Please help find out code?

A long spring is attached to a wall at one end. A pulse is sent down the spring. When the pulse reaches the wall it bounces back

slega [8]

Answer: they will meet and make a biger wave then seperat

Explanation: When two or more waves meet, they interact with each other. The interaction of waves with other waves is called wave interference. Wave interference may occur when two waves that are traveling in opposite directions meet. The two waves pass through each other, and this affects their amplitude.

8 0

3 years ago

What is the molar mass of an unknown gas with a density of 2.00 g/L at 1.00 atm and 25.0 °C?

soldier1979 [14.2K]

Answer:

Explanation:Explanation:

Your starting point here will be the ideal gas law equation

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

P

V

=

n

R

T

a

∣

−−−−−−−−−−−−−−−

, where

P

- the pressure of the gas

V

- the volume it occupies

n

- the number of moles of gas

R

- the universal gas constant, usually given as

0.0821

atm

⋅

L

mol

⋅

K

T

- the absolute temperature of the gas

Now, you will have to manipulate this equation in order to find a relationship between the density of the gas,

ρ

, under those conditions for pressure and temperature, and its molar mass,

M

.

You know that the molar mass of a substance tells you the mass of exactly one mole of that substance. This means that for a given mass

m

of this gas, you can express its molar mass as the ratio between

m

and

n

, the number of moles it contains

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

M

=

m

n

a

∣

−−−−−−−−−−−−−

(

1

)

Similarly, the density of the substance tells you the mass of exactly one unit of volume of that substance.

This means that for the mass

m

of this gas, you can express its density as the ratio between

m

and the volume it occupies

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

ρ

=

m

V

a

∣

−−−−−−−−−−−

(

2

)

Plug equation

(

1

)

into the ideal gas law equation to get

P

V

=

m

M

⋅

R

T

Rearrange to get

P

V

⋅

M

=

m

⋅

R

T

P

⋅

M

=

m

V

⋅

R

T

M

=

m

V

⋅

R

T

P

Finally, use equation

(

2

)

to write

M

=

ρ

⋅

R

T

P

Convert the temperature of the gas from degrees Celsius to Kelvin then plug in your values to find

M

=

1.02

g

L

⋅

0.0821

atm

⋅

L

mol

⋅

K

⋅

(

273.15

+

37

)

K

0.990

atm

M

=

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

26.3 g mol

−

1

a

∣

−−−−−−−−−−−−−−−−

I'll leave the answer rounded to three

7 0

3 years ago

What is the temperature of 4.5 moles of a gas that occupies 50. mL at 1.35 atm?

devlian [24]

Answer:

$T = 0.182$ Kelvin

Explanation:

As we know that

$PV = nRT\\$

Where P is the pressure in atmospheric pressure

T is the temperature in Kelvin

R is the gas constant

V is the volume in liters

$R = 0.08206$

Substituting the given values in above equation, we get -

$1.35 * \frac{50}{1000}= 4.5 * 0.08206* T\\$

On rearranging, we get

$T = \frac{1.35*50}{1000*0.08206*4.5} \\$

$T = 0.182$ Kelvin

7 0

3 years ago

Calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting tota

Ghella [55]

Answer:

(a) Δ $S_{sys}$ = 2.881 J/K; Δ $S_{sur}$ = -2.881 J/K; total change in entropy = 0

(b)Δ $S_{sys}$ = 2.881 J/K; Δ $S_{sur}$ = 0 ; total change in entropy = 2.881 J/K

(c) Δ $S_{sys}$ = 0 ; Δ $S_{sur}$ = 0 ; total change in entropy = 0

Explanation:

