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3 years ago

What is the concentration of an unknown Mg(OH)2 solution if it took an average of 15.4mL of

Chemistry

1 answer:

vova2212 [387]3 years ago

7 0

Answer:

0.077M

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2HCl + Mg(OH)2 —> MgCl2 + 2H2O

From the balanced equation above,

The mole ratio of the acid (nA) = 2

The mole ratio of the base (nB) = 1

Step 2:

Data obtained from the question.

Concentration of base Cb =...?

Volume of base (Vb) = 10mL

Concentration of acid (Ca) = 0.1M

Volume of acid (Va) = 15.4mL

Step 3:

Determination of the concentration of the base, Mg(OH)2.

The concentration of the base can be obtained as follow:

CaVa/CbVb = nA/nB

0.1 x 15.4 /Cb x 10 = 2/1

Cross multiply to express in linear form

Cb x 10 x 2 = 0.1 x 15.4

Divide both side by 10 x 2

Cb = (0.1 x 15.4) /(10 x 2)

Cb = 0.077M

Therefore, the concentration of the base, Mg(OH)2 is 0.077M

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A and b are two gases that are mixed together: 2.50 mol a is mixed with 0.850 mol b. if the final pressure of the mixture is 1.7

USPshnik [31]

Partial pressure of gas A is 1.31 atm and that of gas B is 0.44 atm.

The partial pressure of a gas in a mixture can be calculated as

Pi = Xi x P

Where Pi is the partial pressure; Xi is mole fraction and P is the total pressure of the mixture.

Therefore we have Pa = Xa x P and Pb = Xb x P

Let us find Xa and Xb

Χa = mol a/ total moles = 2.50/(2.50+0.85) = 2.50/3.35 = 0.746

Xb = mol b/total moles = 0.85/(2.50+0.85) = 0.85/3.35 = 0.254

Total pressure P is given as 1.75 atm

Pa = Xa x P = 0.746 x 1.75 = 1.31atm

Partial pressure of gas A is 1.31 atm

Pb = Xb x P = 0.254 x 1.75 = 0.44atm

Partial pressure of gas B is 0.44 atm.

Learn more about Partial pressure here:

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6 0

2 years ago

Answers to all of these

Paul [167]

Answer:

1. Percent composition of Al = 13.423 %

Percent composition of Zn = 28.02 %

Percent composition of Cl = 30.6 %

Percent composition of O = 41.3 %

3. The empirical formula is C₅O₁₆

4. Molecular Formula= P₄O₆

Explanation:

Part first :

Data Given

Formula of the Molecule = Al₂ (CrO₄)₃

% of Al₂ = ?

> First of all find the atomic masses of each component in a molecule

For Al₂ (CrO₄)₃ atomic masses are given below

Al = 27 g/mol

Cr = 52 g/mol

O = 16 g/mol

> Then find the total masses of each component

2 atoms of Al = 27 g/mol x 2

= 54 g/mol

3 atoms of Cr = 52 g/mol x 3

= 156 g/mol

12 atoms of O = 16 g/mol x 12

= 192 g/mol

> find total Molar Mass of Molecule:

Molar Mass of Al₂ (CrO₄)₃ = [27x2 + 52x3 + 16x12]

Molar Mass of Al₂ (CrO₄)₃ = 402

Now to find the mass percent of Al

Formula used to find the Mass percent of a component

Percent composition of Al = mass of Al in Molecula / molar mass of Al₂(CrO₄)₃ x 100%

Put the values

Percent composition of Al = 54 (g/mol) / 402 (g/mol) x 100%

Percent composition of Al = 13.423 %

_______________________________________

Part 2

Data Given

Formula of the Molecule = Zn(ClO₃)₂

% Zn = ?

% Cl = ?

% O = ?

