9) Observe the halogens. Fluorine and chlorine are gases, bromine is a liquid, and
iodine is a solid all at room temperature.
Distinguish between these differences in phase.
A) From Fluorine to lodine, the number of electrons increases.
This allows the intermolecular forces to strengthen from
temporary dipole (dispersion) forces to permanent dipoles.
Therefore, iodine is a solid because it has the most electrons.
B) From Fluorine to lodine, the nucleus gets larger. More
neutrons, protons and electrons allow the atoms of these
elements to become more attracted to each other. This is also
the reason that the halogens are diatomic naturally.
) From Fluorine to lodine, the molecular mass increases. With
the increase in mass, there is an increase in protons. This causes
the strength of the intermolecular forces between particles to
increase as well.
D) From Fluorine to lodine, the molecular mass increases. The
electron polarizability increases with the mass increase. In turn,
this strengthens the temporary dipole (dispersion) forces
between particles

Answer:

78n

Explanation:

The output force exerted on the nail in the claw is equal to 78 N which has a mechanical advantage of 5.2.

The mechanical advantage can be demonstrated as the ratio of the output force to the Input force. The mechanical advantage of any machine can be expressed in the form of the ratio of the forces utilized to do the work.

The ratio of the resistance to the effort is said to be the actual mechanical advantage which will be less. The efficiency of a machine can be evaluated by equating the ratio of the output to its input.

Given, the input force = 15 N

The mechanical advantage of the hammer = 5.2

Mechanical advantage = Output force/ Input force

5.2 = Output/15

Output force = 15 ×5.2 = 78 N

Therefore, the force is exerted on the nail in the claw is equal to 78 N.

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Answer:

Tetrahedral molecules are normally spy hybridized.

Explanation:

One of the oxygen atoms in the anion O3− is hypervalent and the formal charge on this oxygen atom is -1.

Ozone is a triatomic molecule. The anion formed from ozone is called the ozonide anion. This anion is also triatomic. The resonance structures of the ozonide anion are shown in the image attached to this answer.

We can see that one of the oxygen atoms in the ozonide ion is hypervalent because it contains ten instead of eight electrons. This hypervalent oxygen atom has a formal charge of -1 while the two other oxygen atoms has a formal charge of zero.

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Answer:

Explanation:

The material that is not a mixture; it has the same properties all the way through is called a substance. Thus the material that is not a mixture; it has the same properties all the way through is called a substance.

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Answer:

i dont speak mexican

Explanation:

Answer:

Mass Number

Explanation:

In nuclear physics, the sum of the numbers of protons and neutrons present in the nucleus of an atom.

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Answer:

Model C has more double the oxygen than A. I think it is B

Explanation:

Answer:

Explanation:

You have to find the mass in the mole before the molarity and then in the mole.

Answer:

i can't understand the question

Explanation:

the investigation lasts for 7 days !

hope this helps you.

Explanation:

a+b =ab

answer

ab added all

is give answer

The mass of 5.15 × 10²¹ atoms of chromium is 0.44 g

From Avogadro's hypothesis,

6.02×10²³ atoms = 1 mole of Cr

But:

1 mole of Cr = 52 g

Thus, we can say that:

6.02×10²³ atoms = 52 g of Cr

With the above information, we can obtain the mass of 5.15 × 10²¹ atoms of chromium. This can be obtained as follow:

6.02×10²³ atoms = 52 g of Cr

Therefore,

5.15×10²¹ atoms = (5.15×10²¹ × 52) / 6.02×10²³

5.15×10²¹ atoms = 0.44 g of Cr

Thus, the mass of 5.15 × 10²¹ atoms of chromium is 0.44 g

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Hey there!

It is evident that the problem gives the mass of the bottle with the calcite, with water and empty, which will allow us to calculate the masses of both calcite and water. Moreover, with the given density of water, it will be possible to calculate its volume, which turns out equal to that of the calcite.

