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Oliga [24]
4 years ago
10

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

Chemistry
2 answers:
Scilla [17]4 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]4 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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4 years ago
The pH of a Ba(OH) 2 solution is 10.00. What is the H+ ion concentration of this solution?
I am Lyosha [343]

Answer:

Option D. 1×10¯¹⁰ M

Explanation:

From the question given above, the following data were obtained:

pH = 10

Hydrogen ion concentration, [H+] =?

The hydrogen ion concentration, [H+] of the Ba(OH)2 solution can be obtained as follow:

pH = – Log [H+]

pH = 10

10 = – Log [H+]

Divide both side by – 1

– 10 = Log [H+]

Take the antilog of – 10

[H+] = antilog (– 10)

[H+] = 1×10¯¹⁰ M

Therefore, the hydrogen ion concentration, [H+] of the Ba(OH)2 solution is 1×10¯¹⁰ M

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3 years ago
Is platinum an element or compound
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3 years ago
Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant at for the following reaction. N2(g)H
soldier1979 [14.2K]

<u>Answer:</u> The equilibrium constant for this reaction is 5.85\times 10^{5}

<u>Explanation:</u>

The equation used to calculate standard Gibbs free change is of a reaction is:

\Delta G^o_{rxn}=\sum [n\times \Delta G^o_{(product)}]-\sum [n\times \Delta G^o_{(reactant)}]

For the given chemical reaction:

3H_2(g)+N_2(g)\rightarrow 2NH_3(g)

The equation for the standard Gibbs free change of the above reaction is:

\Delta G^o_{rxn}=[(2\times \Delta G^o_{(NH_3(g))})]-[(1\times \Delta G^o_{(N_2)})+(3\times \Delta G^o_{(H_2)})]

We are given:

\Delta G^o_{(NH_3(g))}=-16.45kJ/mol\\\Delta G^o_{(N_2)}=0kJ/mol\\\Delta G^o_{(H_2)}=0kJ/mol

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (-16.45))]-[(1\times (0))+(3\times (0))]\\\\\Delta G^o_{rxn}=-32.9kJ/mol

To calculate the equilibrium constant (at 25°C) for given value of Gibbs free energy, we use the relation:

\Delta G^o=-RT\ln K_{eq}

where,

\Delta G^o = standard Gibbs free energy = -32.9 kJ/mol = -35900 J/mol  (Conversion factor: 1 kJ = 1000 J )

R = Gas constant = 8.314 J/K mol

T = temperature = 25^oC=[273+25]K=298K

K_{eq} = equilibrium constant at 25°C = ?

Putting values in above equation, we get:

-32900J/mol=-(8.314J/Kmol)\times 298K\times \ln K_{eq}\\\\K_{eq}=e^{13.279}=5.85\times 10^{5}

Hence, the equilibrium constant for this reaction is 5.85\times 10^{5}

5 0
3 years ago
How many joules of energy do you get from eating an apple?
iris [78.8K]
I know I'm days late but why not answer your question anyway, I don't know how many calories your apple has but I believe it would be 95 calories is a normal amount for an apple. So the correct answer to your question is <span>397480 J Hope I'm correct please let me know and mark me Brainliest.
</span>
8 0
3 years ago
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