What is the mass of a neutron close to

Answers

Answer 1

Answer:

Answer:

The free neutron has a mass of 939,565,413.3 eV/c2, or 1.674927471×10−27 kg, or 1.00866491588 u. The neutron has a mean square radius of about 0.8×10−15 m, or 0.8 fm, and it is a spin-½ fermion.

Mass: 1.67492749804(95)×10−27 kg; 939.56542052(54) MeV/c2; 1.00866491588...

Composition: 1 up quark, 2 down quarks

Electric charge: 0 e; (−2±8)×10−22 e (experimental limits)

Magnetic polarizability: 3.7(20)×10−4 fm3

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The vapor pressure of liquid octane, C8H18, is 100 mm Hg at 339 K. A sample of C8H18 is placed in a closed, evacuated 537 mL container at a temperature of 339 K. It is found that all of the C8H18 is in the vapor phase and that the pressure is 68.0 mm Hg. If the volume of the container is reduced to 338 mL at constant temperature, which of the following statements are correct?a. No condensation will occur.
b. Some of the vapor initially present will condense.
c. The pressure in the container will be 100. mm Hg.
d. Only octane vapor will be present.
e. Liquid octane will be present.

Answers

Answer:

the final pressure (108.03 mmHg ) inside the container at 339 K is more than the vapor pressure of liquid octane (100 mmHg) at 339 K.

Hence,

b. Some of the vapor initially present will condense.

e. Liquid octane will be present.

Explanation:

Given that;

The vapor pressure of liquid octane, C8H18, is 100 mm Hg at 339 K

Initial volume of the container, V1 = 537 mL

Initial vapor pressure, P1 = 68.0 mmHg

Final volume of the container, V2 = 338 mL

Let us say that the final vapor pressure = P2

From Boyle's law,

P2V2 = P1V1

P2 * 338 = 68.0 * 537

338P2 = 36516

P2 = 36516 / 338

P2 = 108.03 mmHg

Thus, the final pressure (108.03 mmHg ) inside the container at 339 K is more than the vapor pressure of liquid octane (100 mmHg) at 339 K.

Hence,

b. Some of the vapor initially present will condense.

e. Liquid octane will be present.

When 1 mole of H2(g) reacts with F2(g) to form HF(g) according to the following equation, 542 kJ of energy are evolved H2(g) + F2(g2HF(g) Is this reaction endothermic or exothermic? What is the value of q?

Answers

Answer: The reaction is exothermic. The value of q is -542 kJ.

Explanation:

Endothermic reactions are defined as the reactions in which energy of the product is greater than the energy of the reactants. The total energy is absorbed in the form of heat and for the reaction comes out to be positive.

Exothermic reactions are defined as the reactions in which energy of the product is lesser than the energy of the reactants. The total energy is released in the form of heat and for the reaction comes out to be negative.

Thus $H_2(g)+F_2(g)\rightarrow 2HF(g)$ evolves heat , it is exothermic in nature. The value of q is -542kJ.

You have 2.2 mol Xe and 2.0 mol F2, but when you carry out thereaction you end up with only 0.25 mol XeF4. What is the percent yield

of this experiment?

Xe(g) + 2 F2 (g) - XeF. (g)

Answers

The percentage yield of XeF from the concentration of the given reactants will be 25%.

The reaction states that 1 mole of Xe will give 1 mole of XeF, and 2 moles of fluorine, will gives 1 mole of XeF.

The limiting reactant can be calculated:

1 mole Xe = 2 moles of $\rm F_2$

2.2 moles of Xe = 2.2 $*$ 2 moles of $\rm F_2$

2.2 moles of Xe = 4.4 moles of $\rm F_2$

Since the amount of available $\rm F_2$ has been in the limiting, thus $\rm F_2$ has been the limiting reactant.

So, the yield of XeF in terms of $\rm F_2$ will be:

2 moles $\rm F_2$ = 1 mole XeF

Thus the theoretical yield of XeF is 1 mole.

