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miskamm [114]
4 years ago
14

Jane Doe has a cholesterol (C27H46O) count of 178 mg/dL. How many cholesterol molecules does Jane Doe have in each deciliter of

blood?
Chemistry
1 answer:
Ilya [14]4 years ago
4 0

Answer:

Jane has 2.77 * 10^20 molecules cholesterol in 1 dL of blood

Explanation:

<u>Step 1:</u> Given data

Molar mass of Cholesterol =  386.654 g/mol

Jane has a cholesterol of 178 mg/dL

<u>Step 2:</u> Calculate mass of cholsterol in 1 dL

178 mg/dL means in 1 dL she has 178 mg cholesterol or 0.178 grams

<u>Step 3</u>: Calculate number of moles of cholesterol

Number of moles = mass of cholesterol / molar mass of cholesterol

Number of moles = 0.178 grams / 386.654 g/mol = 4.6 * 10^-4 moles

<u>Step 4: </u>Calculate number of molecules

Number of molecules = 4.6 * 10^-4 mol * 6.022 *10^23 / mol = 2.77 * 10^20 molecules

Jane has 2.77 * 10^20 molecules cholesterol in 1 dL of blood

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Answer:
             3.2 g of O₂

Solution:
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                                           K  +  O₂     →     KO₂

First let us confirm that either the given amount of Potassium produces the given amount of Potassium oxide or not,
So,
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So,
                          3.91 g of K will produce  =  X g of K₂O

Solving for X,
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                      X  =  7.11 g of K₂O

Hence, it is confirmed that we have selected the right equation,
So,
As,
                     39.098 g of K required  =  32 g of O₂
So,
                     3.91 g of K will require  =  X g of O₂

Solving for X,
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At a certain temperature, Kc = 0.0500 and ∆H = +39.0 kJ for the reaction below, 2 MgCl2(s) + O2(g) → 2MgO(s) + Cl2(g) Calculate
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Explanation:

Since, it is shown that the reaction has been reversed. Therefore, value of K_{c} will become \frac{1}{K_{c}}.

Hence, new K_{c'} = \frac{1}{K_{c}}

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Also, the number of moles of each reactant has been halved. So, K_{c''} for the reaction MgO(s) + \frac{1}{2}Cl2(g) → MgCl_{2}(s) + \frac{1}{2} O2(g) will also get halved.

Therefore,     K_{c''}  = K_{c'} = (20)^{0.5}

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As a result, K_{c} will also increase with increase in temperature.

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