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noname [10]
3 years ago
15

How to balance this equation: Cu(s) + O2(g) --> CuO(s)

Chemistry
1 answer:
Lelu [443]3 years ago
3 0
2Cu(s) + O2(g) --> 2CuO(s)
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Glucose is an example of which carbon-based macromolecule?
Blababa [14]

Answer: Glucose is an example of carbon-based macromolecule known as carbohydrates

Explanation:

carbon based macromolecule are important cellular components and they perform a variety of functions necessary for growth and development of living organisms. There are 4 major types of carbon based molecules and these includes;

Carbohydrate

Lipids

Proteins and

Nucleic acids.

Carbon is the primary components of these macromolecules. Carbohydrate macromolecules are made up of monosaccharide, disaccharide and polysaccharides. Glucose is an example of a monosaccharide and it has two important types of functional groups: a carbonyl group and a hydroxyl group. I hope this helps. Thanks

8 0
3 years ago
Which of the following aqueous solutions will have the lowest freezing point? a. 0.75 m (NH4)3PO4 b. 1.0 m CaSO4 c. 1.0 m LiCl d
Nataly [62]

Answer: a. 0.75m (NH4)3PO4 will have the lowest freezing point .

Explanation: Freezing point decreases as the concentration of the solute substance increases. Assuming the same solvent for all of them, for instance water.

∆T= i.Kf.b

∆T= freezing point depression

i= vant Hoff factor

Kf= molality

Assuming water to be the solvent for all Kf=1.86°C/M

VANT HOFF FACTORS :

For (NH4)3PO4

This has 3 ionic bonding and 1 covalent bonding making it 4 bond

Therefore i=4

For CaSO4

This has 1 metallic bond and 1 covalent bond making it 2 bond.

Therefore i=2

For LiCl

This has 1 metallic bond and 1 non metallic bond making it 2 bond.

Therefore i=2

For CH3OH

This has only 1 covalent bond.

Therefore i=1

MOLALITY:

(NH4)3PO4 = 0.75M

CaSO4= 1.0M

LiCl= 1.0M

CH3OH= 1.5M

FREEZING POINT DEPRESSION:

For (NH4)3PO4

∆T= 4×0.75×1.86=5.58°C

For CaSO4

∆T= 2×1.0×1.86=3.72°C

For LiCl

∆T = 2×1.0×1.86= 3.72°C

For CH3OH

∆T= 1×1.5×1.86=2.79°C

REMEMBER THE HIGHER THE FREEZING POINT DEPRESSION THE LOWER THE FREEZING POINT.

FREEZING POINT DEPRESSION IS THE CHANGE IN THE FREEZING POINT PROPORTIONAL TO THE AMOUNT OF SOLUTE ADDED THE THE SOLUTION.

THEREFORE THE ONE WITH THE LOWEST FREEZING POINT IS (NH4)3PO4

4 0
3 years ago
What type of ignition occurs when a mixture of fuel and oxygen encounter an external heat source with sufficient heat?
mars1129 [50]

The ignition occurs when a mixture of fuel & oxygen encounter an external heat source with sufficient heat<u> Piloted ignition.</u>

<h3>What is Piloted ignition?</h3>

When a volatile fuel is close to a nearby local energy source (pilot) and reaches its lower limit of flammability in air, piloted ignition may be possible. The flame that originates in the premixed system spreads outward from the pilot. The first law of thermodynamics for systems with fixed mass only describes the energetics of this process.

<h3>Definition of thermodynamics </h3>

The science of thermodynamics examines how heat, work, temperature, and energy are related. The general topic of thermodynamics is the transfer of energy from one location or form to another. The fundamental idea is that heat is a type of energy that is equivalent to a specific quantity of mechanical labor.

To learn more about ignition

brainly.com/question/6481027

#SPJ4

4 0
2 years ago
If a solid, a liquid, and a gas of a substance are each in separate covered jars, then the amount of kinetic energy of the parti
allsm [11]

Answer:

false

Explanation:

i had the same thing to do

8 0
3 years ago
Determine the ph of a 0.188 m nh3 solution at 25°c. the kb of nh3 is 1.76 × 10-5.
Ber [7]
When NH3 is dissolved in water, it dissociates  partially producing NH4+ ions and OH- ions. It has an equation:
NH3 + H2O → NH4+ + OH- 

<span>We use the Kb expression to determine the [OH-] concentration,
</span>
<span>Kb = [NH4+] [OH-] /* [NH3] </span>

We can write NH4+ as OH- since they are of equal ratio. 
<span>(1.76*10^-5) = [OH-]² / 0.188 
</span><span>[OH-]² = 3.3088*10^-6 </span>
<span>[OH-] = 1.819*10^-3 </span>

We calculate for H+ concentration as follows:

<span>[H+] [OH-] = 10^-14 </span>
<span>[H+] = 10^-14 / [OH-] </span>
<span>[H+] = 10^-14 / (1.819*10^-3) </span>
<span>[H+] = 5.50*10^-12 </span>

<span>pH = -log [H+] </span>
<span>pH = -log (5.5*10^-12) </span>
<span>pH = 11.26</span>
4 0
3 years ago
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