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3 years ago

The______ the forces among the particles in a sample of matter, the more rigid the matter will be.

Chemistry

1 answer:

Alexus [3.1K]3 years ago

3 0

Answer:

Stronger!

Explanation:

The stronger the forces among the particles in a sample of matter, the more rigid the matter will be.

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The initial activity of 37Ar is 8540 disintegrations per minute. After 10.0 days, the activity is 6990 disintegrations per minut

mylen [45]

Answer:

Approximately 3318 disintegrations per minute.

Explanation:

The activity $A$ of a radioactive decay at time $t$ can be found with the following equation:

$\displaystyle A(t) = A_0 \cdot \mathrm{e}^{-\lambda\cdot t}$ .

In this equation,

$\mathrm{e}$ is the natural base. $\mathrm{e}\approx 2.71828$ .
$A_0$ is the initial activity of the decay. For this question, $A_0 = \rm 8540\; min^{-1}$ .
The decay constant $\lambda$ of this sample needs to be found.

The decay constant here can be found using the activity after 10 days. As long as both times are in the same unit (days in this case,) conversion will not be necessary.

$A(10) = A_0\cdot \mathrm{e}^{\rm -\lambda \times 10\;day}= (\mathrm{8540\; min^{-1}})\cdot \mathrm{e}^{\rm -\lambda \times 10\;day}$ .

$A(10) = \rm 6690\; min^{-1}$ .

$\displaystyle \frac{\rm 6690\; min^{-1}}{\mathrm{8540\; min^{-1}}} = \mathrm{e}^{\rm -\lambda \times 10\;day}$

Apply the natural logarithm to both sides of this equation.

$\displaystyle \ln{\left(\frac{\rm 6690\; min^{-1}}{\mathrm{8540\; min^{-1}}}\right)} = \ln{\left(\mathrm{e}^{\rm -\lambda \times 10\;day}\right)}$ .

$\displaystyle \rm -\lambda \cdot (10\;day) = \ln{\left(\frac{\rm 6690\; min^{-1}}{\mathrm{8540\; min^{-1}}}\right)}$ .

$\displaystyle \rm \lambda= \rm \frac{\displaystyle \ln{\left(\frac{\rm 6690\; min^{-1}}{\mathrm{8540\; min^{-1}}}\right)}}{-10 \; day} \approx 0.0200280\; day^{-1}$ .

Note that the unit of the decay constant $\lambda$ is $\rm day^{-1}$ (the reciprocal of days.) The exponent $-\lambda \cdot t$ should be dimensionless. In other words, the unit of $t$ should also be days. This observation confirms that there's no need for unit conversion as long as the two times are in the same unit.

Apply the equation for decay activity at time $t$ to find the decay activity after 47.2 days.

$\displaystyle \begin{aligned}A(t)& = A_0 \cdot \mathrm{e}^{-\lambda\cdot t}\\&\approx \rm \left(8540\; min^{-1}\right)\cdot \mathrm{e}^{-0.0200280\; day^{-1}\times 47.2\;day}\\&\approx \rm 3318\; min^{-1}\end{aligned}$ .

By dimensional analysis, the unit of activity here should also be disintegrations per minute. The activity after 47.2 days will be approximately 3318 disintegrations per minute.

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3 years ago

Functions of the skin other than thermoregulation

german

Answer:

Protection and sensation

Explanation:

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What would happen if there were not enough plants on earth

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For each property listed, identify the type of element it describes.

Juliette [100K]

Answer:

1.Very good electrical conductivity : Metals (Decreacing order of conductivity)

Silver > Copper > Gold > aluminium

2. Amphoteric : Metal elements

Beryllium , Aluminium , Zinc ,

3.Gaseous at room temperature: mostly Nobel gases elements and some non - metal elements.

Helium ,neon , argon , krypton , fluorine , Oxygen , nitrogen

4.Solid at room temperature: Mostly Metals (few non-metals, metalloid elements)

Metals (Sodium , potassium , calcium , gold are solid)

Non- metals(Carbon ,Boron )

Metalloids(antimony)

5. Brittle : non - metals (can't be rolled into wires)

Hydrogen , carbon , sulfur , phosphorus

Explanation:

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2 years ago

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What kind of bonds are hydrogen bonds???

34kurt

Hydrogen bonding is a special type of dipole-dipole attraction between molecules, not a covalent bond to a hydrogen atom. It results from the attractive force between a hydrogen atom covalently bonded to a very electronegative atom such as a N, O, or F atom and another very electronegative atom.

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