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vova2212 [387]
3 years ago
5

Give the Unabbreviated electron configuration for barium

Chemistry
1 answer:
Shalnov [3]3 years ago
8 0

Answer:

Ba: 1s² 2s²2p⁶ 3s²3p⁶ 4s²3d¹⁰4p⁶ 5s²4d¹⁰5p⁶ 6s²

Step-by-step explanation:

Step 1. Locate barium in the Periodic Table.

It's in Period 6, Group 2: Element 56 (highlighted blue in the Periodic Table below).

Step 2. Add 54 electrons to the energy levels

You add then in the order shown in the diagram below.

The complete electron configuration is:

Ba: 1s² 2s²2p⁶ 3s²3p⁶ 4s²3d¹⁰4p⁶ 5s²4d¹⁰5p⁶ 6s²

n =  2 +     8    +    8     +       18       +       18       +  2 = 56

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3.65 gram of hcl is dissolved in 180 gram of water. Find the total number of molecules of hydrogen​
Morgarella [4.7K]

Answer:

Molec_{\ H_{tot}}=1.206x10^{25}molec

Explanation:

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In this case, taking into account that HCl has one molecule of hydrogen per mole of compound which weights 36.45 g/mol, we compute the number of molecules of hydrogen in hydrochloric acid by considering the given mass and the Avogadro's number:

molec_{\ H}=3.65gHCl*\frac{1molHCl}{36.45gHCl} *\frac{1molH}{1molHCl}*\frac{6.022x10^{23}molec_\ H}{1molH}  =6.03x10^{22}molec

Now, from the 180 g of water, we see two hydrogen molecules per molecule of water, thus, by also using the Avogadro's number we compute the molecules of hydrogen in water:

molec_{\ H}=180gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O}*\frac{6.022x10^{23}molec_\ H}{1molH}  =1.20x10^{25}molec

Thus, the total number of molecules turns out:

Molec_{\ H_{tot}}=6.03x10^{22}+1.20x10^{25}\\\\Molec_{\ H_{tot}}=1.206x10^{25}molec

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6 0
3 years ago
Which of the following would happen if Earth's moon were half the size that it is now? O The oceans' tidal variations would be s
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Explanation:

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3 years ago
In ionic bonding, during the transfer of electrons between two neutrally charged atoms, one electron moves from one atom to anot
forsale [732]

The given question is incomplete . The complete question is :

In ionic bonding, during the transfer of electrons between two neutrally charged atoms, one electron moves from one atom to another. What are the new relative charges between the two atoms?

a. The giving atom and receiving atom are both negatively charged

b. The giving atom is now positively charged and the receiving atom is now negatively charged.

c. The giving and receiving atom are both positively charged

d. The giving atom is now negatively charged and the receiving atom is now positively charged.

Answer:  The giving atom is now positively charged and the receiving atom is now negatively charged.

Explanation:

Ionic compounds are formed by transference of electrons between metals and non metals. The bond formed between a metal and a non-metal is always ionic in nature.

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

Thus when  one electron moves from one atom to another, the metal acquires a positive charge and the non metal acquires negative charge.

For example:  NaCl is formed by transfer of one electron from sodium to chlorine , thus forming Na^+ and Cl^-

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3 years ago
What would ne the acceleration in a body moving wit uniform velocity and why​
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3 0
3 years ago
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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
3 years ago
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