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3 years ago

5

Chemistry

1 answer:

Archy [21]3 years ago

6 0

Answer:

<h3>The answer is option C</h3>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

<h3>mass = Density × volume</h3>

From the question

volume of liquid = 15 mL

density = 2.5 g/mL

We have

mass = 15 × 2.5

We have the final answer as

Hope this helps you

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Which inorganic substance taken up by plants is also taken in and used directly by animals, including humans?

masha68 [24]

Answer: water

water can not be organic...but is is took by humans and plants because theur life dpends on it

Explanation:

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3 years ago

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What is the percent composition of aluminum in the ore bauxite (AI2O3)?

Schach [20]

Answer:

Molar mass of Al2O3 = 101.961276 g/mol

This compound is also known as Aluminium Oxide.

Convert grams Al2O3 to moles or moles Al2O3 to grams

Molecular weight calculation:

26.981538*2 + 15.9994*3

Percent composition by element

Element Symbol Atomic Mass # of Atoms Mass Percent

Aluminium Al 26.981538 2 52.925%

Oxygen O 15.9994 3 47.075%

Explanation:

Percent composition by element

Element Symbol Mass Percent

Aluminium Al 52.925%

Oxygen O 47.075%

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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.

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3 years ago

What do you know about a chemical compound by looking at its chemical formula?

coldgirl [10]

The type of bonds present in the compound. and the type of structure it has and the elements that are presents and the number of moles of each element in one mole of the compound.

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3 years ago

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When rolling a die, find the probability of rolling an even number or rolling a

frozen [14]

<h3><em><u>ᎪꪀsωꫀᏒ</u></em></h3>

even no = 3/6 = 1/2

no. less than 5 = 4/6 = 2/3

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2 years ago