A flask at room temperature contains exactly equal amounts (in moles) of nitrogen and xenon.

a. Which of the two gases exerts the greater partial pressure?
b. The molecules or atoms of which gas have the greater average velocity?
c. The molecules of which gas have the greater average kinetic energy?
d . If a small hole were opened in the flask, which gas effuses more quickly?

Answer:

No

No

Yes 25cm

Explanation:

1) mass can't be divided by distance.

2) the measurements are completely different but under the same category of distance. you can only divide it if meters was cm and squared

3) measurements the same and you can multiply it

Answer: [tex]MgS[/tex], [tex]MgBr_2[/tex], [tex]V_2S_5[/tex] and [tex]VBr_5[/tex] are four binary ionic compounds that could be formed.

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

Mg is having an oxidation state of +2 called as [tex]Mg^{2+}[/tex] cation and [tex]S^{2-}[/tex] is an anion with oxidation state of -2. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral [tex]MgS[/tex]. Also [tex]Mg^{2+}[/tex] cation and [tex]Br^{-}[/tex] can combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral [tex]MgBr_2[/tex].

Also Vanadium is having an oxidation state of +5 called as [tex]V^{5+}[/tex] cation and [tex]S^{2-}[/tex] is an anion with oxidation state of -2. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral [tex]V_2S_5[/tex]. Also [tex]V^{5+}[/tex] cation and [tex]Br^{-}[/tex] anion can combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral [tex]VBr_5[/tex].

Answer:

Positive charges: 2

Negative charges: 1

Overall charge: +1

Explanation:

In Diagram B we have the chemical element Helium (He). You can see in the Periodic Table that He has an atomic number of 2 and an atomic mass of approximately 4.

The atomic number is equal to the number of protons, so He has 2 protons. The protons are the positive charges in the atom, so the number of positive charges is 2. That is invariable, if the number of protons - or the atomic number - of an atom changes, it changes its identity.

In the drawing, we can see that the nucleus is composed of 2 protons (2 'p' particles) and 2 neutrons (2 'n' particles). The atomic mass is calculated as the number of protons and neutrons, so 2 protons+ 2 neutrons = 4.

The surrounding particle with an 'e' is an electron. So, we have 1 electron. Electrons are negative charges, so the number of negative charges is 1.

Now, we can calculate the overall charge. We add the positive charges (protons) and negative charges (electrons). Each proton has a +1 charge, whereas each electron has a -1 charge. Finally, we perform the addition:

Overall charge= (number of protons x charge) + (number of electrons x charge) = (2 x (+1)) + (1 x (-1)) = (+2) + (-1) = +1

Answer: The periodic table is a chart where different types of elements organized by what they have in common.

Explanation:

Answer:

sdasdasddsafsadsfds

Explanation:

Answer:

Number three, because precipitation is rain, which comes from cumulonimbus clouds. Precipitation happens when clouds have accumulated enough evaporated water for the cloud to get heavy and need to get rid of some of the water, which it does through condensing the water and then delivering it as rain.

Hope this helps! Have a wonderful day! (Please give me brainliest and thank you!)

:)

Answer:

Pretty sure it's C

Explanation:

Answer:

The group 14 elements all have ns2np2 valence electron configurations. All form compounds in which they formally lose either the two np and the two ns valence electrons or just the two np valence electrons, giving a +4 or +2 oxidation state, respectively

Explanation:

Answer:

The group 14 elements all have ns2np2 valence electron configurations. All form compounds in which they formally lose either the two np and the two ns valence electrons or just the two np valence electrons, giving a +4 or +2 oxidation state, respectively.

Explanation:

Answer:

number 2

Explanation:

even though i have never done this i still got it right

Because metals do not form covalent bonds, when naming covalent bonds, all the given rules apply except write the metal first; option 4.

Covalent bonds are bonds formed by the sharing of valence electrons between the atoms involved in the bonding.

In naming covalent bonds, the following rules are applied:

For example, CO₂, is carbon dioxide.

Therefore, when naming covalent bonds, all the given rules apply except write the metal first because metals do not form covalent bonds.

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Answer:

pH=11.

Explanation:

Hello!