In this case, it turns out possible to solve this problem by firstly calculating the mass of calcite present into the bottle, by using its mass when empty and the mass when having the calcite:

[tex]m_{calcite}=15.4448g-12.4631g=2.9817g[/tex]

Now, we calculate the volume of the calcite, which is the same to that had by water when weights 13.5441 g by using its density:

[tex]V_{calcite}=V_{water}=\frac{13.5441g-12.4631g}{0.997g/mL}=1.084mL[/tex]

Thus, the density of the calcite sample will be:

[tex]\rho _{calcite}=\frac{m_{calcite}}{V_{calcite}}\\\\\rho _{calcite}=\frac{2.9817g}{1.084mL}=2.750g/mL[/tex]

This result makes sense, as it sinks in chloroform but floats on bromoform as described on the last part of the problem, because this density is between 1.444 and 2.89. g/mL

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Regards!

Answer:

310K

Explanation:

Rearrange PV=nRT to get T=PV/nR

T=(2.0L)(0.75atm)/(0.5mol)(0.08206)

=36.5 or 37

add 273 for K to get

310

Answer:

Brass is an alloy, and either a "solid solution".

Alloys in general may be solid solutions or they simply be mixtures

Explanation:

Explanation:

since 5g saturates 12.5g of water at 10°c

so......x would saturate 50g of water at 0°c

then you can cross multiply

The partial pressure of oxygen is equal to 10.2 atm.

Given the following data:

To determine the partial pressure of oxygen:

First of all, we would find the total number of moles of the elements:

[tex]n=3.25+2.75[/tex]

n = 6 moles

Next, we would determine the mole fraction of the oxygen by using this formula:

[tex]Molefraction \;of \;a \;substance =\frac{No.\; of \; moles \;of \;substance}{Total \;no. \;of \; moles \;of \;substances}[/tex]

Substituting the values, we have:

[tex]Molefraction \;of \;a \;substance =\frac{2.75}{6} \\\\Molefraction \;of \;a \;substance =0.4583[/tex]

For oxygen:

[tex]Partial \;pressure = Molefraction \times Total\;pressure\\\\Partial \;pressure = 0.4583 \times 22.4[/tex]

Partial pressure of oxygen = 10.2 atm.

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Answer:

12.1

Explanation:

3.25 + 2.75= 6 moles total

then take: 3.25/6 to get 0.542

multiply 0.542 by 22.4 to get 12.1

the other answer solved for nitrogen, not oxygen.

This problem is describing the equilibrium whereby the formartion of a complex is attained when 1.47 g of nickel(II) chloride is dissolved in 100.0 mL of ammonia so that the latter's equilibium concentration is 0.20 M. Thus, it is asked to calculate the equilibrium concentrations of both nickel(II) ions and that of the complex.

Firstly, we can write out the chemical equation to be considered:

[tex]Ni^{2+}+6NH_3\rightleftharpoons Ni(NH_3)_6^{2+}[/tex]

Next, we can calculate the initial concentration of nickel(II) ions by using the concept of molarity:

[tex][Ni^{2+}]=\frac{1.47gNiCl_2*\frac{1molNiCl_2}{129.6g}*\frac{1molNi^{2+}}{1molNiCl_2} }{0.1000L}=0.113M[/tex]

Afterwards, we set up an equilibrium expression for this chemical reaction:

[tex]Kf=\frac{[Ni(NH_3)_6^{2+}]}{[Ni^{2+}][NH_3]^6}[/tex]

Which can be written in terms of the reaction extent, [tex]x[/tex]:

[tex]Kf=\frac{x}{(0.113-x)(0.2)^6}[/tex]

Now, for the calculation of [tex]x[/tex], we plug in Kf, and solve for it:

[tex]1.2x10^9=\frac{x}{(0.113-x)(0.2)^6}\\\\1.2x10^9=\frac{x}{(0.113-x)(6.4x10^{-5})}\\\\7.68x10^4(0.113-x)=x\\\\x=0.112999 M[/tex]

Which is about the same to the initial concentration of nickel(II) ions because the Kf is too large.

