Which of the following compounds will have the highest boiling point?


MgCl2



CO2


C12H22O12

,

H2O

Answer:

4 meters

Explanation:

4 centimeters and millimeters are too small, while 4 kilometers is too large.

Answer:

the nature of the bond is covalent.

covalent bonds are bonds between 2 NON-METAL elements. since phosphorus and oxygen are both non-metals, the bond formed between them is covalent.

hope this answers your question!

This problem is providing describing a room in which air has a total pressure of 0.987 atm and the partial pressure of nitrogen is asked, turning out to be 0.746 atm.

In such a case, we can start by looking at the provided table, which shows the composition of dry air in mass basis, which means we must convert the composition of nitrogen to mole fraction by knowing that the average molar mass of air is 28.9647 g/mol, and thus, we infer that the mole fraction of nitrogen in air is about 0.755 according to:

[tex]x_{N_2}=\frac{0.781*14.0067g/mol*2}{28.9647g/mol} =0.755[/tex]

Then, we recall the Dalton's law version that relates total pressure, partial pressure and mole fraction written for nitrogen:

[tex]P_{N_2}=x_{N_2}P[/tex]

Finally, we plug in the previously calculated mole fraction and the total pressure of air to obtain:

[tex]P_{N_2}=0.755*0.987 atm\\\\P_{N_2}=0.746atm[/tex]

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Answer:

Hey mate.....

Explanation:

This is ur answer.....

A triple bond is made of two pi bonds and one sigma bond. Examples of compounds with triple bonds include nitrogen gas, the cyanide ion, acetylene and carbon monoxide.

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It requires 60 mL of 0.20 M sodium hydroxide [tex](NaOH)[/tex] to neutralize 25 mL of carbonic acid [tex](H_2CO_3)[/tex], hence the carbonate ions concentration is [tex]0.24M[/tex].

Given:

Reaction:

[tex]\to \bold{2NaOH + H_2CO_3 \to Na_2CO_3 + 2H_20 }[/tex]

[tex]NaOH[/tex] volume [tex](V_B) = 60 \ ml[/tex]  

[tex]H_2CO_3[/tex] Volume [tex](V_A) = 25\ ml[/tex]  

[tex]NaOH[/tex] Molarity [tex](C_B) = 0.20\ M[/tex]

[tex]H_2CO_3[/tex] moles [tex](n_A) = 1[/tex]

[tex]NaOH[/tex] moles [tex](n_B) = 2[/tex]

To find:

[tex]H_2CO_3[/tex] Molarity [tex](C_A) =?[/tex]

Solution:

Using the neutralization reaction:  

[tex]\to \frac{C_AV_A}{C_BV_B} =\frac{n_A}{n_B} \\\\[/tex]

[tex]\to C_B = 0.2\ M \\\\ \to n_A = 1 \\\\ \to n_B = 2 \\\\ \to V_B = 60\ ml \\\\ \to V_A = 25\ ml[/tex]

Calculating the [tex]C_A[/tex]:

 [tex]\to \frac{C_A \times 25}{0.2 \times 60} =\frac{1}{2} \\\\\to C_A =\frac{1 \times 0.2 \times 60}{2 \times 25} \\\\\to C_A =\frac{1 \times 2 \times 12}{2 \times 5 \times 10} \\\\ \to C_A =\frac{ 12}{ 5 \times 10} \\\\\to C_A = \frac{6}{5 \times 5} \\\\\to C_A = \frac{6}{25} \\\\\to C_A=0.24\ M[/tex]

Therefore, the concentration of carbonic acid is "0.24M".

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The concentration of carbonic acid( H₂CO₃) is 0.24 M.

The given reaction:

2 NaOH + H₂CO₃ → Na₂CO₃ + 2H₂O

Volume of NaOH  = 60mL

Volume of H₂CO₃ = 25 mL

Molarity of NaOH= 0.20M

To find:

Molarity of H₂CO₃=?

