All categories

3 years ago

Can some one please help to understand this problem??

Chemistry

1 answer:

Rama09 [41]3 years ago

6 0

Answer:

mercury, magnesium, water and gasoline.

Explanation:

Density:

Density is equal to the mass of substance divided by its volume.

Units:

SI unit of density is Kg/m3.

Other units are given below,

g/cm3, g/mL , kg/L

The liquid with higher density will sink in bottom while the liquid with lower density will float.

In given samples,

density of water = 1 g/mL

density of mercury = 13.6 g/mL

density of gasoline = 0.70 g/mL

density of magnesium = 1.7 g/mL

So on the basis of given density values mercury will sink in bottom then above it magnesium will present than water and above on these three gasoline will present.

You might be interested in

A fictitious element Z has an average atomic mass of 223.06 u.223.06 u. Element Z has two naturally occuring isotopes. The more

zhannawk [14.2K]

Answer:

221.37 u

Explanation:

The formula for the calculation of the average atomic mass is:

$Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})$

Given that:

For first isotope:

% = 65.51 %

Mass = 223.95 u

For second isotope:

Since the element has only 2 isotopes, so the percentage of second is 100 - first percentage.

% = 100 % - 65.51 % = 34.49 %

Let, Mass = x u

Given, Average Mass = 223.06 u

Thus,

$223.06=\frac {65.51}{100}\times {223.95}+\frac {34.49}{100}\times {x}$

Solving for x, we get that:

x = 221.37 u

<u>Thus mass of second isotope = 221.37 u</u>

5 0

3 years ago

A system of gas at low density has an initial pressure of 1.90 × 10 5 1.90×105 Pa and occupies a volume of 0.18 m³. The slow add

Bad White [126]

Answer:

ΔU = -6.2 × 10⁴ J

Explanation:

The system absorbs 965 J of heat, that is, q = 965 J.

The work (w) can be calculated using the following expression.

w = -P . ΔV

where,

P is the external pressure

ΔV is the change in the volume

w = - (1.90 × 10⁵ N/m²) × (0.51 m³ - 0.18 m³) = -6.3 × 10⁴ J

The change in the internal energy (ΔU) is:

ΔU = q + w = 965 J + (-6.3 × 10⁴ J) = -6.2 × 10⁴ J

8 0

3 years ago

An allylic carbocation is an example of a(n) _____ system. (select all that apply.)

DerKrebs [107]

An allylic carbocation is an example of a system that is:

delocalized
conjugated.

<h3>What is allylic carbocation?</h3>

An allylic carbocation is a resonance-stabilized carbocation with a formal charge of +1 on an allylic carbon in each of the two resonance forms.

Thus, it is correct to state that:

An allylic carbocation is an example of a system that is:

delocalized
conjugated.

Learn more about allylic carbocation:
brainly.com/question/16258022
#SPJ4

Full Question:

An allylic carbocation is an example of a(n) _______ system. (Select all that apply.)

- localized

- isolated

- delocalized

- conjugated

6 0

1 year ago

The ratio of atoms of potassium to ratio of atoms of oxygen is

DanielleElmas [232]

The ratio of atoms of potassium to ratio of atoms of oxygen is 2:1

5 0

3 years ago

Consider a solution that contains 60.0% r isomer and 40.0% s isomer. if the observed specific rotation of the mixture is –43.0°,

pychu [463]

The solution for this problem is:
Let x denote the specific rotation, R; andLet y denote the specific rotation, S = -x
Solution:60 x - 40 x/100 = - 43
20x = - 4300Divide both sides by 20The answer is:x = - 215 is the specific rotation of the pure r isomer.

5 0

3 years ago

Can some one please help to understand this problem??​

Can some one please help to understand this problem??