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2 years ago

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Students in a science class were studying the physical properties of various elements and compounds. One lab group observed that

the element they were given was shiny, brittle, and when placed in the gap shown by A and B of an open circuit caused the light bulb to glow dimly. Based on the students’ observations, what is the correct classification of this element?

1 answer:

katrin2010 [14]2 years ago

7 0

Answer:

A:metal

Explanation:

because if a light bulb was placed on metal with force it would light

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A compounds empirical formula is H.O. If the formula mass is 34 amu what is the molecular formula?

BigorU [14]

Answer:

C

Explanation:

1. The molecule has to have a ratio of 1 hydrogen to 1 oxygen.

2. Hydrogen = 1amu, Oxygen = 16amu. (2*2)+(16*2) = 34amu

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2 years ago

Define the term compound

dalvyx [7]

Answer:

A compound is a substance which contains two or more elements chemically combined together.

A compound is formed as a result of a chemical change.

Explanation:

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2 years ago

In a certain chemical process, water reacts with a compound, and the water is converted into another compound during the reactio

gtnhenbr [62]

Answer:

hydration reaction

Explanation:

The type of reaction would be hydration reaction.

<u>Hydration reaction generally involves a chemical reaction of water with another reactant and in which the water ends up being converted to another product entirely. </u>

A good example of hydration reaction is the reaction between alkene and water leading to the production of alcohol.

$CH_2=CH_2(g) + H_2O(g)$ ⇄ $CH_3CH_2OH(g)$

8 0

3 years ago

Read 2 more answers

58. Which of the following aqueous solutions has the lowest freezing point? A. 0.2 M NaCl B. 0.2 M FeCl3 C. 0.2M sucros D. 0.2 M

m_a_m_a [10]

The correct option is D. 0.2 M CaCl2 is has the lowest freezing point.

<h3>What is aqueous solution?</h3>

When one significance liquefies into another, a solution is formed. A solution is a homogenous mixture consisting of a solute dissolved into a solvent. The solute is the essence that is being dissolved, while the solvent is the dissolving medium. Solutions can be formed with multiple different classifications and forms of solutes and solvents. In this branch, we will focus on a resolution where the solvent is water.

An aqueous solution is a moisture that contains one or more dissolved essence. The dissolved importance in an aqueous solution may be solids, gases, or different liquids.
In directive to be a true solution, an assortment must be stable. When sugar is fully dissolved into moisture, it can stand for an undetermined amount of time, and the sugar will not recompense out of the solution. Further, if the sugar-water solution is passed through a filter, it will stay with the water.
This is because the liquefied particles in a resolution are very small, usually less than 1nm in diameter. Solute particles can be atoms, ions, or molecules, counting on the type of essence that has been dissolved.

To learn more about aqueous solution, refer to:

brainly.com/question/14469428

#SPJ9

6 0

1 year 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:

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