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oksano4ka [1.4K]
2 years ago
14

I need help pleaseee

Chemistry
1 answer:
Likurg_2 [28]2 years ago
3 0
The answer is d!!!!!!!
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Explanation:

nuclear fusion yields more energy than nuclear fission and the products of the reaction are not radioactive

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If 5.0 moles of C3H8 react, how many molecules of water are formed?
artcher [175]
Balance Chemical Equation for combustion of Propane is as follow,

                         C₃H₈  +  5 O₂    →    3 CO₂  +  4 H₂O

According to equation,

              1 mole of C₃H₈ on combustion gives  =  4 moles of H₂O
So,
        5 moles of C₃H₈ on combustion will give  =  X moles of H₂O

Solving for X,
                                    X  =  (5 mol × 4 mol) ÷ 1 mole

                                    X  =  20 moles of H₂O

Calculating number of molecules for 20 moles of H₂O,
As,
                        1 mole of H₂O contains  =  6.022 × 10²³ molecules
So,
                20 moles of H₂O will contain  =  X molecules

Solving for X,
                                 X  =  (20 mole × 6.022 × 10²³ molecules) ÷ 1 mol

                                 X  =  1.20 ×10²⁵ Molecules of H₂O
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13. What physical property is characteristic of all of the elements in the group located in the
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3 years ago
The vapor pressure of pure water at 25 °c is 23.8 torr. What is the vapor pressure (torr) of water above a solution prepared by
maxonik [38]

The vapour pressure of the solution is 23.4 torr.

Use <em>Raoult’s Law</em> to calculate the vapour pressure:  

<em>p</em>₁ = χ₁<em>p</em>₁°  

where  

χ₁ = the mole fraction of the solvent  

<em>p</em>₁ and <em>p</em>₁° are the vapour pressures of the solution and of the pure solvent  

The formula for vapour pressure lowering Δ<em>p</em> is  

Δ<em>p</em> = <em>p</em>₁° - <em>p</em>₁  

Δ<em>p</em> = <em>p</em>₁° - χ₁<em>p</em>₁° = p₁°(1 – χ₁) = χ₂<em>p</em>₁°  

where χ₂ is the mole fraction of the solute.  

<em>Step 1</em>. Calculate the <em>mole fraction of glucose </em>

<em>n</em>₂ = 18.0 g glu × (1 moL glu/180.0 g glu) = 0.1000 mol glu  

<em>n</em>₁ = 95.0 g H_2O × (1 mol H_2O/18.02 g H_2O) = 5.272 mol H_2O  

χ₂ = <em>n</em>₂/(<em>n</em>₁ + n₂) = 0.1000/(0.1000 + 5.272) = 0.1000/5.372 = 0.018 62  

<em>Step 2</em>. Calculate the <em>vapour pressure lowering</em>  

Δ<em>p</em> = χ₂<em>p</em>₁° = 0.018 62 × 23.8 torr = 0.4430 torr  

<em>Step 3</em>. Calculate the <em>vapour pressure</em>  

<em>p₁</em> = <em>p</em>₁° - Δ<em>p</em> = 23.8 torr – 0.4430 torr = 23.4 torr

3 0
3 years ago
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