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2 years ago

1. How many times will a mole of paperclips (think giant paperclip chain) stretch around the equator of the

Chemistry

1 answer:

max2010maxim [7]2 years ago

3 0

Yea, it’s either earth (1 paperclip=0.030 m earths circumference = 4x10m)

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A mineral that breaks into irregular pieces is said to show which of the following?

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3 years ago

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A certain radioactive isotope decays at a rate of 0.2% annually. Determine the half-life of this isotope, to the nearest year

pychu [463]

Answer:

The half-life of the radioactive isotope is 346 years.

Explanation:

The decay rate of the isotope is modelled after the following first-order linear ordinary differential equation:

$\frac{dm}{dt} = -\frac{m}{\tau}$

Where:

$m$ - Current isotope mass, measured in kilograms.

$t$ - Time, measured in years.

$\tau$ - Time constant, measured in years.

The solution of this differential equation is:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Where $m_{o}$ is the initial mass of the isotope. It is known that radioactive isotope decays at a yearly rate of 0.2 % annually, then, the following relationship is obtained:

$\%e = \frac{m(t)-m(t+1)}{m(t)}\times 100\,\% = 0.2\,\%$

$1 - \frac{m(t+1)}{m(t)} = 0.002$

$1 - \frac{m_{o}\cdot e^{-\frac{t+1}{\tau} }}{m_{o}\cdot e^{-\frac{t}{\tau} }}=0.002$

$1 - e^{-\frac{1}{\tau} } = 0.002$

$e^{-\frac{1}{\tau} } = 0.998$

$-\frac{1}{\tau} = \ln 0.998$

The time constant associated to the decay is:

$\tau = -\frac{1}{\ln 0.998}$

$\tau \approx 499.500\,years$

Finally, the half-life of the isotope as a function of time constant is given by the expression described below:

$t_{1/2} = \tau \cdot \ln 2$

If $\tau \approx 499.500\,years$ , the half-life of the isotope is:

$t_{1/2} = (499.500\,years)\cdot \ln 2$

$t_{1/2}\approx 346.227\,years$

The half-life of the radioactive isotope is 346 years.

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2 years ago

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It is measured with a barometer

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3 years ago

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3 years ago

What is the enthalpy change when 49.4 ml of 0.430 m sulfuric acid reacts with 23.3 ml of 0.309 m potassium hydroxide?

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Potassium hydroxide commonly known as caustic potash is an inorganic compound with a chemical formula KOH which is a strong base that is marketed in several forms.

H2SO4(aq) + 2KOH(aq) ---> K2SO4(aq) + 2H2O(l) H°= –111.6 kJ/mol

The answer in this question is -0.402 kJ

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3 years ago