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3 years ago

Determine the mass in grams of 4.69 x 1021 atoms of barium. (The

Chemistry

1 answer:

Virty [35]3 years ago

5 0

Answer:

1.07 g Ba

Explanation:

Hello there!

In this case, according to the definition of the Avogadro's number and the molar mass, it is possible to say that 6.022x10^{23} atoms of barium equal one mole, and at the same time, 1 mole equals 137.327 grams of this element; thus, it is possible to say that 6.022x10^{23} atoms of barium have a mass of 137.327 grams; therefore, it i possible for us to calculate the required mass in grams as shown below:

$4.69x10^{21}atoms*\frac{137.327gBa}{6.022x10^{23} atoms} \\\\=1.07gBa$

Best regards!

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4 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.

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3 years ago

Which are characteristics of a prokaryotic cell? Select three options. contains DNA lacks DNA contains ribosomes lacks ribosomes

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Contains DNA, contains ribosomes, lacks a nucleus

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3 years ago

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3 years ago

The representative particles of a covalent compound is called

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