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Goshia [24]
3 years ago
9

The molecular geometry consists of ________.

Chemistry
1 answer:
elena-s [515]3 years ago
6 0
The molecular geometry consist of A SINGLE BOND AND A MULTIPLE BOND.
The molecular geometry refers to the three dimensional arrangement of the atoms that constitute a molecule. The molecular geometry of a molecule influences the properties of the substance.
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A sample of argon gas has a volume of 73 mL at a pressure of 1.20 atm and a temperature of 112 degrees Celsius. What is the fina
NeX [460]

Answer:

V₂ → 106.6 mL

Explanation:

We apply the Ideal Gases Law to solve the problem. For the two situations:

P . V = n . R . T

Moles are still the same so → P. V / R. T = n

As R is a constant, the formula to solve this is: P . V / T

P₁ . V₁ / T₁ = P₂ .V₂ / T₂   Let's replace data:

(1.20 atm . 73mL) / 112°C = (0.55 atm . V₂) / 75°C

((87.6 mL.atm) / 112°C) . 75°C = 0.55 atm . V₂

58.66 mL.atm = 0.55 atm . V₂

58.66 mL.atm / 0.55 atm = V₂ → 106.6 mL

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3 years ago
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Answer:

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Explanation:

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Protons are positive
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When an electric field is applied to a metallic crystal, the movement of electrons is _____
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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.

6 0
3 years ago
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