How many atoms of each element are in one molecule of this product.

A tank contains 200 gallons of water in which 300 grams of salt is dissolved. A brine solution containing 0.4 kilograms of salt

Nadusha1986 [10]

Answer:

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

Explanation:

given amount of salt at time t is A(t)

initial amount of salt =300 gm =0.3kg

=>A(0)=0.3

rate of salt inflow =5*0.4= 2 kg/min

rate of salt out flow =5*A/(200)=A/40

rate of change of salt at time t , dA/dt= rate of salt inflow- ratew of salt outflow

$dA/dt=2-(A/40)\\\\dA=2dt-(A/40)dt\\\\dA+(A/40)dt=2dt$

integrating factor

$=e^{\int\limits (1/40) \, dt}$

integrating factor $=e^{(1/40)t}$

multiply on both sides by $=e^{(1/40)t}$

$dAe^{(1/40)t}+(A/40)e^{(1/40)t} dt =2e^{(1/40)t}t\\\\(Ae^{(1/40)t})=2e^{(1/40)t}t$

integrate on both sides

$\int\limits(Ae^{(1/40)t})=\int\limits2e^{(1/40)t}dt\\\\(Ae^{(1/40)t})=2*40e^{(1/40)t}+C\\\\A=80+(C/e^{(1/40)t})\\\\A(0)=0.3\\\\0.3=80+(C/e^{(1/40)t}^*^0)\\\\0.3=80+(C/1)\\\\C=0.3-80\\\\C=-79.7\\\\A(t)=80-(79.7/e^{(1/40)t})$

b)

after long period of time means t - > ∞

${t \to \infty}\\\\ \lim_{t \to \infty} A_t \\\\ \lim_{t \to \infty} (80)-(79/{e^{(1/40)t}}\\\\=80-(0)\\\\=80$

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

6 0

2 years ago

Kc for the reaction N2O4 <=> 2NO2 is 0.619 at 45 degrees C If 50.0g of N2O4 is introduced into an empty 2.10L container, w

Nadya [2.5K]

Answer:

p(N2O4) = 0.318 atm

p(NO2) = 7.17 atm

Explanation:

Step 1: Data given

Kc = 0.619

Temperature = 45.0 °C

Mass of N2O4 = 50.0 grams

Volume = 2.10 L

Molar mass N2O4 = 92.01 g/mol

Step 2: The balanced equation

N2O4 ⇔ 2NO2

Step 3: Calculate moles N2O4

Moles N2O4 = 50.0 grams / 92.01 g/mol

Moles N2O4 = 0.543 moles

Step 4: The initial concentration

[N2O4] = 0.543 moles/2.10 L = 0.259 M

[NO2]= 0 M

Step 5: Calculate concentration at the equilibrium

For 1 mol N2O4 we'll have 2 moles NO2

[N2O4] = (0.259 -x)M

[NO2]= 2x

Step 6: Calculate Kc

Kc = 0.619= [NO2]² / [N2O4]

0.619 = (2x)² / (0.259-x)

0.619 = 4x² / (0.259 -x)

x = 0.1373

Step 7: Calculate concentrations

[N2O4] = (0.259 -x)M = 0.1217 M

[NO2]= 2x = 0.2746 M

Step 8: The moles

Moles = molarity * volume

Moles N2O4 = 0.1217 M * 2.10 = 0.0256 moles

Moles NO2 = 0.2746 M * 2.10 = 0.577 moles

Step 9: Calculate partial pressure

p*V = n*R*T

⇒ with p = the partial pressure

⇒ with V = the volume = 2.10 L

⇒ with n = the number of moles

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 45 °C = 318 K

p = (nRT)/V

p(N2O4) = (0.0256 *0.08206 * 318)/ 2.10

p(N2O4) = 0.318 atm

p(NO2) = (0.577 *0.08206 * 318)/ 2.10

p(NO2) = 7.17 atm

6 0

3 years ago

Two different atoms have four protons each and the same mass. However, one has a positive charge while the other is neutral. Des

seraphim [82]

The possible number and location of all subatomic are one of them is electrically neutral, while the other has a stable electronic configuration.

<h3>What are subatomic particles?</h3>

Subatomic particles are those particles that are present inside the atoms. They are electron, neutron, and proton. They are charged particles, protons are positively charged, electrons are negatively charged and neutrons are neutral.

The protons and electrons totally contribute to the atomic mass of the elements.

Thus, the subatomic particles are electrically neutral and stable to electronic configurations.

To learn more about subatomic particles, refer to the below link:

brainly.com/question/13303285

#SPJ1

8 0

1 year ago

An example of a physical property of an element is the element's ability to

FromTheMoon [43]

Answer:

dissolve

Explanation:

When it dissolves , no chemicals are formed and it is considered as a physical property.

4 0

2 years ago

Read 2 more answers

Consider the reaction for the decomposition of hydrogen disulfide: 2H2S(g)⇌2H2(g)+S2(g), Kc = 1.67×10−7 at 800∘C A 0.500 L react

trasher [3.6K]

Answer:

Molar concentration of S₂ is 1.77×10⁻⁶M

Explanation:

For the reaction:

2H₂S(g) ⇄ 2H₂(g) + S₂(g)

The equilibirum constant, K, is defined as:

$K = \frac{[S_2][H_2]^2}{[H_2S]^2}$ <em>(1)</em>

Concentrations in equilibirum are:

[H₂S] : 0,163/0.500L - X

[H₂] : 0,0500/0.500L + X

[S₂] : X

Replacing the concentrations and the equilibrium value in (1):

$K = \frac{[X][0.1+X]^2}{[0326-X]^2}$

1.67x10⁻⁷ = X (X² + 0.2X + 0.01) / (X² -0.652X + 0.106)

1.67x10⁻⁷X² - 1.09x10⁻⁷X + 1.77x10⁻⁸ = X³ + 0.2X² + 0.01X

0 = X³ + 0.2X² + 0.01X - 1.77x10⁻⁸

Solving for X:

X = 1.77×10⁻⁶

As [S₂] = X, <em>molar concentration of S₂ is 1.77×10⁻⁶M</em>

I hope it helps!

5 0

3 years ago