In this case, since the data is not given, it is possible to use a similar problem like:

"An analytical chemist is titrating 185.0 mL of a 0.7500 M solution of ethylamine(C2HNH2) with a 0.4800 M solution of HNO3.ThepK,of ethylamine is 3.19. Calculate the pH of the base solution after the chemist has added 114.4 mL of the HNO3 solution to it"

Thus, for the reaction:

[tex]C_2H_5NH_2+H^+\rightleftharpoons C_2H_5NH_4^+[/tex]

Tt is possible to compute the remaining moles of ethylamine via the following subtraction:

[tex]n_{ethylamine}=0.1850L*0.7500mol/L=0.1365mol\\\\n_{acid}=0.1144L*0.4800mol/L=0.0549mol\\\\n_{ethylamine}^{remaining}=0.1365mol-0.0549mol=0.0816mol[/tex]

Thus, the concentration of ethylamine in solution is:

[tex][ethylamine]=\frac{0.0816mol}{0.1850L+0.1144L}=0.2725M[/tex]

Now, we can also infer that some salt is formed, and has the following concentration:

[tex][salt]=\frac{0.0549mol}{0.1850L+0.1144L}=0.1834M[/tex]

Therefore, we can use the Henderson-Hasselbach equation to compute the resulting pOH first:

[tex]pOH=pKb+log(\frac{[salt]}{[base]} )\\\\pOH=3.19+log(\frac{0.1834M}{0.2725M})\\\\pOH=3.0[/tex]

Finally, the pH turns out to be:

[tex]pH=14-pOH=14-3\\\\pH=11[/tex]

NOTE: keep in mind that if you have different values, you can just change them and follow the very same process here.

Best regards!

Answer:

First, let's review the ideal gas law, PV = nRT. In this equation, 'P' is the pressure in atmospheres, 'V' is the volumes in liters, 'N' is the number of particles in moles, 'T' is the temperature in Kelvin and 'R' is the ideal gas constant (0.0821 liter atmosphere per moles Kelvin).

Explanation:

I majored in Chemistry

Answer:

0.43 atm

Explanation:

From the question given above, the following data were obtained:

Initial pressure (P₁) = 0.95 atm

Initial volume (V₁) = 0.55 L

Final volume (V₂) = 1.22 L

Final pressure (P₂) =?

The final pressure of the gas can be obtained by using the Boyle's law equation as follow:

P₁V₁ = P₂V₂

0.95 × 0.55 = P₂ × 1.22

0.5225 = P₂ × 1.22

Divide both side by 1.22

P₂ = 0.5225 / 1.22

P₂ = 0.43 atm

Therefore, the final pressure of the gas is 0.43 atm

The final pressure of the gas is 0.43 atm.

Pressure is the process of force exerted perpendicular to any surface.

Given, the Initial pressure (P₁) is 0.95 atm

The initial volume (V₁) is 0.55 L

The final volume (V₂) is 1.22 L

To find the final pressure (P₂)

By the Boyle's Law

[tex]\rm P_1V_2= P_2V_2[/tex]

[tex]\rm 0.95\times 0.55 = P_2\times1.22\\\\0.5225 = P_2 \times 1.22\\\\Divided\;both \;side\;by\;1.22\\\\P_2=\dfrac{0.5225}{1.22} = 0.43 atm.[/tex]

Thus, the final pressure of the gas is 0.43 atm.

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ANSWER:

Nonpolar

sorry if I'm wrong

The question is missing some parts. Here is the complete question.

Carbon monoxide is a colorless, odorless gas that binds irreversibly to hemoglobin in our blood, causing suffocation and death. CO is formed during incomplete combustion of carbon. One way to represent this equilibrium is:

[tex]2CO_{(g)}[/tex] ⇄ [tex]2C_{(s)}+O_{2}_{(g)}[/tex]

we could also write this reaction three other ways listed below. The equilibrium constant for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above.

1) [tex]2C_{(s)}+O_{2}_{(g)}[/tex] ⇄ [tex]2CO_{(g)}[/tex]       K₁ =

2) [tex]C_{(s)}+1/2O_{2}_{(g)}[/tex] ⇄ [tex]CO_{(g)}[/tex]    K₂ =

3) [tex]CO_{(g)}[/tex] ⇄ [tex]C_{(s)}+1/2O_{2}_{(g)}[/tex]    K₃ =

Answer: 1) [tex]K_{1}=\frac{1}{K}[/tex]

              2) [tex]K_{2}=\frac{1}{K^{1/2}}[/tex]

             3) [tex]K_{3}=K^{1/2}[/tex]

Explanation: A chemical reaction can be reversible, i.e., can proceed in both directions: to the right of the arrow (forward) or towards the left of the arrow (backward).

When the rates of forward and backward reactions are the same, the reaction is in equilibrium. In that state, we can determine the equilibrium constant, [tex]K_{c}[/tex].