2 moles of NaOH and 1 mol of H₂CO₃ reacts to give 1 mol of Na₂CO₃.

Using the neutralization reaction:  

Consider A to be NaOH and B to be  H₂CO₃

[tex]\frac{\text{Number of moles of A}}{\text{Number of moles of B}} =\frac{\text{Molarity of A*Volume of A}}{\text{Molarity of B*Volume of B}}[/tex]

On substituting the values in order to calculate molarity of H₂CO₃:

[tex]\text{Molarity of B}=\frac{1*0.2*60}{2*25} \\\\\text{Molarity of B}=0.24M[/tex]

Therefore, the concentration of H₂CO₃ is 0.24 M.

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The statement is true in this situation is C. The size of Ffric is the same as the size of Fapp:

From the diagram, since the body is in equilibrium, the sum of vertical forces equals zero. Also, the sum of horizontal forces equal zero.

So, ∑Fx = 0 and ∑Fy = 0

Since Fapp acts in the negative x - direction and Ffric acts in the positive x - direction,

∑Fx = -Fapp + Ffric = 0

-Fapp + Ffric = 0

Fapp = Ffric

Also, since Fgrav acts in the negative y - direction and Fnorm acts in the positive y - direction,

∑Fy = Fnorm + (-Fgrav) = 0

Fnorm - Fgrav = 0

Fnorm = Fgrav

So, we see that the size of Fapp equals size of Ffric and the size of Fnorm equals the size of Fgrav.

So, the correct option is C

The statement which is true in this situation is C. The size of Ffric is the same as the size of Fapp.

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Considering the rule of three and molar mass of O, the mass of O²⁻ ions in 3 mol of alumina Al₂O₃ is 96 grams.

By multiplying the number of moles by the subscripts next to each atom, you get the amount of each element present in the compound. IIf the same atom appears more than once in the compound, you must add its amounts.

So, in 1 mole of Al₂O₃, you have 2 ions of O²⁻.

Then you can apply the following rule of three: if 1 mole of the compound contains 2 ions of O²⁻, 3 moles of the compound contains how many ions of O²⁻?

[tex]amount of ions of O^{2-} =\frac{3 moles of Al_{2} O_{3} x2 ions of O^{2-}}{1mole of Al_{2} O_{3} }[/tex]

amount of ions of O²⁻= 6 ions

Being the molar mass of O 16 g/mole, this is the amount of mass that a substance contains in one mole, then the mass of O²⁻ can be calculated as:

6 x 16 g/mole= 96 grams

Finally, the mass of O²⁻ ions in 3 mol of alumina Al₂O₃ is 96 grams.

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When perfume or cologne is sprayed into the air, it turns into a gas and its particles mix with other air particles that quickly move around the room. This is because gas is the least dense of the three states of matter. It moves quickly into any space or volume because it has the least density. This is called diffusion, when molecules or particles move from areas with high concentration to low concentration areas. 

Answer: Semiconductors

Explanation:

Answer:

O it's not

Explanation:

Have a great day!

A GEL is a semisolid preparation that contains very small solid particles that are suspended in a liquid. A gel always contains an agent (e.g., agarose) that provides stiffness to the preparation.

A gel is a semisolid preparation that contains a gelling agent which provides stiffness to the preparation.

The gelling agent can be, for example, agarose (this gelling agent is used to prepare gels in electrophoresis).

In an agarose gel, agarose molecules are organized into three-dimensional (3D) structures similar to pores, which allow the passage of DNA fragments during electrophoresis.

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Answer:

5 significant figures

Explanation:

The zeros as considered when they are in between or after numbers greater than zero.

Ex1: 0.005 = 1 significant figure

Ex2: 0.0050 = 2 significant figure

Ex3: 0.05500= 4 significant figure

Ex4: 5.00 = 1 significant figure

The intermolecular bonding for HF is van der Waals, whereas for HCL, the intermolecular bonding is hydrogen. Since the van der Waals bond is stronger than hydrogen, HF will have a higher boiling temperature. Since the covalent bond is stronger than van der Waals, HF will have a higher boiling temperature.

