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3 years ago

Which choice best describes what would happen if a cube with a density of 0.2 g/mL were placed in a container of water?

Chemistry

1 answer:

Andrew [12]3 years ago

5 0

Answer:

A. It would float with about 80% of the cube below the surface of the water and 20% above the surface.

Explanation:

The choice that best describes what happens to cube of the given density value is that it would float with about 80% of the cube would be below the surface of the water and 20% above the surface.

Density is the mass per unit volume of a substance. The more mass a body has relative to volume, the great it's density. In short, density is directly proportional to mass and inversely related to volume.

The density of water is 1g/mL

If the density of the cube were to be the same with that of water, the substance will just mix up with water .

Here the density is less than that of water.

The density is 0.2g/mL

Therefore, 20% will stay afloat and 80% will be below the surface of the water.

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The answer to your question is HNO3

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Table salt is a refined salt containing about 97 to 99% sodium Chloride iodized salt containing potassium iodide is widely avail

Free_Kalibri [48]

Answer:

Explanation:

Sodium chloride is ionic compound. It is formed by the transfer of electron from one atom to the atom of another element.

Both bonded atoms have very large electronegativity difference. The atom with large electronegativity value accept the electron from other with smaller value of electronegativity.

The electronegativity of chlorine is 3.16 and for sodium is 0.93. There is large difference is present. That's why electron from sodium is transfer to the chlorine. Sodium becomes positive and chlorine becomes negative ion. Both atoms are bonded together electrostatic attraction occur between anion and cations.

Sodium atom have one valance electron by losing this one valance electron sodium atom get the complete octet. Chlorine atom has seven valance electrons and needed to lose seven valance electrons or to get one electron and thus complete the octet. It is very easy for chlorine atom to get one electrons instead of losing all seven electron. Thus when it react with sodium it gain the valance electron of sodium and form ionic compound.

That's why only one atom of sodium combine with one atom chlorine.

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3 years ago

Explain the change in mass that occurs when the following substances are separately heated in open crucibles

aleksandrvk [35]

Answer:

your answer is (a) Copper Metal

Explanation:

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3 years ago

You are given a 1.55 g mixture of calcium nitrate and calcium chloride. You dissolve this mixture in 20 mL of water and add an e

irina [24]

Answer:

13.4 (w/w)% of CaCl₂ in the mixture

Explanation:

All the Cl⁻ that comes from CaCl₂ (Calcium chloride) will be precipitate in presence of AgNO₃ as AgCl.

To solve this problem we must find the moles of AgCl = Moles of Cl⁻. As 2 moles of Cl⁻ are in 1 mole of CaCl₂ we can find the moles of CaCl₂ and its mass in order to find mass percent of calcium chloride in the original mixture.

<em>Moles AgCl - Molar mass: 143.32g/mol -:</em>

0.535g * (1mol / 143.32g) = 3.733x10⁻³ moles AgCl = Moles Cl⁻

<em>Moles CaCl₂:</em>

3.733x10⁻³ moles Cl⁻ * (1mol CaCl₂ / 2mol Cl⁻) = 1.866x10⁻³ moles CaCl₂

<em>Mass CaCl₂ -Molar mass: 110.98g/mol-:</em>

1.866x10⁻³ moles CaCl₂ * (110.98g/mol) = 0.207g of CaCl₂ in the mixture

That means mass percent of CaCl₂ is:

0.207g CaCl₂ / 1.55g * 100 =

<h3>13.4 (w/w)% of CaCl₂ in the mixture</h3>

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3 years ago

The activation energy of an uncatalyzed reaction is 95kJ/mol. The addition of a catalyst lowers the activation energy to 55kJ/mo

notka56 [123]

Answer:

a) at 25°C the rate of reaction increases by a factor of 1,027*10^7

b) at 25°C the rate of reaction increases by a factor of 1,777*10^5

Explanation:

using the Arrhenius equation

k= ko*e^(-Ea/RT)

where

k= reaction rate

ko= collision factor

Ea= activation energy

R= ideal gas constant= 8.314 J/mol*K

T= absolute temperature

for the uncatalysed reaction

k1= ko*e^(-Ea1/RT)

for the catalysed reaction

k2= ko*e^(-Ea2/RT)

dividing both equations

k2/k1= e^(-(Ea2-Ea1)/RT)

a) at 25°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,027*10^7

therefore at 25°C , k2/k1 = 1,027*10^6

b) at 125°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,777*10^5

therefore at 125°C , k2/k1 = 1,777*10^5

Note:

when the catalysts is incorporated, the catalysed reaction and the uncatalysed one run in parallel and therefore the real reaction rate is

k real = k1 + k2 = k2 (1+k1/k2)

since k2>>k1 → 1+k1/k2 ≈ 1 and thus k real ≈ k2

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3 years ago