                                -0.771

<u>2IO₃⁻ + 12H⁺ + 10e⁻ ⇌ I₂ + 6H₂O                   </u>   <u>  1.195 </u>

10Fe²⁺ + 2IO₃⁻ + 12H⁺ ⇌ 10Fe³⁺ + I₂ + 6H₂O     0.424

The ℰ° values for the cells are \boxed{\textbf{(a) 0.00 V; (b) 0.424 V}}

4 0
4 years ago
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