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Dafna1 [17]
3 years ago
13

Massive stars have lower absolute brightness true or false

Chemistry
2 answers:
Juliette [100K]3 years ago
6 0
<span>The answer should be false, because mass should not have to do with how bright a star is </span>
Inessa [10]3 years ago
6 0

The brightness of a star is measured as luminosity. It is a simple factor of energy emitted by a star in a unit time or generally per second.

the following factors may affect the brightness of a star

a) Temperature of star

b) size of star: larger the size more the brightness

c) distance from observer : closer the star brighter it will appear

so massive stars or stars with large mass have lower absolute brightness will be false.

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Which statement about protons is false?
Alekssandra [29.7K]

<u>Answer:</u> The correct answer is Option 4.

<u>Explanation:</u>

There are three sub-atomic particles present in an atom. They are: electrons, protons and neutrons.

Protons constitute in each and every atom.

The charge on proton is of equal magnitude as that of electron but having opposite sign. Proton carry a positive charge and electron carry a negative charge.

Protons and neutrons, both determine the mass of an atom.

Mass of 1 proton = 1.007276 u

Mass of 1 neutron = 1.008664 u

Mass of 1 electron = 0.00054858 u

Mass of proton is almost same as that of neutron but is more than the mass of electron.

Hence, the correct answer is Option 4.

6 0
3 years ago
If a buffer solution is 0.220 M in a weak acid ( Ka=7.4×10−5) and 0.540 M in its conjugate base, what is the pH?
valkas [14]

Answer: the pH of the solution is 4.52

Explanation:

Consider the weak acid as Ha, it is dissociated as expressed below

HA     H⁺  +  A⁻

the Henderson -Haselbach equation can be expressed as;

pH = pKa + log( [A⁻] / [HA])

the weak acid is dissociated into H⁺ and A⁻ ions in the solution.

now the conjugate base of the weak acid HA is

HA(aq) {weak acid}     H⁺(aq)  +  A⁻(aq) {conjugate base}

so now we calculate the value of Kₐ as well as pH value by substituting the values of the concentrations into the equation;

pKₐ = -logKₐ

pKₐ = -log ( 7.4×10⁻⁵ )

pKₐ = 4.13

now thw pH is

pH = pKₐ  + log( [A⁻] / [HA])

pH = 4.13 + log( [0.540] / [0.220])

pH = 4.13 + 0.3899

pH = 4.5199 = 4.52

Therefore the pH of the solution is 4.52

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Answer:

H2C=CH2 + H2O + CO → CH3CH2CO2H. It is also produced by the aerobic oxidation of propionaldehyde.

5 0
3 years ago
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Answer:

2C₃H₇BO₃ + 8O₂ → 6CO₂ + 7H₂O + B₂O₃.

Explanation:

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So, the balanced equation:

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