Answer:

It is strongly recommended that students do NOT wear contacts lenses in the laboratory. Chemical vapors may penetrate the contact lens material and cause the lens to adhere to one's eye, which could be detrimental to the eye. Goggles are not vapor tight and do not completely eliminate this absorption.

Answer: the answer is  b

Explanation:I took the test and got it

pls mark as brainliest if correct for you

Answer:

A

Explanation:

I think it's the answer

Answer:

it's a chemistry change

Explanation:

this is because heat is causing permanent changes and can no longer be changed back to its original atate

Answer:

cellular respiration

Explanation:

hope this helps :)

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Answer:

the answer is C

Explanation:

This is because group6 elements are diatomic and when they are chemically combined their subscript 2 cancels out

Answer:

[tex]FeCl_{2} +H_{2} S->FeS + 2HCl[/tex]

Explanation:

I am assuming that we have to balance this equation. On the left side, we have one Fe, 2 H, 2 Cl, and 1 S. On the right side, we have 1 Fe, 1 H, 1 Cl, and 1 S. Adding a 2 as a coefficient in front of the HCl on the right side will make 2 H and 2 Cl instead, balancing the overall equation.

FeCl_{2} +H_{2} S- > FeS + 2HClFeCl 2 +H 2 S−>FeS+2HCl

a.

The atomic radius of iridium is 1.355 Å

For a face-centered cubic unit cell, with edge length, a and atomic radius r,

we have a² + a² = (4r)²

2a² = 16r²

r² = a²/8

r = a/√8

Since a = 3.833 Å,

Substituting this into the equation, we have

r = a/√8

r = 3.833 Å/√8

r = 3.833 Å/2.828

r = 1.355 Å

So, the atomic radius of iridium is 1.355 Å

b.

The density of iridium metal is 22.67 g/cm³

To find the density of the iridium metal, we need to find the mass of atoms per unit cell.

So, for the face centered cubic iridium atom, there are 6 faces per unit cell × 1/2 atoms per face + 4 corners per unit cell × 1/4 atoms per corner = 3 atoms + 1 atom = 4 atoms per unit cell.

So, the mass of the atoms in the unit cell is m = number of atoms/cell × number of moles/atom × molar mass of Iridium

= 4 atoms/cell × 1 mol/6.022 × 10²³ atoms × 192.22 g/mol

= 768.88/6.022 × 10²³ g/cell

= 127.68 × 10⁻²³ g/cell

= 1.2768 × 10⁻²¹ g/cell

Next, we need to find the volume of the unit cell which is V = a³

= (3.833 Å)³

= (3.833 × 10⁻¹⁰ m)³

= 56.314 × 10⁻³⁰ m

= 5.6314 × 10⁻²⁹ m

So, the density of iridium = mass of unit cell/volume of unit cell

= m/V

= 1.2768 × 10⁻²¹ g/cell ÷ 5.6314 × 10⁻²⁹ m

= 0.2267 × 10⁸ g/m³

= 0.2267 × 10⁸ g/m³ × 10⁻⁶ m³/cm³

= 0.2267 × 10² g/cm³

= 22.67 g/cm³

So, the density of iridium metal is 22.67 g/cm³

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the release of energy

Answer:

a hydrogen atom contains only one proton in it's nucleus, while atoms of all other elements contain more than one proton

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Answer:

pl

saponification palmitate

Answer:

we do the best possible in our

Answer:

In the picture A kettle is boiling water and evaporating it we can see the water vapours coming out of the kettle and there is a paper on top of the kettle from where the water vapours are coming out .

Explanation:

is this right ?? i hope it helps

Answer:

10.C

11.B

15.A

Explanation: