Can someone help me find the 5 digits numbers in here?

Answers

Answer 1

Answer:

Answer:

20322

Explanation:

Answer 2

Answer: Could it have something to deal with the teeth or mouths? The first pumpkin you have two teeth, second pumpkin looks like a 0, the third pumpkin I’m unsure of. The forth pumpkin has two teeth again same with the fifth.

So first pumpkin - 2
The second pumpkin - 0
The third pumpkin I’m unsure of
The forth pumpkin - 2
The fifth pumpkin - 2
So 20?22

Just an idea for ya. Hopefully it helps

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How does the number of molecules in 1 mol of oxygen compare with the number of molecules in 1 mol of nitrogen?1) 1 mol of oxygen has fewer molecules.2) 1 mol of oxygen has more molecules.3) Each sample has the same number of molecules.4) The molecules cannot be compared without knowing the mass.

Answers

The number of molecules in 1 mol of oxygen compare with the number of molecules in 1 mol of nitrogen is that each sample has the same number of molecules. The answer is number 3

Answer:

3) Each sample has the same number of molecules

Explanation:

1 mol are 6,022x10²³ particles. Moles of a chemical substance represent the number of molecules that this chemical has.

Thus, one mol of oxygen are 6,022x10²³ molecules of oxygen and one mol of nitrogen are 6,022x10²³ molecules of nitrogen.

Right answer is:

3) Each sample has the same number of molecules.

I hope it helps!

Which of the following sets represents a pair of isotopes? 14C and 14N 206Pb and 208Pb O2 and O3 32S and 32S2-

Answers

Isotopes are substances that have the same number of protons but differ in the number of neutrons. Hence, the pair of isotopes above should be of the same element. In the given choices, 14C is not an isotope of 14N. 206Pb is an isotope of 208 Pb. O2 and O3 differ in molecular formula but still made up of same kind of atom, hence they are allotropes, while 32S and 32S2- are not isotopes.

A 1.00 g sample of a hydrogen peroxide (H2O2) solution is placed in an Erlenmeyer flask and diluted with 20 mL of 1 M aqueous sulfuric acid. To this solution is added 0.0200 M KMnO4 solution via a buret, until a pale purple color persists. This requires 22.50 mL of KMnO4 solution. What is the percent by mass of hydrogen peroxide in the original solution? (A) 0.613% (B) 1.53% (C) 3.83% (D) 7.65%

Answers

the percent by mass is0.613

When 70.4 g of benzamide (C,H,NO) are dissolved in 850. g of a certain mystery liquid X, the freezing point of the solution is 2.7 C lower than the freezing point of pure X. On the other hand, when 70.4 g of ammonium chloride (NH CI) are dissolved in the same mass of X, the freezing point of the solution is 9.9 °C lower than the freezing point of pure X. Calculate the van't Hoff factor for ammonium chloride in X. Be sure your answer has a unit symbol, if necessary, and round your answer to 2 significant digits.

Answers

Answer:

1.60 is the van't Hoff factor for ammonium chloride in X.

Explanation:

$\Delta T_f=iK_f* m$

$Delta T_f=K_f* \frac{\text{Mass of solute}}{\text{Molar mass of solute}* \text{Mass of solvent in Kg}}$ ...(1)

where,

$\Delta T_f$ =Elevation in boiling point =

i = van't Hoff factor

$K_f$ = Freezing point constant

m = molality

1) When 70.4 g of benzamide are dissolved in 850. g of a certain mystery liquid X.

Mass of benzamide = 70.4 g

Molar mass of benzamide = 121 g/mol

i = 1 (organic molecule)

Mass of liquid X = 850 g = 0.850 kg

$K_f$ = Freezing point constant of liquid X= ?

$\Delta T_f=2.7^oC$

Putting all value in a (1):

$2.7^oC=K_f* (70.4 g)/(121 g/mol* 0.850 kg)$

$K_f=3.944 ^oC kg/mol$

2) When 70.4 g of ammonium chloride are dissolved in 850. g of a certain mystery liquid X.

Mass of ammonium chloride= 70.4 g

Molar mass of ammonium chloride = 53.5 g/mol

i = ? (ionic molecule)

Mass of liquid X = 850 g = 0.850 kg

$K_f=3.944 ^oC kg/mol$

$\Delta T_f=9.9^oC$

Putting all value in a (1):

$9.9^oC=i* 3.944^oC kg/mol* (70.4 g)/(53.5 g/mol* 0.850 kg)$

i = 1.6011 ≈ 1.60

1.60 is the van't Hoff factor for ammonium chloride in X.

Final answer:

The van't Hoff factor, which measures ionization, for ammonium chloride in the mysterious liquid X can be calculated to be approximately 1.01. This is calculated by first determining the cryoscopic constant from the observed depression of the freezing point by benzamide (which does not ionize), and then utilizing this value to calculate the theoretical freezing point depression for ammonium chloride (pretending it does not ionize either). Since the observed depression was 9.9℃ and the calculated was 9.8℃, the van't Hoff factor is their quotient, or approximately 1.01.

Explanation:

To solve this problem, we need to understand that the van't Hoff factor (i) is a measure of the extent of ionization in solution. It can be calculated using the formula i = ΔTf observed / ΔTf calculated, where ΔTf observed is the observed freezing point depression and ΔTf calculated is the theoretical freezing point depression if no ionization occurs.

First, we calculate the theoretical freezing point depression for ammonium chloride. We know that this is given by the benzamide that reduces the freezing point of the same amount of liquid X by 2.7℃. Therefore we assume the van't Hoff factor of benzamide is 1 (since it does not ionize) and we get the cryoscopic constant (Kf) of X from ΔTf = Kf * m * i. Substituting into the formula and rearranging gives Kf = ΔTf / (m * i) = 2.7 ℃/(70.4 g/850 g) = 2.7 ℃/0.082824 = 32.6 ℃ kg/mol.

We then use this Kf to calculate the ΔTf calculated for ammonium chloride: ΔTf calculated = Kf * m * i (where we again assume i=1) = 32.6 ℃ kg/mol * (70.4 g/850 g) = 9.8 ℃. Finally we can calculate the van't Hoff factor for ammonium chloride using the original formula: i = ΔTf observed / ΔTf calculated = 9.9 ℃ / 9.8 ℃ = 1.01.

Learn more about Van't Hoff Factor here:

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What happens to the particles when a pure substance melts

Answers

When a pure substance melts, the particles are separated. That is because a liquid tends to have particles that are loosely held, while a solid's particles are squished together.

As this reaction takes place at higher temperatures, it is observed that the equilibrium shifts toward the products. The reaction is

Answers

As this reaction takes place at higher temperatures, it is observed that the equilibrium shifts toward the products. The reaction is an endothermic reaction since it requires energy for the reaction to proceed which describes an endothermic reaction.

The answer is: endothermic reaction.

This reaction is endothermic (enthalpy is higher than zero), which means that heat is added.

According to Le Chatelier's principle when the reaction is endothermic heat is included as a reactant and when the temperature increased, the heat of the system increase, so the system consume some of that heat by shifting the equilibrium to the right, producing more products.