Which of the following stars is a B class main sequence star?

Answer: The correct answer is 5.6atm.

Explanation: Since we only have information on the pressure and temperature, we will use Gay-Lussac's Law: [tex]\frac{P_{1} }{T_{1} } =\frac{P_{2} }{T_{2} }[/tex]. Now that we have the formula, let's plug in our given information!

[tex]\frac{3.8atm}{372K} =\frac{P_{2} }{550K}[/tex]

We must find the value [tex]P_{2}[/tex]. To do this, we will write a stoichiometric equation to solve this.

[tex]P_{2} =\frac{3.8atm*550K}{372K}[/tex]

Let's input the numbers into a scientific calculator to get our answer!

[tex]P_{2} =\frac{3.8atm*550K}{372K}=5.6 atm[/tex]

Given all of this work, the value [tex]P_{2}[/tex] equals 5.6 atm.

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The  portion of the electromagnetic spectrum do we need to observe is the red part of the This is further explained below.

Generally, Spectrum refers to the range of wavelengths or frequencies that electromagnetic radiation may cover.

In conclusion, We need to keep an eye on the red part of the electromagnetic spectrum since a large positive redshift causes an increase in wavelength and a corresponding fall in frequency and photon energy (such as light).

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The weaker bond is the bond between the two hydrogen atoms since it has a lower bond energy.

The term bond energy has to do with the energy that is required to break a bond. It is also the energy that must be supplied when a bond is formed.

The lower the bond energy of  bond, the weaker the bond. Hence, it follows that the weaker bond here is the bond between the two hydrogen atoms since it has a lower bond energy.

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Answer:

2 g/mL

Explanation:

Density can be calculated using the ratio:

Density = mass (g) / volume (mL)

Since you have been given both the mass and volume of the liquid, you can calculate the density.

Mass = 6 grams (g)

Volume = 3 milliliters (mL)

Density = mass / volume

Density = 6 g / 3 mL

Density = 2 g/mL

Answer:

When the temperature increases, the molecules of the gas gain energy. Therefore, they move faster.

This causes the molecules to hit the walls of the container more frequently and with greater force. Hence the pressure inside the container increases.

The beginning and ending energy levels of the electron during the emission of energy that leads to spectral line n(i)= 2 and n(f)=2

wavelength(λ) = 656.3 nm = 656.3 x 10^-9

E= hc / λ

E = energy of the photon

h=Planck's constant

c=speed of light

λ= wavelength of the photon

E= (6.626 x 10^-34)(3 x 10^8)/( 656.3 x 10^-9)

E = 3.028 x 10^-37J

It's a red line which means it falls in a visible region which is known as the Balmer series so;

n (f) = 2

ΔE= RH (1/n(f)^2 - 1/n(i)^2)

3.028 x 10^-37 =-2.18 x 10^-18(1/2^2 - 1/n(i))

n(i)= 2

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The mole of the substance, given the data from the question is 0.0014 mole

This is simply defined as the mole of solute per kilogram of water. Mathematically, it is expressed as

Molality = mole / mass (Kg) of water

Mole = molality × mass of water

Mole of substance = 0.056 × 0.0254

Mole of substance = 0.0014 mole

Thus, the mole of the substance is 0.0014 mole

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[tex]density = \frac{mass}{volume} [/tex]

[tex]d = \frac{25.7}{22.6} = 1.13717 \: g/ml[/tex]

[tex]pay \: attention \: to \: the \: unit \: used \: to \\ express \: the \: density \\ since \: the \: mass \: is \: given \: in \: grams \\ and \: volume \: in \: milliliter \: then \\ the \: density \: is \: expressed \: in \\ grams \: per \: \: milliliter[/tex]

When 400g each of water, oil, and sand are taken and heated from room temperature to 700⁰C on identical Bunsen burners. In this case, Substances that take more time to get heated will take more time to cool Hence, Option (A) is correct.

Heat transfer is a form of energy transfer and can occur by conduction, convection, and/or radiation.

Heat transfer occurs any time there is a temperature difference between two objects and occurs in the direction of decreasing temperature, meaning from a hot object to a cold object.

