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3 years ago

A solution is prepared by dissolving a

Chemistry

2 answers:

goldfiish [28.3K]3 years ago

7 0

Answer:

119g

Explanation:

Just multiply

2.38% of 500 =

119 g

Levart [38]3 years ago

4 0

solution = solute + solvent

% mass of solute:

$\tt \dfrac{mass~solute}{mass~solution}\times 100\%$

for % mass of solute = 2.38%, then mass of solute:

$\tt =2.38\%\times 500~g=\boxed{\bold{11.9~g}}$

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A 50.0 mL solution of 0.133 M KOH is titrated with 0.266 M HCl . Calculate the pH of the solution after the addition of each of

shtirl [24]

Answer:Calculate The PH Of The Solution After The Addition Of The Following Amounts Of HCl. PLEASE HELP! SHOW ALL STEPS!! This problem has been solved!

Explanation:

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3 years ago

Given the two reactions PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq), K3 = 1.84×10−10, and AgCl(aq)⇌Ag+(aq)+Cl−(aq), K4 = 1.14×10−4, what is the

Nimfa-mama [501]

Answer: The value of equilibrium constant for reaction is, $1.42\times 10^{-2}$

Explanation:

The given chemical equations are:

(1) $PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$ ; $K_3=1.84\times 10^{-10}$

(2) $AgCl(aq)\rightleftharpoons Ag^{+}(aq)+Cl^-(aq)$ ; $K_4=1.14\times 10^{-4}$

Now we have to calculate the equilibrium constant for chemical equation as:

$PbCl_2(aq)+2Ag^{+}(aq)\rightleftharpoons 2AgCl(aq)+Pb^{2+}(aq)$ ; $K=?$

We are reversing reaction 2 and multiplying reaction 2 by 2 and then adding both reaction, we get the final reaction.

The equilibrium constant for the reverse reaction will be the reciprocal of that reaction.

If the equation is multiplied by a factor of '2', the equilibrium constant of that reaction will be the square of the equilibrium constant.

If we are adding equations then the equilibrium constants will be multiplied.

The value of equilibrium constant for reaction is:

$K=(\frac{1}{K_4})^2\times K_3$

Now put all the given values in this expression, we get:

$K=(\frac{1}{1.14\times 10^{-4}})^2\times (1.84\times 10^{-10})$

$K=1.42\times 10^{-2}$

Hence, the value of equilibrium constant for reaction is, $1.42\times 10^{-2}$

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3 years ago

A rectangle solid of unknown density is 5 m long two meters high and 4 meters wide. The mass of this solid is 300 grams. Find th

aksik [14]

D= M/V so, D=300/40 wich is 7.5 so D= 7.5g/m3

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3 years ago

Main job is to sort and package proteins and other substances in a plant cell

melomori [17]

Answer:

Main job of golgi bodies is to sort and package proteins and other substances in a plant cell.

Explanation:

Golgi bodies are also called post office of the cell because it modify and distribute proteins for the cell. First, proteins are made in the organelle of the cell i. e. endoplasmic reticulum. From here, it is send to the Golgi apparatus for modification. Golgi bodies add some special structures with the protein and this protein leaves golgi bodies which is used by the cell where it is needed.

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4 years ago

I need help with 1,2,3, and 4

Schach [20]

Answer:

Problem 1: 1.85atm

Problem 2: 110mL

Problem 3: 290 mL

Problem 4: 1.14 atm

Explanation:

Problem 1

1. Data

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

2. Formula

Since the temeperature is constant you can use Boyle's law for idial gases:

$PV=constant\\\\P_1V_1=P_2V_2$

3. Solution

Solve, substitute and compute:

$P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2$

$P_2=3.25atm\times755mL/1325mL=1.85atm$

Problem 2

1. Data

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

2. Formula

You assume that the temperature does not change, and then can use Boyl'es law again.

$P_1V_1=P_2V_2$

3. Solution

This time, solve for V₂:

$P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2$

Substitute and compute:

$V_2=548mmHg\times 125mL/625mmHg=109.6mL$

You must round to 3 significant figures:

$V_2=110mL$

Problem 3

1. Data

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

2. Formula

At constant pressure, Charle's law states that volume and temperature are inversely related:

$V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

The temperatures must be in absolute scale.

3. Solution

a) Convert the temperatures to kelvins:

T₁ = 25 + 273.15K = 298.15K

T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

$\dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml$

You must round to two significant figures: 290 ml

Problem 4

1. Data

a) P = 865mmHg

b) Convert to atm

2. Formula

You must use a conversion factor.

1 atm = 760 mmHg

Divide both sides by 760 mmHg

$\dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}$

3. Solution

Multiply 865 mmHg by the conversion factor:

$865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer$

3 0

3 years ago