In the given problem, we need to calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting total change in entropy, when a sample of nitrogen gas of mass 14 g at 298 K and 1.00 bar doubles its volume. We have the following variable:

mass (m) = 14 g

Temperature = 298 K

Pressure = 1.00 bar

Initial volume = $V_{1}$

Final volume = $V_{2}$ = $2V_{1}$

(a) Change in entropy of the system Δ $S_{sys}$ = $nRIn\frac{V_{2} }{V_{1} }$

where R = 8.314 J/(mol*K)

n = number of moles = mass/molar mass = 14/ 28 = 0.5 moles

Δ $S_{sys}$ = 0.5*8.314*ln2 = 2.881 J/K

Change in entropy of the surrounding Δ $S_{sur}$ = -2.881 J/K

Therefore, for a reversible process, the total change in entropy = Δ $S_{sys}$ +Δ $S_{sur}$ = 2.881 - 2.881 = 0

(b) Because entropy is a state function, we use the same procedure as in part (a). Thus, Δ $S_{sys}$ = 2.881 J/K

Since surrounding does not change in this process Δ $S_{sur}$ = 0.

total change in entropy = Δ $S_{sys}$ +Δ $S_{sur}$ = 2.881 - 0 = 2.88 J/K

(c) For an adiabatic reversible expansion, q(rev) = 0, thus:

Δ $S_{sys}$ = 0

Since heat energy is not transferred from the system to the surrounding

Δ $S_{sur}$ = 0

total change in entropy = Δ $S_{sys}$ +Δ $S_{sur}$ = 0

6 0

3 years ago

Use the half-reaction method to balance the equation for the conversion of ethanol to acetic acid in acid solution:CH₃CH₂OH + Cr

garik1379 [7]

Balanced chemical equation is 3CH₃CH₂OH + 2Cr₂O₇²⁻ + 16H⁺ → 3CH₃COOH + 4Cr³⁺ + 11H₂O.

<h3>What is Balanced Chemical Equation ?</h3>

The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.

The chemical equation

CH₃CH₂OH + Cr₂O₇²⁻ → CH₃COOH + Cr³⁺

First assign the oxidation number for each atom in the equation.

$\overset{+3}{C}\overset{+1}{H_3} \overset{-1}{C} \overset{+1}{H_2} \overset{-2}{O} \overset{+1}{H} + \overset{+6}{Cr_2} \overset{-2}{O_7} + \overset{+1}{H} \rightarrow \overset{+3}{C}\overset{+1}{H_3} \overset{+3}{C}\overset{-2}{O} \overset{-2}{O} \overset{+1}{H} + \overset{+3}{Cr}$

Oxidation: C₂H₆O → C₂H₄O₂ + 4e⁻

Reduction: Cr₂O₇ + 6e⁻ → 2Cr⁺³

Now, balance the charge

Oxidation: C₂H₆O → C₂H₄O₂ + 4e⁻ + 4H⁺

Reduction: Cr₂O₇ + 6e⁻ + 14H⁺ → 2Cr⁺³

Now balance the oxygen atoms

Oxidation: C₂H₆O + H₂O → C₂H₄O₂ + 4e⁻ + 4H⁺

Reduction: Cr₂O₇ + 6e⁻ + 14H⁺ → 2Cr⁺³ + 7H₂O

Now, make electron gain equivalent to lost

Oxidation: C₂H₆O + H₂O → C₂H₄O₂ + 4e⁻ + 4H⁺ } × 3

Reduction: Cr₂O₇ + 6e⁻ + 14H⁺ → 2Cr⁺³ + 7H₂O } × 2

Now,

Oxidation: 3C₂H₆O + 3H₂O → 3C₂H₄O₂ + 12e⁻ + 12H⁺

Reduction: 2Cr₂O₇ + 12e⁻ + 28H⁺ → 4Cr⁺³ + 14H₂O

Now, add the both equations

3CH₃CH₂OH + 2Cr₂O₇²⁻ + 16H⁺ → 3CH₃COOH + 4Cr³⁺ + 11H₂O

Thus from the above conclusion we can say that the balanced chemical equation is 3CH₃CH₂OH + 2Cr₂O₇²⁻ + 16H⁺ → 3CH₃COOH + 4Cr³⁺ + 11H₂O.

Learn more about the Balanced chemical equation here: brainly.com/question/26694427

#SPJ4

7 0

1 year ago