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Oliga [24]
3 years ago
10

Magnesium bromide is a binary ionic compound. From its formula, MgBr2, how do you know that

Chemistry
2 answers:
Scilla [17]3 years ago
8 0
You that Mg is the metal if in the periodic table it lands between group 1-12. So yeas Mg is a metal because it lands in group 2.

andrew-mc [135]3 years ago
8 0

Answer: Magnesium loses electrons to form Mg^{2+}

Explanation:

An ionic bond is formed when an element completely transfers its valence electron to another element. Metals donate the electron and forms a positively charged ion called as cation. Non metals accept the electrons and forms a negatively charged ion called as anion.

Electronic configuration of magnesium:

[Mg]:12: 1s^22s^22p^63s^2

Magnesium atom will loose one electron to gain noble gas configuration and form magnesium cation with +2 charge.

[Mg^{2+}]:10:1s^22s^22p^63s^0

Electronic configuration of bromine

[Br]:35:1s^22s^22p^63s^23p^64s^23d^{10}4p^5

Bromine atom will gain one electron to gain noble gas configuration and form bromide ion with -1 charge.

[Br^-]=1s^22s^22p^63s^23p^64s^23d^{10}4p^6

Thus as magnesium forms a cation , it is the metal.

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Read 2 more answers
CH3OH can be synthesized by the following reaction.
puteri [66]

Answer:

A: There is 43.12 Liter of H2 needed

B: There is 21.56 liter of CO needed

Explanation:

Step 1: Data given

CO(g)+2H2(g)?CH3OH(g)

For 1 mol of CO consumed, we need 2 moles of H2 to produce 1 mol of CH3OH

Molar mass of CO = 28.01 g/mol

Molar mass of H2 = 2.02 g/mol

Molar mass of CH3OH = 32.04 g/mol

Step 2: What volume of H2 gas (in L), measured at 754mmHg and 90?C, is required to synthesize 23.0g CH3OH?

Pressure = 754 mmHg = 0.992 atm

Temperature = 90°C = 363 Kelvin

mass of CH3OH produced = 23.0 grams

Step 3: Calculate moles of CH3OH

Moles CH3OH = mass CH3OH / Molar mass CH3OH

Moles CH3OH = 23.0 grams / 32.04 g/mol

Moles CH3OH = 0.718 moles

Step 4: Calculate moles of H2

For 1 mol of CO consumed, we need 2 moles of H2 to produce 1 mol of CH3OH

For 0.718 moles CH3OH produced, we have 2*0.718 moles =1.436 moles of H2 and 0.718 moles of CO

Step 5: Calculate volume of H2

p*V = n*R*T

with p = the pressure = 0.992 atm

with V the volume = TO BE DETERMINED

with n = the number of moles = 1.436 moles H2

with R = the gasconstant = 0.08206 L*atm/ K*mol

with T = the temperature = 363 Kelvin

V = (n*R*T)/p

V = (1.436*0.08206*363)/0.992

V = 43.12 L

Step 6: Calculate volume of CO

p*V = n*R*T

with p = the pressure = 0.992 atm

with V the volume = TO BE DETERMINED

with n = the number of moles = 0.718  moles CO

with R = the gasconstant = 0.08206 L*atm/ K*mol

with T = the temperature = 363 Kelvin

V = (n*R*T)/p

V = (0.718*0.08206*363)/0.992

V = 21.56 L

A: There is 43.12 Liter of H2 needed

B: There is 21.56 liter of CO needed

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Answer:

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