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Dmitry [639]
3 years ago
14

Which element has three electron shells and 4 valence electrons?

Chemistry
2 answers:
Tcecarenko [31]3 years ago
4 0

Answer: Silicon

Explanation:

For something to have 4 valence electrons, it must be in the 14th column, which elimates gallium and argon for the possible answers. For somethign to have 3 electron shells, it must be in the third row of the periodic table which means the answer is silicon.

jolli1 [7]3 years ago
3 0

Answer:

Gallium

Explanation:

I believe this is the answer if not then the only other option is boron but anyways the answer is Gallium

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etermine the number of valence electrons for each of the atoms. Enter each answer as a numeral. For example, if an atom has two
Sav [38]

Answer:

1. 1

2. 2

3. 6

4. 8

Explanation:

Valence electrons can be defined as the number of electrons present in the outermost shell of an atom. Valence electrons are used to determine whether an atom or group of elements found in a periodic table can bond with others. Thus, this property is typically used to determine the chemical properties of elements.

1. Li is the symbol for the chemical element known as Lithium: it has 1 valence electron. The electronic configuration of lithium is 1s²2s¹

2. Mg is the symbol for the chemical element known as Magnesium: it has 2 valence electrons. The electronic configuration of magnesium is 1s²2s²2p63s²

3. O is the symbol for the chemical element known as Oxygen: it has 6 valence electrons. The electronic configuration of magnesium is 1s²2s²2p⁴.

4. Ne is the symbol for the chemical element known as Neon: it has 8 valence electrons.

7 0
3 years ago
How many moles are in 2.0 grams of Na?
lisov135 [29]

Answer:

46

Explanation:

Sodium metal has a molar mass of  

22.99

3 0
3 years ago
One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate
Ugo [173]

Answer:

\delta H_{rxn} = -66.0  \ kJ/mole

Explanation:

Given that:

3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \  \ \delta H = -47.0 \ kJ/mole  -- equation (1)  \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)}  \ \ \delta H = -25.0 \ kJ/mole  -- equation (2)  \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole  -- equation (3)

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

3FeO_{(s)} + CO_{2(g)}    \to    Fe_3O_4_{(s)} + CO_{(g)}   \ \delta H = -19.0 \ kJ/mole  -- equation (4)

Multiplying  (2) with equation (4) ; we have:

6FeO_{(s)} + 2CO_{2(g)}    \to    2Fe_3O_4_{(s)} + 2CO_{(g)}   \ \delta H = -38.0 \ kJ/mole  -- equation (5)

From equation (1) ; multiplying (-1) with equation (1); we have:

2Fe_3O_4_{(s)} +CO_{2(g)} \to     3FeO_3_{(s)}+CO_{(g)}   \  \ \delta H = 47.0 \ kJ/mole  -- equation (6)

From equation (2); multiplying (3) with equation (2); we have:

3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)}  \ \ \delta H = -75.0 \ kJ/mole  -- equation (7)

Now; Adding up equation (5), (6) & (7) ; we get:

6FeO_{(s)} + 2CO_{2(g)}    \to    2Fe_3O_4_{(s)} + 2CO_{(g)}   \ \delta H = -38.0 \ kJ/mole  -- equation (5)

2Fe_3O_4_{(s)} +CO_{2(g)} \to     3FeO_3_{(s)}+CO_{(g)}   \  \ \delta H = 47.0 \ kJ/mole  -- equation (6)

3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)}  \ \ \delta H = -75.0 \ kJ/mole  -- equation (7)

<u>                                                                                                                      </u>

FeO  \ \ \ +  \ \ \ CO   \ \  \to   \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \  \delta H = - 66.0 \ kJ/mole

<u>                                                                                                                     </u>

<u />

\delta H_{rxn} = \delta H_1 +  \delta H_2 +  \delta H_3    (According to Hess Law)

\delta H_{rxn} = (-38.0 +  47.0 + (-75.0)) \ kJ/mole

\delta H_{rxn} = -66.0  \ kJ/mole

8 0
3 years ago
A sample of oxygen gas occupies a volume of 250 mL at .8 atm of pressure. What volume will it occupy at 1.2 atm of pressure?
musickatia [10]
167 mL

P1V1 = P2V2
P1 = .8 atm
V1 = 250 mL
P2 = 1.2 atm

Solve for V2 —> V2 = P1V1/P2

V2 = (0.8 atm)(250 mL) / (1.2 atm) = 167 mL
7 0
3 years ago
A student has the following data recorded: final readings: 760. mm Hg, 6.0L, 197 'C. initial readings:
irina [24]
Is that Geometry or Algebra?
7 0
3 years ago
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