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Semenov [28]
2 years ago
12

How much volume (in cm3) is gained by a person who gains 11.1 lbs of pure fat?

Chemistry
1 answer:
kirill [66]2 years ago
6 0

The density of human fat is 0.918 g/cm^{3}. The mass of the pure fat is 11.1 lbs.

First convert mass from lb to g as follows:

1 lb=453.592 g

Thus,

11.1 lb\rightarrow 11.1\times 453.592 g=5034.875 g

Density of a substance is defined as mass per unit volume, thus volume can be calculated as:

V=\frac{m}{d}

Putting the values,

V=\frac{5034.875 g}{0.918 g/cm^{3}}=5484.61 cm^{3}

Therefore, volume gained by person will be 5484.61 cm^{3}.

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podryga [215]

100m/26s=50m/13s

50m/13s=3.846m/s

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How many moles of KCl will be<br> produced from 15.0 g of KClO3?
slega [8]

Answer:

0.12 mol KCl

Explanation:

2 KClO3 (s)   2 KCl (s) + 3 O2 (g)

15 g                x mol

x g KCl = 15 g KClO3 x[ (1 mol KClO3)/ (122.5 g KClO3) ] x [(2 mol KCl)/ (2 mol KClO3)]

x g KCl = 0.12 mol KCl

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A 50.0 mL sample of 0.600 M calcium hydroxide is mixed with 50.0 mL sample of 0.600 M hydrobromic acid in a Styrofoam cup. The t
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Explanation:

The reaction equation will be as follows.

     Ca(OH)_{2}(aq) + 2HBr(aq) \rightarrow CaBr_{2}(aq) + 2H_{2}O(l)

So, according to this equation, 1 mole Ca(OH)_{2} = 2 mol HBr = 1 mol CaBr_{2}

Therefore, calculate the number of moles of calcium hydroxide as follows.

     No. of moles of Ca(OH)_{2} = V \times Molarity

                                    = 50 \times 0.6

                                    = 30 mmol

Similarly, calculate the number of moles of HBr as follows.

        No. of moles of HBr = M \times V

                                          = 50 \times 0.6

                                          = 30 mmol

This means that the limiting reactant is HBr.

So, no. of moles of CaBr_{2} = 30 \times \frac{1}{2}

                                                     = 15 mmol

Hence, calculate the amount of heat released as follows.

                Heat released in the reaction(q) = m \times s \times \Delta T

as,    m = mass of solution

and,             Density = \frac{mass}{volume}

or,                  mass = Density × Volume

                               = 1.08 g/ml \times (50 + 50) ml

                               = 108 g

where,    s = specific heat of solution = 4.18 j/g.k

and,        change in temperature \Delta T = (26 - 23)^{o}C

                                                                 = 3&#10;^{o}C

Hence, the heat released will be as follows.

                   q = m \times s \times \Delta T

                        q = 108 \times 4.18 \times 3^{o}C

                           = 1354.32 joule

or,                        = 1.354 kJ       (as 1 kJ = 1000 J)    

Also,          \Delta H_{rxn} = \frac{-q}{n}

                              = \frac{-1.354}{15 \times 10^{-3}}

                              = -90.267 kJ/mol

Thus, we can conclude that the enthalpy change for the given reaction is -90.267 kJ/mol.

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Answer : Both solutions contain 3.011 X 10^{23} molecules.

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Here, only molarity values are given; where molarity is a measurement of concentration in terms of moles of the solute per liter of solvent.

Since each substance has the same concentration, 0.5 M, each will have the same number of molecules present per liter of solution.

Addition of molar mass for individual substance is not needed. As if both are considered in 1 Liter they would have same moles which is 0.5.

We can calculate the number of molecules for each;

Number of molecules  = N_{A} X M;

∴  Number of molecules =  6.022 X 10^{23} X 0.5 mol/L X 1 L which will be  = 3.011 X 10^{23}

Thus, these solutions compare to each other in that they have not only the same concentration, but they will have the same number of solvated sugar molecules. But the mass of glucose dissolved will be less than the mass of sucrose.

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Answer:D

Explanation:

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