Tests show that the transferred heat relies upon three factors—the adjustment of temperature, the mass of the framework, and the substance and period of the substance.

One of the significant impacts of heat transfer is temperature change: warming builds the temperature while cooling diminishes it.

We expect no stage change and no work is done on or by the framework.

In this case, Substances that take more time to get heated will take more time to cool Hence, Option (A) is correct.

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Answer:

8.81 atm

Explanation:

If we let the unknown pressure be x,

[tex] \frac{713 \times 2.81}{664} = \frac{126x}{368} \\ \\ x = 8.81[/tex]

Answer: The new pressure is 2.8 atm.

Explanation:

Since our given information only involves pressure and volume, we will use Boyle's Law: [tex]P_{1} V_{1} =P_{2} V_{2}[/tex]. Let's input the data into the formula.

[tex]1.4atm*500mL=P_{2} *250mL[/tex]

We have to rearrange our data a bit now because the volume is supposed to be in liters. To convert from milliliters(mL) to liters(L), you must divide the number by 1000. That is what we will do with both of our given volumes.

[tex]\frac{500mL}{1000} =0.5L[/tex]     [tex]\frac{250mL}{1000} =0.25L[/tex]

Now that our volumes are in liters, we can correctly plug in our data.

[tex]1.4atm*0.5L=P_{2} *0.25L[/tex]

Since we do not know the value [tex]P_{2}[/tex], we must rearrange the formula into a stoichiometric equation to find out the answer.

[tex]P_{2}=\frac{1.4atm*0.5L}{0.25L}[/tex]    [tex]P_{2}=2.8 atm[/tex]

Our final answer is 2.8 atmospheres.

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637,980 Joules is needed to raise the temperature of 2.50 kg of this oil from 23 ∘C to 191 ∘C.

The heat in calories required to raise the temperature of one gram of a substance one degree Celsius.

Q = cmΔT

where

Q = heat energy needed for that material to get desired temperature change (in Joules)

M = mass (in grams) so you have to convert from kilograms.

c = specific heat constant for the material being heated

ΔT = change in temperature

Q = (1.75)(2.17 x 1000)(191 - 23)

Q = (1.75)(2170)(168)

Q = 637,980 Joules

Hence, 637,980 Joules is needed to raise the temperature of 2.50 kg of this oil from 23 ∘C to 191 ∘C.

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Answer: 6 valence electrons

Explanation: the atomic number for oxygen is 8. the first shell takes 2, the second 8. so, the outer shell is the last shell, which takes 6 from oxygen because oxygen has only 8.  6 is the number for outer or valence electron for oxygen

It will take less 1.8 seconds for the mass of a sample of magnesium-20 to decay from 65.6 micrograms to 8.20 micrograms.

The half-life of a radioactive element is the time for half the amount of a sample of the substance to decay.

After 0.6 seconds 31.25 remains

After 1.2 seconds, 15.625 remains

After 1.8 seconds, 7.9 micrograms remains.

In conclusion, it will take less 1.8 seconds for the sample to decay to 8.20 micrograms.

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The time taken for the radioactive sample to decay is 1.8 s.

The half life is the time taken for only half of the number of the radioactive isotopes to remain. Now we have;

No = 65.6 micrograms

N = 8.20 micrograms

t1/2 = 0.600 seconds

t = ?

Hence;

N/No = (1/2)^t/t1/2

8.20/ 65.6  = (1/2)^t/0.600

0.125 =  (1/2)^t/0.600

1/8 =  (1/2)^t/0.600

(1/2)^3 =  (1/2)^t/0.600

3 = t/0.600

t = 3 * 0.600

t = 1.8 s

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1) The theoretical yield of SnS2 will be 4.20 grams

2) The percent yield will be 7.93%

From the equation of the reaction:

[tex]SnBr_4(aq)+2Na_2S(aq)-- > 4NaBr(aq)+SnS_2(s)[/tex]

The mole ratio of the reactant is 1:2.

