What are 2 substances that will NOT conduct electricity? Plz explain.

Answer: the answer is  b

Explanation:I took the test and got it

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Answer: Semiconductors

Explanation:

Answer:

The process is to divide the molecules by Avogadro's number (6.02 x 10^23) to get the moles. And since it's specified at STP (standard temperature and pressure) we know that the molar volume would be 22.4L. So we can just multiply the moles by the molar volume in order to get the volume of H2S.

(2.18 x 10^24) / (6.02 x 10^23) = 3.62 moles (rounded)

Now just multiply the moles by the molar volume of STP, 22.4.

3.62 x 22.4 = 81 Liters (rounded) of H2S

Answer:

the answer is C

Explanation:

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

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:

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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Answer:

10.C

11.B

15.A

Explanation:

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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If 25.0ml neutralizes 35.0 ml of a 0.200M HCl solution, the molarity of KOH solution is 0.28M

The reaction between KOH solution and HCl is written below.

[tex]HCl + KOH \rightarrow KCl + H_2O[/tex]

Volume of KOH, [tex]V_B[/tex] = 25.0 ml

Volume of HCl, [tex]V_A=35.0 ml[/tex]

Molarity of HCl, [tex]C_A=0.200 M[/tex]

Molarity of KOH, [tex]C_B=?[/tex]

Number of moles of HCl, [tex]n_A=1[/tex]

Number of moles of KOH, [tex]n_B=1[/tex]

The mathematical equation for neutralization reaction is:

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

Substitute  [tex]C_A=0.200 M[/tex],  [tex]n_A=1[/tex],  [tex]n_B=1[/tex],  [tex]V_B[/tex] = 25.0 ml, and  [tex]V_A=35.0 ml[/tex] into the equation above in order to solve for [tex]C_B[/tex]

[tex]\frac{0.2 \times 35}{C_B \times 25}=\frac{1}{1} \\\\25C_B=7\\\\C_B=\frac{7}{25} \\\\C_B=0.28M[/tex]

Therefore, the molarity of KOH solution is 0.28M

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Answer:

The answer is designing an experiment

Explanation:

Since we already have a hypothesis, we need to test the hypothesis and this is done by designing an experiment that will do this test.

the release of energy

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:

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

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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Answer: E.) C2N2H8

Explanation:

Answer:

E ) C2N2H8

Explanation:

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:

the answers are given in the photo

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:

matter cannot be created or destroyed in chemical reactions. This is the law of conservation of mass. In every chemical reaction, the same mass of matter must end up in the products as started in the reactants. Balanced chemical equations show that mass is conserved in chemical reactions.

Explanation:

Answer:

a species

Explanation:

Answer:

pl

saponification palmitate

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:

A reaction that has the same number and type of atoms on each side of the equation

Explanation:

A pex

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!

Answer:

A

Explanation:

I think it's the answer

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:

[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

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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