Thus, the required concentrations at equilibrium are about:

[tex][Ni(NH_3)_6^{2+}]=0.113M[/tex]

[tex][Ni^{2+}]=0M[/tex]

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In order to extract the compound in the aqueous layer, a strong acid must be added to the system.

Liquid - Liquid extraction is a common method for obtaining substances that can partition between two layers. In this case, compound A is neutral and compound B is acidic.

When the both compounds are dissolved in dichloromethane and extracted using an aqueous base, the acid substance will form a salt in the aqueous layer. In order to extract the compound in the aqueous layer, a strong acid must be added to the system.

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This  problem provides the molar mass and radius of a metal that has an BCC unit cell and the density is required.

Firstly, we consider the formula that relates molar mass and also includes the Avogadro's number and the volume of the unit cell:

[tex]\rho =\frac{Z*M}{V*N_A}[/tex]

Whereas Z stands for the number of atoms in the unit cell, M the molar mass, V the volume and NA the Avogadro's number. Next, since BCC is able to hold 2 atoms and M and NA are given, we calculate the volume of the atom in the unit cell given the radius in meters:

[tex]V=a^3=(\frac{4R}{\sqrt{3} } )^3=(\frac{4*1.74x10^{-10}m}{\sqrt{3} } )^3=6.49x10^{-29}m^3[/tex]

And finally the required density in g/cm³:

[tex]\rho =\frac{2*341.1g/mol}{6.49x10^{-29}m^3\frac{m^3}{atom} *6.022x10^{23}\frac{atom}{mol} } =17455257.8g/m^3\\\\\rho=17.5g/cm^3[/tex]

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Answer: the answer is 0.074Hz

Explanation:

Given, (In air)

Velocity V=340m/s

Frequency f=20,000Hz

Wavelegth λ=?

V=f.λ

λ=

F

V

​

=

20,000

340

​

=0.017Hz

Also, Given (in Water)

Velocity, V=1480m/s

Frequencyf=20,000Hz

wavelength, λ=?

V=F.λ

λ=

F

V

​

=

20,000

1480

=​0.074Hz

Answer:

frequency

Explanation:

frequency is velocity/ wavelength

340/1.36

250

From the information available in the question, the concentration of the acid is 0.21 M.

H3PO4(aq) + 3NaOH(aq) -----> Na3PO4(aq) + 3H2O(l)

Volume of acid(VA) = 51.0 mL

Concentration of acid (CA) = ?

Volume of base (VB) = 26.2 mL

Concentration of base (CB) = 1.25 M

Number of moles of acid (NA) = 1

Number of moles of base (NB) = 3

Using the formula;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CA =  CBVBNA/VANB

CA =  1.25 M × 26.2 mL × 1/51.0 mL × 3

CA = 0.21 M

The concentration of the acid is 0.21 M.

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Answer:

Fe is the excess reactant

Reaction yields are the amount of the reactant and the products of a chemical reaction. In the reaction between iron and oxygen, iron is the excess agent.

The reactant present in an extra quantity than the other reactant in a chemical reaction which reacts with the limiting reactant is called an excess reactant.

In a reaction mixture, the excess reactant is present even when the limiting agent is completely consumed.

The chemical reaction between iron and oxygen is shown as,

[tex]\rm 4 Fe(s) + 3 O_{2}(g) \rightarrow 2 Fe_{2}O_{3}[/tex]

From the reaction, it can be said that oxygen is a limiting reagent that limits the formation of iron (iii) oxide.

Therefore, iron is an excess reactant.

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Answer:

16.7%

Explanation:

40 ml of salt water + 200 ml of solution = 240 ml

40/240 = 4/24 = 1/6=16.7%