7.25 x 10-6 What is the scientific notation

A brine solution of salt flows at a constant rate of 9 L/min into a large tank that initially held 100 L of brine solution in w

Valentin [98]

Answer:

a) C = 0.02 - ((e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾)/9)

b) The concentration of salt in the tank attains the value of 0.01 kg/L at time, t = 0.0713 min = 4.28s

Explanation:

Taking the overall balance, since the total Volume of the setup is constant, then flowrate in = flowrate out

Let the concentration of salt in the tank at anytime be C

Let the Concentration of salt entering the tank be Cᵢ

Let the concentration of salt leaving the tank be C₀ = C (Since it's a well stirred tank)

Let the flowrate in be represented by Fᵢ

Let the flowrate out = F₀ = F

Fᵢ = F₀ = F

Then the component balance for the salt

Rate of accumulation = rate of flow into the tank - rate of flow out of the tank

dC/dt = FᵢCᵢ - FC

Fᵢ = 9 L/min, Cᵢ = 0.02 kg/L, F = 9 L/min

dC/dt = 0.18 - 9C

dC/(0.18 - 9C) = dt

∫ dC/(0.18 - 9C) = ∫ dt

(-1/9) In (0.18 - 9C) = t + k

In (0.18 - 9C) = -9t - 9k

-9k = K

In (0.18 - 9C) = K - 9t

At t = 0, C = 0.1/100 = 0.001 kg/L

In (0.18 - 9(0.001)) = K

In 0.171 = K

K = - 1.766

So, the equation describing concentration of salt at anytime in the tank is

In (0.18 - 9C) = -1.766 - 9t

In (0.18 - 9C) = - (9t + 1.766)

0.18 - 9C = e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾

9C = 0.18 - (e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾)

C = 0.02 - ((e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾)/9)

b) when C = 0.01 kg/L

0.01 = 0.02 - ((e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾)/9)

0.09 = (e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾)

- (9t + 1.766) = In 0.09

- (9t + 1.766) = -2.408

(9t + 1.766) = 2.408

9t = 2.408 - 1.766 = 0.642

t = 0.642/9 = 0.0713 min = 4.28s

4 0

3 years ago

A two-liter soft drink bottle can withstand apressure of

larisa86 [58]

Explanation:

According to the ideal gas equation, PV = nRT.

where, P = pressure, V = volume

n = no. of moles, R = gas constant

T = temperature

Also, density is equal to mass divided by volume. And, no. of moles equals mass divided by molar mass.

Therefore, then formula for ideal gas could also be as follows.

P = $\frac{mass}{volume \times molar mass} \times RT$

or, P = $\frac{density}{\text{molar mass}} \times RT$

Since, density is given as 0.789 g/ml which is also equal to 789 g/L (as 1000 mL = 1 L). Hence, putting the given values into the above formula as follows.

P = $\frac{density}{\text{molar mass}} \times RT$

= $\frac{789 g/l}{46.06 g/mol} \times 0.0821 L atm/mol K \times 373 K$

= 525 atm

As two-liter soft drink bottle can withstand a pressure of 5 atm and the value of calculated pressure is 525 atm which is much greater than 5 atm.

Therefore, the soft drink bottle will obviously explode.

4 0

3 years ago

How would you prepare 250 mL of a 0.100 M solution of fluoride ions from solid CaF2?

andrey2020 [161]

In 250 mL of volumetric flask add 0.975875 grams of $CaF_2$ and dissolve it in the 250 mL of water.

Given:

The solid of calcium fluoride.

To prepare:

The 250 mL solution of 0.100 M of fluoride ions from solid calcium fluoride.

Method:

Molarity of the fluoride ion solution needed = M = 0.100 M

The volume of the fluoride ion solution needed = V = 250 mL

$1 mL = 0.001L\\V=250 mL=250\times 0.001 L=0.250 L$

The moles of fluoride ion needed = n

According to the definition of molarity:

$M=\frac{n}{V}\\0.100M=\frac{n}{0.250 L}\\n=0.100M\times 0.250 L=0.025 mol$

Moles of fluoride ion = 0.025 mol

We know that solid calcium fluoride dissolves in water to give calcium ions and fluoride ions.

$CaF_2(s)\rightarrow Ca^{2+}(aq)+2F^-(aq)$

According to reaction, 2 moles of fluoride ions are obtained from 1 mole of calcium fluoride, then 0.025 moles of fluoride ions will be obtained from:

$=\frac{1}{2}\times 0.025 mol=0.0125 \text{mol of } CaF_2$

Moles of calcium fluoride = 0.0125 mol

Mass of calcium fluoride needed to prepare the solution :

$=0.0125 mol\times 78.07 g/mol=0.975875 g$

Preparation:

Weight 0.975875 grams of calcium fluoride
Add weighed calcium fluoride to a volumetric flask of the labeled volume of 250 mL.
Now add a small amount of water to dissolve the calcium fluoride completely.
After this add more water up to the mark of the volumetric flask of volume 250 mL.

Learn more about molarity of solution ere:

brainly.com/question/10053901?referrer=searchResults

brainly.com/question/10270173?referrer=searchResults

3 0

3 years ago

If 3.8 moles of zinc metal react with 6.5 moles of silver nitrate, how many moles of silver metal can be formed, and how many mo

bulgar [2K]

Answer: 6.5 moles of silver metal is formed in the given chemical reaction. The moles of excess reagent left are 0.55 moles.

Explanation:

To calculate the moles of silver formed and the moles of excess reagent left after the reaction, we need to balance the equation first and need to find the limiting and excess reagent.

The balanced chemical equation is:

$Zn+2AgNO_3\rightarrow Zn(NO_3)_2+2Ag$

By Stoichiometry:

2 moles of Silver nitrate reacts with 1 mole of Zinc metal

So, 6.5 moles of silver nitrate will react with = $\frac{1}{2}\times 6.5=3.25moles$ of zinc metal

The required amount of zinc metal is less than the given amount of zinc metal, hence, it is considered as an excess reagent.

Therefore, silver nitrate is the limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of silver nitrate produces 2 moles of silver metal

So, 6.5 moles of silver nitrate will produce = $\frac{2}{2}\times 6.5=6.5moles$ of silver metal.

Number of moles of excess reagent left after the completion of reaction = (3.8 - 3.25)moles = 0.55 moles

Hence, 6.5 moles of silver metal is formed in the given chemical reaction. The moles of excess reagent left are 0.55 moles.

7 0

3 years ago

Answer the question

Ostrovityanka [42]

It is an adaptation

3 0

3 years ago