Mole of 48.1 mL, 0.478 M SnBr4  = 0.478 x 48.1/100 = 0.023 mols

Mole of 48.8 mL, 0.160 M Na2S = 0.160 x 48.8/1000 = 0.0078 moles

Thus, Na2S is in excess while SnBr4  is limiting.

Mole ratio of SnBr4  and SnS2 = 1:1

Equivalent mole of SnS2 = 0.023 moles

Mass of 0.023 noles SnS2 = 0.023 x 182.81 = 4.20 grams

2) With 0.0333 g of SnS2 recovered, percent yield = 0.333/4.2 x 100 = 7.93%

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Here is the complete question:

Tin(IV) sulfide, SnS2, a yellow pigment, can be produced using the following reaction.

SnBr4(aq)+2Na2S(aq)⟶4NaBr(aq)+SnS2(s)

Suppose a student adds 48.1 mL of a 0.478 M solution of SnBr4 to 48.8 mL of a 0.160 M solution of Na2S.

1) Calculate the theoretical yield of SnS2. ;

2) The student recovers 0.333 g of SnS2. Calculate the percent yield of SnS2 that the student obtained.

The correct rate law for the reaction is given as Rate = -0.208 hr-1 [sucrose].

A first order reaction is one in which a plot of the logarithm of the concentration against time gives a straight line graph. Now we know that this is a decomposition reaction thus the slope is negative.

Thus, in this case, the correct rate law for the reaction is given as Rate = -0.208 hr-1 [sucrose].

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Answer:

[tex]\huge\boxed{\sf 102.5\ cm}[/tex]

Explanation:

Given that,

Circumference = C = 41 inches

Also,

Multiply 41 to both sides

41 inches = 2.5 × 41 cm

41 inches = 102.5 cm

So, The tree has a circumference of 102.5 cm.

[tex]\rule[225]{225}{2}[/tex]

1) There are only single bonds between the carbon atoms, so the ending is 'ane'. Thus, the answer is 2-methylbutane.

2) There is one triple bond between one pair of carbon atoms, so the ending is 'yne'. So, the answer is 2-butyne.

3) There is one double bond between one pair of carbon atoms, so the ending is 'ene'. So, the answer is 2-butene.

Answer:

Neutron does not contain any charge because the charge of the quarks that made up the neutron balances each other out.

Hope it helps.

Answer:

+3

Explanation:

The phosphate ion has a charge of -3, so for the compound to have an overall change of 0, Fe needs to have a charge of +3.

The maximum amount of CaCO3 we can expect is 0.0180 mole x 100 g/mole = 1.80 g.

Given that:

There would be a precipitate of calcium carbonate from the reaction between sodium carbonate and calcium chloride.

The reaction is:

Na2CO3 (aq) + CaCl2(aq) → CaCO3 (s) + 2 NaCl (aq)

2.00 g CaCl2 x 1 mole/ 111 g CaCl2

= 0.0180 mole CaCl2

Mix it with 2.00 g Na2CO3 x 1 mole/ 106 g

= 0.0189 mole Na2CO3

The maximum amount of CaCO3 we can expect is 0.0180-mole x 100 g/mole = 1.80 g

Therefore, the 1.80 g is the theoretical (calculated) yield of CaCO3 in this example.

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Answer:

An electron in an excited energy level of an atom will always drop to lower energy levels, emitting very specific wavelengths of light in the process.

because if you take 42-10 = 32

32 is the answer to this question 14x3-2x5.

According to the BODMAS rule, if an expression contains brackets ((), {}, []) we have first to solve or simplify the bracket followed by 'order' (that means powers and roots, etc.), then division, multiplication, addition and subtraction from left to right.

14x3-2x5

42 -10

32

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Answer:

yes it's a thing, milk too

Answer:

Entropy is a measurement of how much the atoms in a substance are free to spread out, move around, and arrange themselves in random ways

Answer: 2.23

Explanation:

The dissociation of acetic acid is as follows:

HCOOH(aq) ⇋ HCOO− (aq) + H+(aq)

The ICE table for the concentrations of ions is given below. From the table, the concentration of HCOO- and H+ can be found out.