For the first way to represent equilibrium of CO formed, the [tex]K_{c}[/tex] is calculated

[tex]2CO_{(g)}[/tex] ⇄ [tex]2C_{(s)}+O_{2}_{(g)}[/tex]

[tex]K=\frac{[O_{2}]}{[CO]^{2}}[/tex]

in which the symbol [ ] is concentration of the compound.

In equilibrium constant, solids are not included.

Equilibrium constants for the other reactions:

1) [tex]2C_{(s)}+O_{2}_{(g)}[/tex] ⇄ [tex]2CO_{(g)}[/tex]  

[tex]K_{1}=\frac{[CO]^{2}}{[O_{2}]}[/tex]

Comparing K₁ and K, the first one is the inverse of K, so writing in terms of K

[tex]K_{1}=\frac{1}{K}[/tex]

2) [tex]C_{(s)}+1/2O_{2}_{(g)}[/tex] ⇄ [tex]CO_{(g)}[/tex]

[tex]K_{2}=\frac{[CO]}{[O_{2}]^{1/2}}[/tex]

In terms of K, K₂ is

[tex]K_{2}=\frac{1}{K^{1/2}}[/tex]

3) [tex]CO_{(g)}[/tex] ⇄ [tex]C_{(s)}+1/2O_{2}_{(g)}[/tex]

[tex]K_{3}=\frac{[O_{2}]^{1/2}}{[CO]}[/tex]

This constant in terms of K will be

[tex]K_{3}=K^{1/2}[/tex]

In conclusion, K₁, K₂ and K₃ in terms of K is [tex]\frac{1}{K}[/tex],[tex]\frac{1}{K^{1/2}}[/tex] and [tex]K^{1/2}[/tex], respectively.

Answer:

13.9kj/mol

Explanation:

[tex]h = 891 -(393.5 +2*241.8)=1768.1=13.9kj/mol[/tex]

Answer:  [tex]X_3Z_4[/tex]

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

Here metal X is having an oxidation state of +4 called as [tex]X^{4+}[/tex] cation and [tex]Z^{3-}[/tex] is an anion with oxidation state of -3. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral [tex]X_3Z_4[/tex]

[tex]X_3Z_4[/tex] is the formula of the resulting compound.

Answer:

option D is correct answer of this question

Answer:

Explanation:

As the temperature of a liquid rises, the average speed of its molecules rises as well. The molecules in a liquid move faster as the temperature rises, increasing the liquid's kinetic energy. Heating liquids will demonstrate this property in experiments.

Answer:

MnSO4.5H2O(s) ---------> MnSO4(s) + 5H2O(g)

Explanation:

The dehydration of a hydrate implies that the water of crystallization is lost. Water of crystallization is included in the crystal structure as it is formed and is incorporated into the structure of the compound.

Now the equation for the dehydration of manganese(II) sulfate pentahydrate is;

MnSO4.5H2O(s) ---------> MnSO4(s) + 5H2O(g)

The  MnSO4 is now said to be anhydrous.

Answer:

There are 0.5 moles

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number  of particles.

To calculate the moles, we use the equation:

Thus there are 0.5 moles

Explanation:

Answer:

She's right that there's a difference, and you are right that it's all just electromagnetic waves!

The key to this is that there is no such thing as "white light" when you really get down to it. Each light emits a range of wavelengths of light. If they have a sufficiently even distribution of wavelengths, we tend to call that light "white," but we can only use that term informally.

Both the sun and the light bulb emit so-called "Blackbody radiation." This is the particular spectrum of light that's associated with the random thermal emissions of a hot object. Cool objects tend to emit more of their energy in the longer wavelengths like reds and IRs, while hotter objects emit more energy in the shorter wavelengths like blues and UV.

Blackbody graphic

(Note, there are other possible emission spectra, but those are associated with different materials doing the emissions and, for the purposes of this discussion, they aren't too important. We can just claim the emissions are all blackbody)

If you notice, as you get hotter, a larger portion of the energy is emitted in the blue, violet, and ultraviolet. That's how you get a sunburn from the sun. It's harder to get a sunburn from an artificial light, not because it's artificial, but because those lights are almost always cooler than the sun. They don't have as much UV content. Instead, they have more red and yellow, which incidentally is why pictures taken indoors look very yellow. If you use a strobe, however, all those yellow hues go away because a strobe light is very warm, with lots of blues.

You can get a sunburn from artificial light, of course. Tanning beds are the obvious example, but there are other interesting ones. When you're a jeweler working in platinum, for instance, you need to wear UV protective gear (like glasses or even sunscreen). Platinum's melting point is so hot that it actually emits quite a lot of UV light and can give you a sunburn!

