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Ostrovityanka [42]

3 years ago

5

List 4 factors that suggest a chemical reaction has taken place

1 answer:

Degger [83]3 years ago

3 0

The change in smell, The change in taste, If there is rust, i cant think of any other on the spot but if something were to catch fire then there is a chemical change.

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HURRY FAST!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

miss Akunina [59]

B. The sand increases friction by increasing roughness.

8 0

3 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:

1)

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 $1.2\times 10^{-4}$ .

2)

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 $1.6\times 10^{-10} M$ .

3.

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

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

Which compound are ionic and which are covalent? (N2) (CCl4) (SiO2) (AlCl3) (CaCl2) (LiBr)

muminat

Covalent compounds: N2, CCl4, SiO2 and AlCl3.

Ionic compounds: CaCl2 and LiBr.

Hope this helps!

3 0

3 years ago

2. What group is the atom in?<br> 2 or 2A<br> 6<br> 14 <br> 1

Juliette [100K]

Answer:

$\Large \boxed{\mathrm{Group \ 4}}$

Explanation:

The number of valence electrons tells us the group number of the neutral atom.

The atom has 4 valence electrons.

The atom is in group 4.

5 0

3 years ago

Read 2 more answers

What element is represented by the following electron configuration:

Alecsey [184]

For better representation, let me rewrite the electronic configuration:

<span>1s</span>²<span>2s</span>²<span>2p</span>⁶<span>3s</span>²<span>3p</span>⁶<span>4s</span>²<span>3d</span>⁴

The exponents represent the number of electrons in the designated subshell. Thus, the total number of electrons are:
# of electrons = 2+2+6+2+6+2+4 = 24

Assuming this is in neutral state, the element with an atomic number of 24 is Chromium. Thus, the answer is Cr.

6 0

3 years ago