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3 years ago

Solid potassium chloride is obtained by the reaction of solid potassium and chlorine gas.

Chemistry

1 answer:

andreyandreev [35.5K]3 years ago

4 0

Answer:

2 K(s) + Cl₂ (g) --> 2 KCl(s)

Explanation:

Potassium will just be K

Chlorine gas is part of BrIClHOF, which are diatomic gasses. So Cl₂

Looking at the periodic table potassium K has an ion charge of +1 and chlorine Cl has an ion charge of -1, so in a balanced compound they will be written as KCl

Balancing the amounts of each will lead to 2 K(s) + Cl₂ (g) --> 2 KCl(s)

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Rank and explain how the freezing point of 0.100 m solutions of the following ionic electrolytes compare. List from lowest freez

Elden [556K]

Answer:

The solutions are listed from the lowest freezing point to highest freezing point.

Mg₃(PO₄)₂ - AlBr₃ - BeBr₂ - KBr

Explanation:

Colligative property of freezing point, is the freezing point depression.

Formula is:

ΔT = Kf . m . i

Where ΔT = T° freezing pure solvent - T° freezing solution

Kf is the cryoscopic constant

m is molality and i is the Van't Hoff factor (number of ions, dissolved in solution)

In this case, T° freezing pure solvent is 0° because it's water, so the Kf is 1.86 °C/m, and m is 0.1 molal.

The i modifies the T° freezing solution. The four salts, are ionic compounds, so for each case:

BeBr₂ → Be²⁺ + 2Br⁻ i = 3

AlBr₃ → Al³⁺ + 3Br⁻ i = 4

Mg₃(PO₄)₂ → 3Mg²⁺ + 2PO₄⁻³ i = 5

KBr → K⁺ + Br⁻ i = 2

Solution of Mg₃(PO₄)₂ will have the lowest temperature of freezing and the solution of KBr, the highest (always lower, than 0°).

8 0

3 years ago

1.a)How many types of particles are formed when a strong alkali is dissolved in water??

sveticcg [70]

Answer:

I didn't know how to explane it

but this should help you get your answer

4 0

3 years ago

How long it takes a 100 watt light bulb to use 10,000 J of electrical energy?

vlabodo [156]

Answer:

A 100-watt light bulb takes 100 seconds to use 10,000 J of electrical energy.

Explanation:

Power refers to the amount of work done (or energy consumed) per unit of time. It is usually measured in watts (W), which is equal to 1 Joules per second ( $\frac{J}{s}$ )

So, by definition, to calculate the power consumed or supplied in a time interval the following expression is used:

$P=\frac{W}{t}$

where P is the power measured in Watts, W is the work measured in Joules, and t is the time measured in seconds.

In this case:

P=100 Watt
W= 10,000 J
t= ?

Replacing:

$100 Watt=\frac{10,000 J}{t}$

Solving:

$t=\frac{10,000 J}{100 Watt}$

t=100 seconds

<u><em> A 100-watt light bulb takes 100 seconds to use 10,000 J of electrical energy.</em></u>

7 0

3 years ago

How much aluminum oxide are produced when 46.5g of Al react with 165.37g of MnO?

solong [7]

Aluminum oxide produced : = 79.152 g

<h3>Further explanation</h3>

Given

46.5g of Al

165.37g of MnO

Required

Aluminum oxide produced

Solution

Reaction

2 Al (s) + 3 MnO (s) → 3 Mn (s) + Al₂O₃ (s)

mol Al(Ar = 27 g/mol) :

mol = mass : Ar

mol = 46.5 : 27

mol = 1.722

mol MnO(Ar=71 g/mol) :

mol = 165.37 : 71

mol = 2.329

mol : coefficient ratio Al : MnO = 1.722/2 : 2.329/3 = 0.861 : 0.776

MnO as a limiting reactant(smaller ratio)

So mol Al₂O₃ based on MnO as a limiting reactant

From equation , mol Al₂O₃ :

= 1/3 x mol MnO

= 1/3 x 2.329

= 0.776

Mass Al₂O₃ (MW=102 g/mol) :

= 0.776 x 102

= 79.152 g

7 0

3 years ago

Help me get this right no links

ira [324]

Answer:

the rise and fall is the tides.

6 0

3 years ago