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3 years ago

14

Look at the diagram below, which shows an atom of an element. How many valence electrons does it have? Based on this, would the

atom be reactive or unreactive?

1 answer:

Cloud [144]3 years ago

4 0

Answer:

unreactive

Explanation:

because look at the picture the picture will show you easily

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Relate the following structures to your body organs_ walls: _houses_: bricks_:_ a room_: rooms

alexgriva [62]

Answer:

what is this I am not getting which type of this question you have taken from where

4 0

2 years ago

What is the density of chlorine gas (MM = 71.0glmol) at 1.50 atm and 25.0C

IrinaVladis [17]

Answer: D=4.35g/L

Explanation:

The formula for density is $D=\frac{M}{V}$ . M is mass in grams and V is volume in liters.

Since we are give pressure and temperature, we can use the ideal gas law to find moles/volume. FInding moles/volume would give us the base for density. All we would have to do is convert moles to grams.

Ideal Gas Law: PV=nRT

$\frac{n}{V} =\frac{P}{RT}$

$\frac{n}{V}=\frac{1.50 atm}{(0.08206Latm/Kmol)(25+274.15K)}$

$\frac{n}{V} =\frac{0.061309mol}{L}$

Now that we have moles, we can use molar mass of chlorine gas to find grams.

$0.061309mol*\frac{71.0g}{mol} =4.3529g$

With our grams, we can find our density.

$D=\frac{4.3529g}{L}$

We need correct significant figures so our density is:

$D=\frac{4.35g}{L}$

5 0

3 years ago

The work function for metallic potassium is 2.3 eV. Calculate the velocity in km/s for electrons ejected from a metallic potassi

mote1985 [20]

Answer:

932.44 km/s

Explanation:

Given that:

The work function of the magnesium = 2.3 eV

Energy in eV can be converted to energy in J as:

1 eV = 1.6022 × 10⁻¹⁹ J

So, work function = $2.3\times 1.6022\times 10^{-19}\ J=3.68506\times 10^{-19}\ J$

Using the equation for photoelectric effect as:

$E=\psi _0+\frac {1}{2}\times m\times v^2$

Also, $E=\frac {h\times c}{\lambda}$

Applying the equation as:

$\frac {h\times c}{\lambda}=\psi _0+\frac {1}{2}\times m\times v^2$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

m is the mass of electron having value $9.11\times 10^{-31}\ kg$

$\lambda$ is the wavelength of the light being bombarded

$\psi _0=Work\ function$

v is the velocity of electron

Given, $\lambda=260\ nm=260\times 10^{-9}\ m$

Thus, applying values as:

$\frac{6.626\times 10^{-34}\times 3\times 10^8}{260\times 10^{-9}}=3.68506\times 10^{-19}+\frac{1}{2}\times 9.11\times 10^{-31}\times v^2$

$3.68506\times \:10^{-19}+\frac{1}{2}\times \:9.11\times \:10^{-31}v^2=\frac{6.626\times \:10^{-34}\times \:3\times \:10^8}{260\times \:10^{-9}}$

$\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=7.64538\times 10^{-19}-3.68506\times 10^{-19}$

$\frac{1}{2}\times 9.11\times 10^{-31}\times v^2=3.96032462\times 10^{-19}$

$v^2=0.869446\times 10^{12}$

v = 9.3244 × 10⁵ m/s

Also, 1 m = 0.001 km

<u>So, v = 932.44 km/s</u>

5 0

3 years ago

Which of the following gas samples would contain the same amount of gas as 200 mL of helium, He(g), at 25° C and 1 atm?

monitta

Answer:

$200\; \rm mL$ of neon $\rm Ne\, (g)$ at $25^{\circ} \rm C$ and $1\; \rm atm$ (the second choice) would contain an equal number of gas particles as $200\; \rm mL\!$ of $\rm He \, (g)$ at $25^{\circ} \rm C\!$ and $1\; \rm atm\!$ (assuming that all four gas samples behave like ideal gases.)

Explanation:

By Avogadro's Law, if the temperature and pressure of two ideal gases is the same, the number of gas particles in each gas would be proportional to the volume of that gas.

All four gas samples in this question share the same temperature and pressure. Hence, if all these gases are ideal gases, the number of gas particles in each sample would be proportional to the volume of that sample. Two of these samples would contain the same number of gas particles if and only if the volume of the two samples is equal to one another.

The second choice, $200\; \rm mL$ of neon $\rm Ne\, (g)$ at $25^{\circ} \rm C$ and $1\; \rm atm$ , is the only choice where the volume of the sample is also $200\; \rm mL \!$ . Hence, that choice would be the only one with as many gas particles as $200\; \rm mL\!$ of $\rm He \, (g)$ at $25^{\circ} \rm C\!$ and $1\; \rm atm\!$ .

7 0

3 years ago

Gold is the [...] of a valuable element.<br> A. Chemical formula. <br> B. Symbol. <br> C. Name

pychu [463]

Answer:

C. Name

Explanation:

3 0

4 years ago

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