Draw Molecular diagrams of solid, liquid, gas and plasma phases of matter.

What is the lightest particle in the following 1)electron 2)proton 3)neutron 4)photon

Anuta_ua [19.1K]

Answer:

1) Electron

Explanation:

It carries a negative charge of 1.602176634 × 10−19 coulomb, which is considered the basic unit of electric charge. The rest mass of the electron is 9.1093837015 × 10−31 kg, which is only 1/1,836the mass of a proton.

5 0

2 years ago

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Which states of matter only appear in the hydrosphere

Oliga [24]

I would say the answer is liquids

7 0

2 years ago

"acid is responsible for the odor in rancid butter. a solution of 0.25 m butyric acid has a ph of 2.71. what is the ka for"

Salsk061 [2.6K]

Answer:- The Ka for the acid is $1.53*10^-^5$ .

Solution:- In general, monoprotic acid could be represented by HA. The dissociation equation for the ionization of HA is written as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

Now, we make the ice table for this equation as:

HA(aq)\rightarrow H^+(aq) + A^-(aq)

I 0.25 0 0

C -X +X +X

E (0.25 - X) X X

where, I stands for initial concentration, C stands for change in concentration and E stands for equilibrium concentration.

X is the change in concentration and from ice table it's same as the concentration of hydrogen ion that is calculated from given pH.

$Ka = [H^+][A^-]\frac{1}{HA}$

Where, Ka is the acid ionization constant. Let's plug in the values.

$Ka = \frac{X^2}{0.25-X}$

Let's calculate the value of X first using the equation:

$pH = -log[H^+]$ [/tex]

on taking antilog ob above equation we get:

$[H^+]=10^-^p^H$

$[H^+]=10^-^2^.^7^1$

$[H^+]$ = 0.00195

So, X = 0.001195

Let's plug in this value of X in the equation:-

$Ka=\frac{(0.00195)^2}{0.25-0.00195}$

$Ka=1.53*10^-^5$

So, the value of Ka for butyric acid is $1.53*10^-^5$ .

8 0

3 years ago

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An exhaled air bubble underwater at 290.

Vinil7 [7]

The answer for the following problem is mentioned below.

Therefore the final volume of the gas is 52.7 ml.

Explanation:

Given:

Initial pressure ( $P_{1}$ ) = 290 kPa

Final pressure ( $P_{2}$ ) = 104 kPa

Initial volume ( $V_{1}$ ) = 18.9 ml

To find:

Final volume ( $V_{2}$ )

We know;

From the ideal gas equation;

P × V = n × R × T

where;

P represents the pressure of the gas

V represents the volume of gas

n represents the no of the moles

R represents the universal gas constant

T represents the temperature of the gas

So;

P × V = constant

P ∝ $\frac{1}{V}$

From the above equation;

$\frac{P_{1} }{P_{2} } = \frac{V_{2} }{V_{1} }$

$P_{1}$ represents the initial pressure of the gas

$P_{2}$ represents the final pressure of the gas

$V_{1}$ represents the initial volume of the gas

$V_{2}$ represents the final volume of the gas

Substituting the values of the above equation;

$\frac{290}{104}$ = $\frac{V_{2} }{18.9}$

$V_{2}$ = 52.7 ml

Therefore the final volume of the gas is 52.7 ml.

6 0

2 years ago

Since helium is lighter than air, it is difficult to measure the mass of a sample, so to find out the mass of helium in a birthd

d1i1m1o1n [39]

Answer:

0.7457 g is the mass of the helium gas.

Explanation:

Given:

Pressure = 3.04 atm

Temperature = 25.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (25.0 + 273.15) K = 298.15 K

Volume = 1.50 L

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

3.04 atm × 1.50 L = n × 0.0821 L.atm/K.mol × 298.15 K

⇒n = 0.1863 moles

Molar mass of helium = 4.0026 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.1863\ mole= \frac{Mass}{4.0026\ g/mol}$

$Mass_{He}= 0.7457\ g$

0.7457 g is the mass of the helium gas.

6 0

2 years ago