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3 years ago

Some compounds are described as binary compounds what does this mean

Chemistry

1 answer:

zubka84 [21]3 years ago

4 0

Answer:

It means these compounds are only made up of two elements.

Explanation:

Binary compounds:

The compounds which are made up of the atoms of only two elements are called binary compounds.

For example:

The following compounds are binary:

HCl

H₂O

NH₃

HCl is binary because it is composed of only hydrogen and chlorine. Ammonia is also binary compound because it is made up of only two elements nitrogen and hydrogen.

water is also binary because it is also made up of only two elements hydrogen and oxygen.

While all polyatomic compounds are not binary. for example,

H₂SO₄

It is made up of hydrogen, sulfur and oxygen. There are three elements are present in this compound. So, it is not binary compound.

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A fossil was analyzed and determined to have a carbon-14 level that is 70 % that of living organisms. The half-life of C-14 is 5

Citrus2011 [14]

Answer: 2948

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

$t_{\frac{1}{2}}=\frac{0.693}{k}$

$k=\frac{0.69}{t_{\frac{1}{2}}}=\frac{0.693}{5730}=1.21\times 10^{-4}years^{-1}$

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $1.21\times 10^{-4}years^{-1}$

t = age of sample = ?

a = let initial amount of the reactant = 100

a - x = amount left after decay process = $\frac{70}{100}\times 100=70$

$t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{100}{70}$

$t=2948years$

Thus the fossil is 2948 years old.

5 0

2 years ago

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

3 years ago

Balance the equation and write the reaction-quotient expression, qc. u(s) + f2(g) ----> uf6(g)

Alona [7]

The equation presented above is that of uranium reacting with fluorine forming uranium fluoride.

The chemical reaction can be balanced by carefully studying the equation and balancing the number of atoms of each of the element in both sides of the chemical reaction. That is,

<em> U(s) + 3F₂(g) --> UF₆(g)</em>

4 0

2 years ago

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Balance these chem problems

nikklg [1K]

These are hard sorry

4 0

2 years ago

Would precipitation occur when 500 mL of a 0.02M solution of AgNO3 is mixed with 500 mL of a 0.001M solution of NaCl? Show your

Oksanka [162]

We know,
AgNO3 + NaCl ⇒ NaNO3 + AgCl(s)
The moles of Na+ present:
0.5 L * 0.001 mol/L
= 5 x 10⁻⁴ mol
Moles of Ag+ present:
0.5 * 0.02
= 0.01 mol
The limiting reactant is Na
Therefore, the moles of Ag reacted:
5 x 10⁻⁴
AgCl is insoluble in water; therefore, the AgCl formed will precipitate

7 0

2 years ago