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2 years ago

8

Esters are most likely to be encountered in what department in a large department store?

1 answer:

algol [13]2 years ago

4 0

Answer: i do not now

Explanation:

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Carbon-12 and carbon-13 are isotopes of carbon.

ohaa [14]

Answer:

Isotopes of an element share the same number of protons but have different numbers of neutrons. Let's use carbon as an example. There are three isotopes of carbon found in nature – carbon-12, carbon-13, and carbon-14. All three have six protons, but their neutron numbers - 6, 7, and 8, respectively - all differ.

Explanation:

4 0

3 years ago

Read 2 more answers

Is the following statement true or false solar panels collect energy from the sun and power a lamp and home Is the following sta

Katen [24]

The following statement is True.

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3 years ago

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What is the mole fraction of NaCl in a mixture containing 7.21 moles NaCl, 9.37 moles KCL, and 3.42

GrogVix [38]

Answer : The mole fraction of NaCl in a mixture is, 0.360

Explanation : Given,

Moles of $NaCl$ = 7.21 mole

Moles of $KCl$ = 9.37 mole

Moles of $LiCl$ = 3.42 mole

Now we have to calculate the mole fraction of $NaCl$ .

$\text{Mole fraction of }NaCl=\frac{\text{Moles of }NaCl}{\text{Moles of }NaCl+\text{Moles of }KCl+\text{Moles of }LiCl}$

Now put all the given values in this formula, we get:

$\text{Mole fraction of }NaCl=\frac{7.21}{7.21+9.37+3.42}=0.360$

Therefore, the mole fraction of NaCl in a mixture is, 0.360

3 0

3 years ago

What is the energy released in this β − β − nuclear reaction 40 19 K → 40 20 C a + 0 − 1 e 19 40 K → 20 40 C a + − 1 0 e ? (The

Effectus [21]

<u>Answer:</u> The energy released in the given nuclear reaction is 1.3106 MeV.

<u>Explanation:</u>

For the given nuclear reaction:

$_{19}^{40}\textrm{K}\rightarrow _{20}^{40}\textrm{Ca}+_{-1}^{0}\textrm{e}$

We are given:

Mass of $_{19}^{40}\textrm{K}$ = 39.963998 u

Mass of $_{20}^{40}\textrm{Ca}$ = 39.962591 u

To calculate the mass defect, we use the equation:

$\Delta m=\text{Mass of reactants}-\text{Mass of products}$

Putting values in above equation, we get:

$\Delta m=(39.963998-39.962591)=0.001407u$

To calculate the energy released, we use the equation:

$E=\Delta mc^2\\E=(0.001407u)\times c^2$

$E=(0.001407u)\times (931.5MeV)$ (Conversion factor: $1u=931.5MeV/c^2$ )

$E=1.3106MeV$

Hence, the energy released in the given nuclear reaction is 1.3106 MeV.

6 0

2 years ago

What is a flammable mixture of many different substances

Wittaler [7]

The correct answer you are looking for is napalm

Hope this helps!

5 0

3 years ago