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3 years ago

Does the density of water change based on the volume of water? Explain please!

1 answer:

4 0

Hey there! Hello!

The best answer I can give to you is that it depends. It depends on the way that the volume is being increased and if it's being increased along with the mass.

Volume represents the amount of space a substance, object, etc. takes up. Density represents the amount mass found in a given measurement of volume. Taking these definitions into account, we can see that volume directly correlates with the measurement of density.

Think of two buckets, one with one gallon of a substance and the other with two gallons of a substance. Both samples of the substance will have the same density, since mass was being added in addition to volume. Take a look at the example:

Bucket 1:
Mass of Substance = 1kg
Volume of Substance = 3cm^3
Density of Substance= 1/3 kg/cm^3

Bucket 2:
Mass of Substance = 3kg
Volume of Substance = 6cm^3
Density of Substance= 3/6 kg/cm^3 = 1/3 kg/cm^3

If we simply add substance, the mass and volume will increase together at a fixed rate, meaning that the most simplified version of the density fraction will always be the same.

But there's another part to this. You can increase or decrease the temperature of a sample, making volume increase/decrease and the mass stay the same. When you evaporate water, it turns to steam, which takes up much more space than it did as water. But nothing has been created or destroyed, so the mass stays the same. This would make the density change along with the volume.

I hope this helped you out! Feel free to ask me any additional questions if you need further clarification. :-)

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A ____usually requires calculations that are too complicated to do by<br>hand.

liraira [26]

Answer:

computer model

Explanation:

Computer models are cheaper to set up than alternative methods that could be used to predict what will happen in a system, ex. building a prototype. Other benefits include being able to: make alterations and quickly see the outcomes.

8 0

3 years ago

Please help make sure its correct thanks

Wewaii [24]

The empirical formula of metal iodide : CoI₃(Cobalt(III) Iodide)

<h3>Further explanation</h3>

13.02 g sample of Cobalt , then mol Co(MW=58.933 g/mol) :

$\tt mol=\dfrac{mass}{Ar}=\\\\mol=\dfrac{13.02~g}{58.933}\\\\mol=0.221$

Mass of metal iodide formed : 97.12 g, so mass of Iodine :

$\tt =mass~metal~iodide-mass~Cobalt\\\\=97.12-13.02\\\\=84.1~g$

Then mol iodine (MW=126.9045 g/mol) :

$\tt \dfrac{84.1}{126.9045}=0.663$

mol ratio of Cobalt and Iodine in the compound :

$\tt 0.221\div 0.663=1\div 3$

5 0

3 years ago

For the reaction N2O4(g) ⇋ 2NO2(g), Kc = 0.25 at 98°C. At a point during the reaction, the concentration of N2O4 = 0.50 M and th

Scilla [17]

Answer:

Q = 0.50

Left

Explanation:

At a generic reversible equation

aA + bB ⇄ cC + dD

The reaction coefficient (Q) is the ratio of the substances concentrations:

$Q = \frac{[C]^c*[D]^d}{[A]^a*[B]^b}$

Solids and liquid water are not considered in this calculus.

When the reaction achieves equilibrium (concentrations are constant), the Q value is named as Kc, which is the equilibrium constant of the reaction. If Q > Kc, it indicates that the concentration of the products is higher, so, the reaction must progress to the left and form more reactants; if Q < Kc, than the concentrations of the reactants, are higher, so, the reaction progress to the right.

In this case:

Q = $\frac{[NO_2]^2}{[N_2O_4]}$

$Q = \frac{0.50^2}{0.50}$

Q = 0.50

So, Q > Kc, the reaction is not at equilibrium and it progresses to the left.

6 0

3 years ago

Write a balanced equation from each cell notation:(b) Fe(s) | Fe²⁺(aq) | NO₃⁻(aq) | NO(g) | Pt(s)

tatiyna

The balanced redox reaction is

3Fe + 2NO₃⁻+ 8H+ → 3Fe²⁺ + 2NO + 4H2O

<h3>What is Galvanic cell? </h3>

It is a device that is used for the conversion of the chemical energy which is produce in the redox reaction into the electrical energy. It is also termed as the voltaic cell or electrochemical cell.

In the voltaic cell, at anode which is negative electrode of voltaic cell oxidation occurs and at the cathode which is positive electrode, reduction occurs

<h3>What is redox reaction? </h3>

The reaction in which both reduction and oxidation reaction takes place simultaneously is known as redox reaction.

<h3>What is oxidation reaction? </h3>

The reaction in which a substance or compound or species looses its electrons. In this reaction, oxidation state of an element increases.

<h3>What is reduction reaction? </h3>

The reaction in which a substance or compound or species accept the electrons. In this reaction, oxidation state of an element decreases.

The given two-half reactions are:

At anode:

3Fe → 3Fe²⁺ + 6e-

At cathode:

8H(+) + 2NO₃⁻ → 2NO + 4H2O

Redox reaction:

3Fe + 2NO₃⁻+ 8H+ → 3Fe²⁺ + 2NO + 4H2O

According to the voltaic cell, the half-cell reaction (1) shows oxidation reaction therefore, it occurs at anode and the half-cell reaction (2) shows reduction reaction therefore, it occurs at cathode.

Thus, we concluded that the redox reaction is

3Fe + 2NO₃⁻+ 8H+ → 3Fe²⁺ + 2NO + 4H2O

learn more about voltaic cell:

brainly.com/question/4052514

#SPJ4

7 0

2 years ago

My sister is a peach" is a metaphor.<br><br><br> TrueFalse

dolphi86 [110]

Yes that is a metaphor

6 0

3 years ago