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3 years ago

Bromine, a liquid at room temperature, has a boiling point of 58°C and a melting point of –7.2°C. Bromine can be classified as a

Chemistry

2 answers:

Svetllana [295]3 years ago

7 0

Answer: D

Explanation:

WARRIOR [948]3 years ago

4 0

Bromine can be classified as pure substance

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A material has a volume of 63.0 cm3 and a mass of 28 grams. What is the density of the material in g/cm3 to the correct number

dimaraw [331]

Answer:

0.4444 g/cm³ ≅ 0.44 g/cm³ (2 significant figures).

Explanation:

Knowing that:

d = m/V,

where, d is the density of the material (g/cm³).

m is the mass of the material (m = 28 g).

V is the volume of the material (V = 63.0 cm³).

∴ d = m/V = (28 g)/(63.0 cm³) = 0.4444 g/cm³ ≅ 0.44 g/cm³ (2 significant figures).

7 0

3 years ago

An example of a physical property of an element is the element’s ability to(1) react with an acid(2) react with oxygen(3) form a

Thepotemich [5.8K]

4: Form An Aqueous Solution

This is the only answer that can be observed without testing gear and with the naked eye.... Hope I helped ^-^

5 0

4 years ago

Read 2 more answers

A 100 g sample of potassium chlorate, KCIO3(s), is completely decomposed by heating:

Mama L [17]

Explanation:
In order to be able to calculate the volume of oxygen gas produced by this reaction, you need to know the conditions for pressure and temperature.
Since no mention of those conditions was made, I'll assume that the reaction takes place at STP, Standard Temperature and Pressure.
STP conditions are defined as a pressure of
100 kPa
and a temperature of
0
∘
C
. Under these conditions for pressure and temperature, one mole of any ideal gas occupies
22.7 L
- this is known as the molar volume of a gas at STP.
So, in order to find the volume of oxygen gas at STP, you need to know how many moles of oxygen are produced by this reaction.
The balanced chemical equation for this decomposition reaction looks like this
2
KClO
3(s]
heat
×
−−−→
2
KCl
(s]
+
3
O
2(g]
↑
⏐
⏐
Notice that you have a
2
:
3
mole ratio between potassium chlorate and oxygen gas.
This tells you that the reaction will always produce
3
2
times more moles of oxygen gas than the number of moles of potassium chlorate that underwent decomposition.
Use potassium chlorate's molar mass to determine how many moles you have in that
231-g
sample
231
g
⋅
1 mole KClO
3
122.55
g
=
1.885 moles KClO
3
Use the aforementioned mole ratio to determine how many moles of oxygen would be produced from this many moles of potassium chlorate
1.885
moles KClO
3
⋅
3
moles O
2
2
moles KClO
3
=
2.8275 moles O
2
So, what volume would this many moles occupy at STP?
2.8275
moles
⋅
22.7 L
1
mol
=
64.2 L

6 0

3 years ago

Ethanol, C2H6O, is most often blended with gasoline - usually as a 10 percent mix - to create a fuel called gasohol. Ethanol is

weeeeeb [17]

Answer:

This means 463 grams of ethanol would provide less amount of energy

Explanation:

Step 1: Data given

Heat of combustion of ethanol = 326.7 kcal/mol

The heat of combustion of octane = 1.308*10³ kcal/mol

Mass of octane = 463 grams

Molar mass octane = 114.23 g/mol

Molar mass ethanol = 46.07 g/mol

Step 2: Calculate moles octane

Moles octane = mass octane / molar mass octane

Moles octane = 463 grams / 114.23 g/mol

Moles octane = 4.05 moles

Step 3: Calculate energy of combustion of 4.05 moles octane

Combustion of 1 mol octane gives us: 1.308 * 10³ kcal/mol

Combustion of 4.05 moles octane gives us 4.05 * 1.308 * 10³ kcal/mol = 5.30 * 10³ kcal

This means the combustion reaction of 463 grams of octane gives us 5.30 * 10³ kcal

Step 4:

Heat of combustion of ethanol = 326.7 kcal/mol

OR in words: combustion of 1 mol ethanol gives us 326.7 kcal energy

Moles ethanol = 463 grams / 46.07 g/mol

Moles ethanol = 10.05 moles

Since combustion of 1 mol ethanol gives us 326.7 kcal

10.05 moles ethanol will give us = 10.05 * 326.7 = 3283.3 kcal = 3.28 * 10³ kcal

5.30 * 10³ kcal > 3.28 * 10³ kcal

This means 463 grams of ethanol would provide less amount of energy

3 0

4 years ago

Why are the oxidation and reduction half-reactions separated in an electrochemical cell?

Aleksandr [31]

Answer:

<h2>It makes the current viable enough to pass through an exterior wire.</h2>

Explanation:

Electrochemical cells primarily comprise of two half-cells. These half-cells assist in isolating the oxidation and reduction half-reactions. These two reactions are linked by a wire which allows the current to move from one edge to the other. The oxidation at the anode and the reduction take place at the cathode and the addition of a salt bridge helps in completing the circuit and permits the current to flow and leads to the generation of electricity.

7 0

3 years ago