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3 years ago

Which of these particles are classified as hadrons? a. leptons c. mesons b. electrons d. neutrinos

Chemistry

2 answers:

agasfer [191]3 years ago

5 0

Answer:

C mesons

Explanation:

Hardrons are subatomic particles that are made by quarks.

Mesons is an example of hardon. and belong to this group.

disa [49]3 years ago

4 0

The answer would be mesons.

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An unknown salt is either NaF, NaCl, or NaOCl. When 0.050 mol of the salt is dissolved in water to form 0.500 L of solution, the

victus00 [196]

Answer: The unknown salt is NaF

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of salt = 0.050 moles

Volume of solution = 0.500 L

Putting values in above equation, we get:

$\text{Molarity of salt}=\frac{0.050mol}{0.500L}\\\\\text{Molarity of salt}=0.1M$

To calculate the hydroxide ion concentration, we first calculate pOH of the solution, which is:

pH + pOH = 14

We are given:

pH = 8.08

$pOH=14-8.08=5.92$

To calculate pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$

Putting values in above equation, we get:

$5.92=-\log[OH^-]$

$[OH^-]=10^{-5.92}=1.202\times 10^{-6}M$

The unknown salt given are formed by the combination of weak acid and strong acid which is NaOH

The chemical equation for the hydrolysis of $X^-$ ions follows:

$X^-(aq.)+H_2O(l)\rightleftharpoons HX(aq.)+OH^-(aq.);K_b$

Initial: 0.1

At eqllm: 0.1-x x x

Concentration of $OH^-=x=1.202\times 10^{-6}M$

The expression of $K_b$ for above equation follows:

$K_b=\frac{[OH^-][HX]}{[X^-]}$

Putting values in above expression, we get:

$K_b=\frac{(1.202\times 10^{-6})\times (1.202\times 10^{-6})}{(1-(1.202\times 10^{-6}))}\\\\K_b=1.445\times 10^{-11}M$

To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

$K_w=K_b\times K_a$

where,

$K_w$ = Ionic product of water = $10^{-14}$

$K_a$ = Acid dissociation constant

$K_b$ = Base dissociation constant = $1.445\times 10^{-11}$

Putting values in above equation, we get:

$10^{-14}=1.445\times 10^{-11}\times K_a\\\\K_a=\frac{10^{-14}}{1.445\times 10^{-11}}=6.92\times 10^{-4}$

We know that:

$K_a\text{ for HF}=6.8\times 10^{-6}$

$K_a\text{ for HCl}=1.3\times 10^{6}$

$K_a\text{ for HClO}=3.0\times 10^{-8}$

So, the calculated $K_a$ is approximately equal to the $K_a$ of HF

Hence, the unknown salt is NaF

6 0

2 years ago

During _____________ there are 24 hours of daylight at the North Pole.

Tresset [83]

Answer:

summer

The North Pole stays in full sunlight all day long throughout the entire summer (unless there are clouds)

8 0

2 years ago

How does your Skeletal and Muscular Systems work together?

elena-s [515]

Answer:

Answer would be A, Muscles move your bones

Explanation:

Without the Muscular System, your bones would not be connected properly

3 0

2 years ago

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I’m not sure what the answer is

Sergio [31]

I believe it’s the third option
Chemically combined to make a new pure substance

7 0

3 years ago

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A sample of iron absorbs 67.5 j of heat upon which the temperature of the sample increases from 21.5 °c to 28.5 °c. if the speci

asambeis [7]

Q = mcΔθ

67.5 = m x 0.45 x (28.5 - 21.5)

M = 67.5 / 3.15
= 21.4 g

3 0

2 years ago

Read 2 more answers