Other than these spectra, there is nothing different between light from an artificial source and light from the sun. Photons are photons.

Explanation:

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#STAY SAFE

Answer:

cell membrane

Explanation:

it controls what comes and goes out of a cell

Answer:

Cell membrane

Explanation:

It separates the interior of all cells from the outside environment (the extracellular space) which protects the cell from its environment.

Answer:

[tex]V_2=120L[/tex]

Explanation:

Hello!

In this case, since Helium is undergoing a volume-pressure change, according the Boyle's law, we infer the following equation is used:

[tex]P_1V_1=P_2V_2[/tex]

Thus, since we are not given the volume at 0.25 atm, we can solve for V2 to do so:

[tex]V_2=\frac{P_1V_1}{P_2}[/tex]

Thus, we plug in to obtain:

[tex]V_2=\frac{1atm*30L}{0.25atm}=120L[/tex]

Best regards!

The final volume of the balloon at the given altitude is 120 L.

The given parameters;

The final volume of the balloon at the given altitude is calculated by applying Boyles law as follows;

[tex]P_1 V_1 = P_2 V_2\\\\V_2 = \frac{P_1V_1}{P_2} \\\\V_2 = \frac{30 \times 1 }{0.25} \\\\V_2 = 120 \ L[/tex]

Thus, the final volume of the balloon at the given altitude is 120 L.

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Answer:

B.) P4

Explanation:

Step 1: Convert 25.0 grams of phosphorus into moles of phosphorous:

25 g P4 x 1 mol P4/123.90 g P4 = 0.202 mol P4

Step 2: Convert 0.202 mol P4 into moles of O2 using their stoichiometric ratios:

0.202 mol P4 x 5 mol O2/1 mol P4 = 1.01 mol O2

Step 3: Convert 1.01 mol O2 into grams of O2:

1.01 mol O2 x 31.98 g O2/1 mol O2 = 32.3 g O2

Because 25.0 grams of phosphorous only needs 32.3 grams of oxygen to react, phosphorous will be the limiting reactant, as after the 25.0 grams of phosphorous is used up, there will still be 17.7 grams of oxygen leftover (excess reactant).

Because the reaction cannot occur without more phosphorous to react with the 17.7 grams of oxygen remaining, P4 is the limiting reactant.

Answer:  B.) P4

Explanation:

        25 g P4 x 1 mol P4/123.90 g P4 = 0.202 mol P4

        0.202 mol P4 x 5 mol O2/1 mol P4 = 1.01 mol O2

        1.01 mol O2 x 31.98 g O2/1 mol O2 = 32.3 g O2

Because 25.0 grams of phosphorous only needs 32.3 grams of oxygen to react, phosphorous will be the limiting reactant, as after the 25.0 grams of phosphorous is used up, there will still be 17.7 grams of oxygen leftover (excess reactant).

Because the reaction cannot occur without more phosphorous to react with the 17.7 grams of oxygen remaining, P4 is the limiting reactant.

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Answer:

stratosphere

Explanation:

Answer:

See explanation below

Explanation:

In this case we have the starting reactant which is the ethine, In the first step reacts with NaNH₂, a strong base. This base will substract the hydrogen from one of the carbon of the ethine, and form a carbanion. This will react with the propane bromide, displacing the bromine and forming a 5 carbon chain with the triple bond on the carbon 1 and 2.

In the second step, reacts with the lindlar catalyst to do a reduction, and form a double bond between carbon 1 and 2. In essence, compound A is similar to compound B.

Finally B reacts with water in acid and makes a addition reaction, and form an alcohol.

The whole process can be seen in the picture below.

Hope this helps

Answer:

Sexual Reproduction

Explanation:

In Asexual Reproduction offsprings are know as ramets and many ramets are known as clowns. So, when two parents are involved genetic variations occur in offspring. Pollen possess male gamete (antherozoid/sperm) and ovule possess egg.

A part

Answer:

[tex]m_{H_2}=12.6gH_2[/tex]

Explanation:

Hello there!

In this case, given the reaction:

[tex]3 H_2+ N_2 \rightarrow 2NH_3[/tex]

It is possible to compute the necessary grams of hydrogen which produce 71.1 g of ammonia, given the 3:2 mole ratio and their molar mass as shown below:

[tex]m_{H_2}=71.1gNH_3*\frac{1molNH_3}{17.04gNH_3}*\frac{3molH_2}{2molNH_3} *\frac{2.02gH_2}{1molH_2}\\\\m_{H_2}=12.6gH_2[/tex]

Best regards!