The yield of XeF we get = 0.25 moles.

Thus the percentage yield = $\rm (Obtained)/(Theoretical\;yield)\;*\;100$

Percentage yield = $\rm (0.25)/(1)\;*\;100$

Percentage yield = 25%

Thus the percentage yield of XeF from the concentration of the given reactants will be 25%.

For more information about the percent yield, refer to the link:

brainly.com/question/12809634

Perform the calculation and report the answer using the proper number of significant figures. Make sure the answer is rounded correctly. 1.012×10^-3 J/(0.015456 g)(298.3682−298.3567)K

Answers

Answer:

$=5.694(J)/(g*K)$

Explanation:

Hello,

In this case, since the result of the operation between two magnitudes is shown with the same significant figures of the shortest number, we obtain:

$1.012x10^(-3) J/[(0.015456 g)(298.3682-298.3567)]K$

Next, we proceed as follows:

$=0.065476J/[(g)(20.0115K)]\n\n=5.693582(J)/(g*K)$

Nevertheless, since 1.012 is the shortest number and has four significant figures, the result is rounded to four significant figures, that is until the three but it rounded due to the fact that the next digit is five:

$=5.694(J)/(g*K)$

Regards.

Hich nuclide is most likely to undergo beta decay?which nuclide is most likely to undergo beta decay?co−52ar−35si−22mg−28

Answers

200
Pt
78
is most likely to undergo beta decay.

During the process of beta decay the mass number of the parent nuclide becomes equal to the newly created nuclide. On the other hand, the atomic number of the newly created nuclei increases or decreases by one unit.

Predict whether the compounds are soluble or insoluble in water

Answers

Answer:

The polar compounds are soluble in water while non polar are insoluble in water.

Explanation:

Solvent is the that part of solution which is present in large proportion and have ability to dissolve the solute. In simplest form it is something in which other substance get dissolve. The most widely used solvent is water, other examples are toluene, acetone, ethanol, chloroform etc.

Water is called universal solvent because of high polarity all polar substance are dissolve in it. Hydrogen is less electronegative while oxygen is more electronegative and because of difference in electronegativity hydrogen carry the partial positive charge while oxygen carry partial negative charge.

Water create electrostatic interaction with other polar molecules. The negative end of water attract the positive end of polar molecules and positive end of water attract negative end of polar substance and in this way polar substance get dissolve in it.

Example:

when we stir the sodium chloride into water the cation Na⁺ ions are surrounded by the negative end of water i.e oxygen and anion Cl⁻ is surrounded by the positive end of water i.e hydrogen and in this way all salt is get dissolved.

The chemicals that can dissolve in a certain solvent to create a homogenous mixture known as a solution are said to be soluble chemicals. The compounds that are soluble are: $KNO_3$ , $AgNO_3$ , and $CuBr_2$ .

As per this,

Nitrate ( $NO^{3-$ ) salts are often soluble in water. $KNO_3$ is potassium nitrate.
The majority of nitrate ( $NO^{3-$ ) salts, including silvernitrate, are soluble in water, including $AgNO_3$ .
The majority of bromide ( $Br^-$ ) salts, including copper(II) bromide, are water soluble.

Insoluble:

Lead(II) chloride, or $PbCl_2$ , is an exception and is regarded as being insoluble in water among the chloride ( $Cl^-$ ) salts.
Barium sulphate, also known as BaSO4, is an exception to the rule of most sulphate ( $SO^{4-$ ) compounds being soluble in water.

Thus, these are the classification of the compounds as per their solubility.

For more details regarding solubility, visit:

brainly.com/question/31493083

#SPJ6

Your question seems incomplete, the prpbable complete question is:

Predict whether the following compounds are soluble or insoluble in water. Soluble Insoluble PbCl2, BaSO4, KNO3, AgNO3, and CuBr2.

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