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3 years ago

Which property is shared by solids and liquids?

Chemistry

2 answers:

rewona [7]3 years ago

6 0

Answer:

Explanation:

all solids and liquids have a definite volume

<u>Shape</u> <u>Volume </u> <u>Compressibility</u>

SOLID D D NOT POSSIBLE

LIQUID I D POSSIBLE

GAS I I POSSIBLE

I STANDS FOR INDEFINITE

D STANDS FOR DEFINITE

MA_775_DIABLO [31]3 years ago

3 0

Your answer is B all solids & liquids have definite volume

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A neutral isotope has 45 neutrons and 36 electrons. Identify the element symbol of this isotope and determine the mass number an

Degger [83]

Answer:

Kr, mass number 81, number protons 36.

Explanation:

If the atom has 36 electrons, and it is neutral, it has 36 protons. 36 protons means that this element has atomic number = 36. This is Kr.

Mass number = number protons + number neutrons = 36 + 45 = 81

5 0

4 years ago

Consider 100.0 g samples of two different compounds consisting only of vanadium and oxygen. one compound contains 61.4 g of vana

xxMikexx [17]

Answer:

2:1

Explanation:

The Law of Multiple Proportions states that when two elements A and B combine to form two or more compounds, the masses of B that combine with a given mass of A are in the ratios of small whole numbers.

That is, if one compound has a ratio r₁ and the other has a ratio r₂, the ratio of the ratios r is in small whole numbers.

1. Compound 1

Mass of O = 100.0 - 61.4 = 38.6 g

$r_{1} = \dfrac{\text{mass of O}}{\text{mass of V}} = \dfrac{ 38.6}{61.4} = 0.6287$

2. Compound 2

Mass of O = 100.0 - 76.1 = 23.9 g

$r_{2} = \dfrac{ 23.9}{76.1} = 0.3141$

3. Ratio of the ratios

$r = \dfrac{r_{1}}{r_{2}} = \dfrac{ 0.6287}{0.3141} = \dfrac{2.00}{1} \approx 2\\\\\text{The relative amounts of O per gram of V are in the ratio }\boxed{\mathbf{\dfrac{2}{1}}}$

For example. the compounds might be VO₂ and VO.

7 0

3 years ago

Which methods could you use to calculate the y-coordinate of the midpoint of a vertical line segment with endpoints at (0, 0) an

maria [59]

One of the most convenient ways of finding or calculating for the midpoint of the line segment is to get the average of the coordinates of the points.

Average of abscissa = (0 + 0) / 2 = 0

Average of ordinate = (0 + 15) / 2 = 7.5

Hence, the midpoint of the line segment is equal to (0, 15/2).

The value of the y-intercept is 15/2.

8 0

3 years ago

Read 2 more answers

How many joules of energy are required to melt 423g of water at 0°C?

Mkey [24]

The amount of heat needed to melt 423 g of water at 0°C is 141282 J

The heat required to melt water can be obtained by using the following formula:

Q is the heat required.

L is the latent heat of fusion (334 J/g)

m is the mass.

With the above formula, we can obtain the heat required to melt the water as illustrated below:

Mass of water (m) = 423 g

Latent heat of fusion (L) = 334 J/g

<h3>Heat (Q) required =? </h3>

Q = mL

Q = 423 × 334

Therefore, the amount of heat needed to melt 423 g of water at 0°C is 141282 J

Learn more: brainly.com/question/17084080

6 0

3 years ago

Read 2 more answers

A galvanic cell consists of one half-cell that contains Ag(s) and Ag+(aq), and one half-cell that contains Cu(s) and Cu2+(aq). W

Agata [3.3K]

Answer : The 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

Explanation :

Galvanic cell : It is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy. It is also known as the voltaic cell or electrochemical cell.

In the galvanic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

We are taking the value of standard reduction potential form the standard table.

$E^0_{[Ag^{+}/Ag]}=+0.80V$

$E^0_{[Cu^{2+}/Cu]}=+0.34V$

In this cell, the component that has lower standard reduction potential gets oxidized and that is added to the anode electrode. The second forms the cathode electrode.

The balanced two-half reactions will be,

Oxidation half reaction (Anode) : $Cu(s)\rightarrow Cu^{2+}(aq)+2e^-$

Reduction half reaction (Cathode) : $Ag^{+}(aq)+e^-\rightarrow Ag(s)$

Thus the overall reaction will be,

$Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)$

From this we conclude that, 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

Hence, the 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

7 0

4 years ago