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GarryVolchara [31]

3 years ago

6

If you dilute 19.0 mL of the stock solution to a final volume of 0.310 L , what will be the concentration of the diluted solutio

n?

1 answer:

Dahasolnce [82]3 years ago

6 0

Answer:

$M_2=0.613M_1$

Explanation:

$M_1$ = Concentration of stock solution

$M_2$ = Concentration of solution

$V_1$ = Volume of stock solution = 19 mL

$V_2$ = Volume of solution = 0.31 L= 310 mL

We have the relation

$M_1V_1=M_2V_2\\\Rightarrow M_2=\dfrac{M_1V_1}{V_2}\\\Rightarrow M_2=\dfrac{M_119}{310}\\\Rightarrow M_2=M_1\times\dfrac{19}{310}\\\Rightarrow M_2=0.613M_1$

$\boldsymbol{\therefore M_2=0.613M_1}$

The concentration of the diluted solution will be 0.613 times the concentration of the stock solution.

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Consider the following reaction between calcium oxide and carbon dioxide: CaO(s)+CO2(g)→CaCO3(s) A chemist allows 14.4 g of CaO

sweet-ann [11.9K]

Answer:

Theoretical yield =26.03 g

Percent yield = 87%

Limiting reactant = CaO

Explanation:

Given data:

Mass of CaO = 14.4 g

Mass of CO₂ = 13.8 g

Actual yield of CaCO₃ = 22.6 g

Theoretical yield = ?

Percent yield = ?

Limiting reactant = ?

Solution:

Chemical equation:

CaO + CO₂ → CaCO₃

Number of moles of CaO:

Number of moles = Mass /molar mass

Number of moles = 14.4 g / 56.1 g/mol

Number of moles = 0.26 mol

Number of moles of CO₂:

Number of moles = Mass /molar mass

Number of moles = 13.8 g / 44 g/mol

Number of moles = 0.31 mol

Now we will compare the moles of CO₂ and CaO with CaCO₃ .

CO₂ : CaCO₃

1 : 1

0.31 : 0.31

CaO : CaCO₃

1 : 1

0.26 : 0.26

The number of moles of CaCO₃ produced by CaO are less it will be limiting reactant.

Mass of CaCO₃: Theoretical yield

Mass of CaCO₃ = moles × molar mass

Mass of CaCO₃ =0.26 mol × 100.1 g/mol

Mass of CaCO₃ = 26.03 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 22.6 g/ 26.03 g × 100

Percent yield = 0.87× 100

Percent yield = 87%

Limiting reactant:

The number of moles of CaCO₃ produced by CaO are less it will be limiting reactant.

7 0

3 years ago

Oxygen atoms have six outer electrons<br><br> Write the symbol for an oxide ion.

Shkiper50 [21]

Answer:

$O^{2-}$ is the symbol for an oxide ion

7 0

3 years ago

Air at 7S°F and14.6 though a cfm(cubic ft per minute) and the The flow rate is 48000 psia flows though a rectangular duct of Ix2

IgorLugansk [536]

Explanation:

The given data is as follows.

Air is at $75^{o}F$ and 14.6 psia.

$\varepsilon$ = 0.00015 ft, Flow rate, (Q) = 48000 $ft^{3}/m$

(a) Formula to calculate hydraulic radius $(r_{H})$ is as follows.

$r_{H} = \frac{\text{free flow area}}{\text{wet perimeter}}$

= $\frac{2 \times 1}{2(1) + 2(2)}$

= $\frac{1}{3}$ ft

Formula for equivalent diameter is as follows.

$D_{eq} = 4 \times r_{H}$

= $4 \times \frac{1}{3} ft$

= $\frac{4}{3}$ ft

(b) Formula for velocity floe is as follows.

Q = VA

V = $\frac{Q}{A}$

= $\frac{48000}{2 \times 1}$ ft/min

= 24000 ft/min

(c) Formula to calculate Reynold's number is as follows.

$R_{e}$ = $\frac{D \times V \times \rho}{\mu}$

= $\frac{\frac{4}{3} \times 24000 \times 0.0744}{0.0443}$ (as $\rho = 0.0744 lb/ft^{3}$ and $\mu$ = 0.0443 lb/ft. hr)

= 53742.66 hr/min

As 1 hr = 10 min. So, $53742.66 hr/min \times \frac{60 min}{1 hr}$

= 3224559.6

(d) Formula to calculate pressure drop $(\Delta P)$ is as follows.

$\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}$

Putting the given values into the above formula as follows.

$\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}$

= $\frac{4 \times 0.00015 \times 100 \times 0.0744 \times (24000)^{2}}{2 \times \frac{4}{3} \times {4}{3}}$

= 6.238 $lb/ft^{2}$

5 0

3 years ago

What is the rate constant of a reaction if rate = 1 x 10-2 (mol/L)/s, [A] is 2 M,

boyakko [2]

The rate constant of a reaction : 8.3 x 10⁻⁴

<h3>Further explanation</h3>

Given

rate = 1 x 10⁻² (mol/L)/s, [A] is 2 M, [B] is 3 M, m = 2, and n = 1

Required

the rate constant

Solution

For aA + bB ⇒ C + D

Reaction rate can be formulated:

$\large{\boxed{\boxed{\bold{r~=~k.[A]^a[B]^b}}}$

the rate constant : k =

$\tt k=\dfrac{rate}{[A]^m[B]^n}\\\\k=\dfrac{1.10^{-2}}{2^2\times 3^1}\\\\k=8.3\times 10^{-4}$

8 0

2 years ago

Read 2 more answers

Explain wether or not a reaction takes place when when bromine water is added to sodium chloride solution

CaHeK987 [17]

Answer:

When chlorine (as a gas or dissolved in water) is added to sodium bromide solution, the chlorine takes the place of the bromine. Because chlorine is more reactive than bromine, it displaces bromine from sodium bromide. The solution turns brown. ... The chlorine has gone to form sodium chloride.

5 0

2 years ago