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Elan Coil [88]
2 years ago
12

A stack of 15 pennies is immersed into a 100 mL graduate initially containing 20.6 mL of water. The volume in the graduate cylin

der increases to 26.0 after the pennies are added. Calculate the volume in mL of the stack of 15 pennies
Chemistry
1 answer:
Gekata [30.6K]2 years ago
5 0

Answer: 5.4

Explanation: The Law of Impenetrability says that two objects can't occupy the same space at the same time; therefore, the pennies and the water can't occupy the same space at the same time. Without the pennies the water level read 20.6 mL, dropping in the pennies gives a level of 26.0.

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3 years ago
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The half-life of cesium-137 is 30 years. Suppose we have a 180 mg sample. Find the mass that remains after t years. Let y(t) be
Stels [109]

Explanation:

1) Initial mass of the Cesium-137=N_o= 180 mg

Mass of Cesium after time t = N

Formula used :

N=N_o\times e^{-\lambda t}\\\\\lambda =\frac{0.693}{t_{\frac{1}{2}}}

Half life of the cesium-137 = t_{1/2}=30 years[p/tex]where,
[tex]N_o = initial mass of isotope

N = mass of the parent isotope left after the time, (t)

t_{\frac{1}{2}} = half life of the isotope

\lambda = rate constant

N=N_o\times e^{-(\frac{0.693}{t_{1/2}})\times t}

Now put all the given values in this formula, we get

N=180mg\times e^{-\frac{0.693}{30 years}\times t}

Mass that remains after t years.

N=180 mg\times e^{0.0231 year^{-1}\times t}

Therefore, the parent isotope remain after one half life will be, 100 grams.

2)

t = 70 years

N_o=180 mg

t_{1/2}= 30 yeras

N=180mg\times e^{-\frac{0.693}{30 years}\times 70 years}

N = 35.73 mg

35.73 mg of cesium-137 will remain after 70 years.

3)

N_o=180 mg

t_{1/2}= 30 yeras

N = 1 mg

t = ?

1 mg =180mg\times e^{-\frac{0.693}{30 years}\times t}

\frac{-30 year}{0.693}\times \ln \frac{1 mg}{180 mg}=t

t = 224.80 years ≈ 225 years

After 225 years only 1 mg of cesium-137 will remain.

7 0
3 years ago
A polar covalent bond will form between which two atoms?
Alex_Xolod [135]

Types of Bonds can be predicted by calculating the difference in electronegativity.

If, Electronegativity difference is,

 

                Less than 0.4 then it is Non Polar Covalent

                

                Between 0.4 and 1.7 then it is Polar Covalent 

            

                Greater than 1.7 then it is Ionic

 

For Be and F,

                    E.N of Fluorine          =   3.98

                    E.N of Beryllium        =   1.57

                                                   ________

                    E.N Difference                2.41          (Ionic Bond)


For H and Cl,

                    E.N of Chorine           =   3.16

                    E.N of Hydrogen        =   2.20

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For Na and O,

                    E.N of Oxygen          =   3.44

                    E.N of Sodium          =   0.93

                                                       ________

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For F and F,

                    E.N of Fluorine          =   3.98

                    E.N of Fluorine          =   3.98

                                                        ________

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Result:

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To summarize a complete procedure for separating a mixture ofseveral substances, it is best to prepare a flow chart. A flowchartis a schematic representation of an algorithm or a stepwiseprocess, showing the steps as boxes of various kinds, and theirorder by connecting these with arrows. Flowcharts are used indesigning or documenting a process.

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