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Sergio039 [100]

2 years ago

7

You’ll soon make two pieces of ice with the approximate shapes shown in the image. Both pieces of ice have the same mass and vol

ume. How are the two shapes different from one another?

1 answer:

Montano1993 [528]2 years ago

3 0

Is there an image? I'll stay to help you.

You might be interested in

Arrange the following in order of increasing bond strength of the carbon oxygen bond:

icang [17]

Answer: Option (d) is the correct answer.

Explanation:

It is known that length of a bond is inversely proportional to the bond strength. This also means that a single bond has long length due to which it is weak in nature.

And, a double bond is shorter in length and has more strength as compared to a single bond. Whereas a triple bond has the smallest length and it has high strength as compared to a double or single bond.

For example, carbon monoxide is CO where there is a triple bond between the carbon and oxygen atom.

Carbon dioxide is $CO_{2}$ where there exists a double bond between the carbon and oxygen atom.

A carbonate ion is $CO^{2-}_{3}$ when two oxygen atoms are attached through single bond with the carbon atom and another oxygen atom is attached through a double bond to the carbon atom.

Hence, we can conclude that order of increasing bond strength of the given carbon oxygen bond is as follows.

Carbonate ion < carbon dioxide < carbon monoxide

5 0

2 years ago

german

Order of metals from least reactive to most reactive: B <C <A <D

<h3>Further explanation</h3>

Reducing agents are substances that experience oxidation

Oxidizing agents are substances that experience reduction

The metal activity series is expressed in voltaic series

<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au </em>

The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent

The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing agent

So that the metal located on the left can push the metal on the right in the redox reaction

Let's analyze the statement in the problem

I. Only A, C and D react with 1 mol/L HCl to give H₂(e)

M + HCl ⇒ MCl + H₂(MCl : alkali, MCl₂ : alkaline earth)

A, C and D can react with 1 mol / L HCl, meaning metals A, C and D are located to the left of element H (more reactive), and B in the right of element H

II. When A is added to solutions of the other metal ions, metallic B and C are formed but not D.

This means that metal A is more reactive than metals B and C, while D is more reactive than A, so metal D is the most reactive

3 0

2 years ago

Identify the reactants in the equation: 6C0, + 6H2O - CHUO+ 60,

Effectus [21]

Answer:

The correct answer is d. 6H20 + 6CO2.

The reactant in the chemical reaction 6H2O + 6CO2 ---> C6H12O6 + 6O2 is 6H20 + 6CO2. Remember that the reactant is always at the left side of the equation. So the correct answer is 6H20 + 6CO2 since it's in the left of the equation. I hope this answer helped you.

Explanation:

5 0

2 years ago

What temperature would 3.54 moles of xenon gas need to reach to exert a pressure of 1.57 atm at a volume of 34.6 l

ira [324]

Answer:

186.9Kelvin

Explanation:

The ideal gas law equation is PV = n R T

where

P is the pressure of the gas

V is the volume it occupies

n is the number of moles of gas present in the sample

R is the universal gas constant, equal to 0.0821 atm L /mol K

T is the absolute temperature of the gas

Ensure units of the volume, pressure, and temperature of the gas correspond to R ( the universal gas constant, equal to 0.0821 atm L /mol K )

n = 3.54moles

P= 1.57

V= 34.6

T=?

PV = n R T

PV/nR = T

1.57 x 34.6/3.54 x 0.0821

54.322/0.290634= 186.908620464= T

186.9Kelvin ( approximately to 1 decimal place)

5 0

2 years ago

A 50.0 mg sample of iodine-131 was placed in a container 32.4 days ago. if its half-life is 8.1 days, how many milligrams of iod

sertanlavr [38]

3.124mg of I-131 is present after 32.4 days.

The 131 I isotope emits radiation and particles and has an 8-day half-life. Orally administered, it concentrates in the thyroid, where the thyroid gland is destroyed by the particles.

What is Half life?

The time required for half of something to undergo a process: such as. a : the time required for half of the atoms of a radioactive substance to become disintegrated.

Half of the iodine-131 will still be present after 8.1 days.

The amount of iodine-131 will again be halved after 8.1 additional days, for a total of 8.1+8.1=16.2 days, reaching (1/2)(1/2)=1/4 of the initial amount.

The quantity of iodine-131 will again be halved after 8.1 more days, for a total of 16.2+8.1+8.1=32.4 days, to (1/4)(1/2)(1/2)=1/16 of the initial quantity.

If the original dose of iodine-131 was 50mg, the residual dose will be (50mg)*(1/16)=3.124mg after 32.4 days.

Learn more about the Half life of radioactie element with the help of the given link:

brainly.com/question/27891343

#SPJ4

7 0

8 months ago