                                                     HCOOH → HCOO−  +   H+
Initial Concentration                     0.019 M         0             0
Equilibrium Concentration          (0.019−x) M    x              x

Where,

At equilibrium, dissociation constant can be calculated as follows.

[tex]K_{\mathrm{a}}=\frac{x^{2}}{(0.19-x) \mathrm{M}}[/tex]

At equilibrium, the concentration of x is negligible as compared to that of HCOOH.

Substitute the value of Ka in the above equation.

[tex]\begin{aligned}K_{\mathrm{a}} &=\frac{x^{2}}{0.19-x} \\x &=\sqrt{1.8 \times 10^{-4} \times 0.19} \\&=0.00584 \mathrm{M}\end{aligned}[/tex]

Here, the concentration of hydrogen ion is obtained. From the hydrogen ion concentration, the pH of the solution is found out as follows:

[tex]\begin{aligned}\mathrm{pH} &=-\log \left[\mathrm{H}^{+}\right] \\&=-\log (0.00584 \mathrm{M}) \\&=2.23\end{aligned}[/tex]

Therefore, the pH of 0.19M HCCOH is 2.23

The moles of Na⁺ in 1.30 L is 0.2158 moles of Na⁺.

Concentration of a solution is the amount of solute present in a given volume of solution or solvent.

The concentration of the Na⁺ in the  intravenous saline solution contains 166 mEq/L.

1 mEq/Liter = 1 mmol/L

Moles of Na⁺ in 1.30 L = M

[tex]M = {166 mEq/L} * \frac{1 mmol/L}{1 mEq/L} * \frac{1.30 L}{1000 mL} = 0.2158\:moles[/tex]

Moles of Na⁺ in 1.30 L = 0.2158 moles of Na⁺.

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3.08 is the pOH of a 0.175 M aqueous solution of [tex]NX_3[/tex].

0.215% is the per cent ionization of a 0.325 M aqueous solution of [tex]NX_3[/tex]

pH is a measure of how acidic/basic water is.

A)

[tex]NX_3 + H_2O[/tex] →[tex]NHX_3^+ + OH^-[/tex]

Kb = 4.5 x[tex]10^-6[/tex]

Kb = {concentration of (NH₄⁺) x concentration of (OH⁻)} ÷ concentration of (NH₃).

concentration of (NH₄⁺) = concentration of (OH⁻) = x.

x² = Kb x concentration of (NH₃)

x² = 4.5 × 10⁻⁶ × 0.175 = 7.0 × 10⁻⁷.

x = concentration of (OH⁻) = √(7.0 × 10⁻⁷)

= 8.367 × 10⁻⁴

pOH = -log(c(OH⁻))

=- log ( 8.367 × 10⁻⁴)

= 3.08

B)

Chemical reaction: NX₃ + H₂O ⇄ NX₃H⁺ + OH⁻.

Concentration of (NX₃) = 0.325 M.

Kb = 4.5 x 10⁻⁶.

[NX₃H⁺] = [OH⁻] = x.

[NX₃] = 0.325 M - x.

Kb = [NX₃H⁺] x [OH⁻] ÷  [NX₃].

4.5 x 10⁻⁶ = x² ÷ (0.325 M - x).

x = 0.0007 M.

Per cent of ionization:

α = 0. 0007 M ÷ 0. 325 M x 100%

= 0.215%.

Hence,

3.08 is the pOH of a 0.175 M aqueous solution of [tex]NX_3[/tex].

0.215% is the per cent ionization of a 0.325 M aqueous solution of [tex]NX_3[/tex]

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The molar ratio of (6 moles of NaBr)/(1 mole of Fe₂S₃) is correct based on the balanced chemical equation.

The balanced chemical reaction is given as;

2FeBr₃ + 3Na₂S   →  Fe₂S₃   +  6NaBr

From the balanced chemical reaction above, 6 moles of NaBr will also correspond to 1 mole of  Fe₂S₃.

Thus, the molar ratio of (6 moles of NaBr)/(1 mole of Fe₂S₃) is correct based on the balanced chemical equation.

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