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4 years ago

Element X reacts with bromine to form an ionic compound that has the formula XCl2?

Chemistry

1 answer:

andreyandreev [35.5K]4 years ago

3 0

Explanation:

Bromine is a group 17 element and contains 7 valence electrons. And, in order to completely fill its octet it needs to gain just one electron. When it chemically combines with another substance and forms a compound with chemical formula as $XCl_{2}$ then it means charge on substance X is +2.

For example, compound formed between magnesium and bromine is magnesium bromide and its chemical formula will be $MgBr_{2}$ .

Any element of group 2 can react with bromine to form a compound with chemical formula $XBr_{2}$ .

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What is a logical way that you could begin to investigate why your tire had low air pressure? Look to see if there is a hole in

MrRissso [65]

Answer:

Any of the answers given will work

Explanation:

I literally just did it.

7 0

3 years ago

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Which fire symbol signifies ammunition with a mass explosion?

marissa [1.9K]

The answer is Fire symbol 1

That is Fire symbol 1, signifies ammunition with a mass explosion.

The symbol can be seen in the image attached.

This symbol includes these :

1. Do not use fire fighting unless rescue attempt is planned.

2. If person's safety is in danger, use cover.

This fire symbol 1 signifies ammunition with a mass explosion.

4 0

4 years ago

Write the oxidation and reduction half reactions;

luda_lava [24]

Answer:

$Fe^{2+}$ ⇒ $Fe^{3+}+e^-$

$Br_2+2e^-$ ⇒ $2Br^-$

$Mg$ ⇒ $Mg^{2+}+2e^-$

$Cr^{3+}+e^-$ ⇒ $Cr^{3+}$

Explanation:

Remember that positive number superscripts mean electrons lack and negative numbers mean electrons 'excess' (if we compare it with the neutral element). So, for the case of Fe2+ which is converted to Fe3+, we know that in Fe2+ there is a two electrons lack, while in Fe3+ there is a 3 electrons lack; it means that Fe2+ was converted to Fe3+ but releasing one electron:

$Fe^{2+}$ ⇒ $Fe^{3+}+e^-$

The same analysis is applied to Br2; Br2 is a molecule which is said to have a zero superscript because it is an apolar covalent bond; and it is converted to Br-, which, according to what I wrote above, means that there is a one electron excess. So, Br2 must have received an electron in order to change to Br-; but Br2 can't change to Br- as simple as that because Br2 is a molecule, not an atom; it is a molecule that has two Br atoms, so, Br2 must give two Br- ions as products, but receiving one electron for each one:

$Br_2+2e^-$ ⇒ $2Br^-$

Applying the same, in Mg2+ there is a 2 electrons lack, and in Mg is not electron lack (its superscript is zero), so Mg must have released two electrons in order to change to Mg2+:

$Mg$ ⇒ $Mg^{2+}+2e^-$

Cr3+ has a 3 electrons lack, and Cr2+ a two electrons one, so, Cr3+ must receive an electron to convert to Cr2+:

$Cr^{3+}+e^-$ ⇒ $Cr^{3+}$

3 0

4 years ago

Read 2 more answers

The length of the school's gymnasium is 30 meters long. How many centimeters long is the school's gymnasium

aleksley [76]

Answer:

The answer to your question is: 3000 cm

Explanation:

30 meters long to cm

We can use a rule of three to solve it

We know that 1 meter ------------------ 100 cm

30 meters ---------------- x

x = 30(100)/1 = 3000 cm

5 0

4 years ago

Read 2 more answers

1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

3 0

4 years ago