> First of all find the atomic masses of each component in a molecule

For Zn(ClO₃)₂ atomic masses are given below

Zn = 65 g/mol

Cl = 35.5 g/mol

O = 16 g/mol

> Then find the total masses of each component

1 atoms of Zn= 65 g/mol x 1

= 65 g/mol

2 atoms of Cl = 35.5 g/mol x

= 71 g/mol

6 atoms of O = 16 g/mol x 6

= 96 g/mol

> find total Molar Mass of Molecule:

Molar Mass of Zn(ClO₃)₂ = [65x1 + 35.5x2 + 16x6]

Molar Mass of Zn(ClO₃)₂ = 232g/mol

Now to find the mass percent of of each component one by one

1. Formula used to find the mass percent of Zn

Percent composition of Zn= mass of Zn in Molecular / molar mass of Zn(ClO₃)₂ x 100%

Put the values

Percent composition of Zn = 65(g/mol) / 232 (g/mol) x 100%

Percent composition of Zn = 28.02 %

-------------------

2. Formula used to find the mass percent of Cl

Percent composition of Cl = mass of Cl in Molecular / molar mass of Zn(ClO₃)₂ x 100%

Put the values

Percent composition of Cl = 71 (g/mol) / 232 (g/mol) x 100%

Percent composition of Cl = 30.6 %

---------------------

3. Formula used to find the mass percent of O

Percent composition of O = mass of O in Molecular / molar mass of Zn(ClO₃)₂ x 100%

Put the values

Percent composition of O = 96 (g/mol) / 232 (g/mol) x 100%

Percent composition of O = 41.3 %

________________________________________

Part 3:

Data Given

Percentage of C = 27.3 %

Percentage of O = 72.7 %

Emperical Formula of the compound = ?

Solution:

So the compound has 27.3 % C and 72% O

First, find the mass of each of the elements in 100 g of the Compound.

C = 27.3 g

O = 72 g

Now find how many moles are there for each element in 100 g of compound

For this molar mass are required

That is

C = 12 g/mol

O = 16 g/mol

Formula Used

mole of C = mass of C / Molar mass of C

mole of C = 27.3 / 12 g/mol

mole of C = 2.275

Formula Used

mole of O = mass of O / Molar mass of O

mole of O = 72g / 16 g/mol

mole of O = 7.2

Divide each one by the smallest number of moles

C = 2.275 / 2.275

C = 1

O = 7.2 / 2.275

O = 3.2

Multiply the mole fraction to a number to get the whole number.

C = 1 x 5 = 5

O = 3.2 x 5 = 16

So, the empirical formula is C₅O₁₆

______________________________________

Part 4

Data Given

Percentage of P= 56.38 %

Percentage of O = 43.62%

Molar Mass = 219.9g

Molecular Formula of the compound = ?

Solution:

First, find the mass of each of the elements in 100 g of the Compound.

Mass of P= 56.38g

Mass of O = 43.62g

Now find how many moles are there for each element in 100 g of compound

find the moles in total compounds

Formula Used

mole of P = mass of / Molar mass of P

mole of P = 56.38 g / 31 g/mol

mole of P = 1.818

Formula Used

mole of O = mass of O / Molar mass of O

mole of O = 43. 62 / 16 g/mol

mole of O = 2.7262

Now

first find the Emperical formula

Divide each one by the smallest number of moles

P = 1.818 /1.818

P= 1

for oxygen

O = 2.7262 / 1.818

O = 1.5

Multiply the mole fraction to a number to get the whole number.

P = 1 x 2 = 2

O = 1.5 x 2 = 3

So, the empirical formula is P₂O₃

Now

Find molar mass of the empirical formula P₂O₃

2 (31) + 3 (16) = 62 + 48 = 110

Now find that how many empirical units are in a molecular unit.

(219.9 g/mol) / ( 110 g/mol) = empirical units per molecular unit

empirical units per molecular unit = 1.999 =2

A here we get two empirical units in a molecular unit,

So the molecular formula is:

2 (P₂O₃) = P₄O₆

7 0

2 years ago

Biacetyl, the flavoring that makes margarine taste "just like butter," is extremely stable at room temperature, but at 200°C it

Sholpan [36]

Answer:

3.91 minutes

Explanation:

Given that:

Biacetyl breakdown with a half life of 9.0 min after undergoing first-order reaction;

As we known that the half-life for first order is:

$t__{1/2}}= \frac{0.693}{k}$

where;

k = constant

The formula can be re-written as:

$k = \frac{0.693}{t__{1/2}}$

$k = \frac{0.693}{9.0 min}$

$k = 0.077 min^{-1}$

Let the initial amount of butter flavor in the food be $(N_0)$ = 100%

Also, the amount of butter flavor retained at 200°C $(N_t)$ = 74%

The rate constant $k = 0.077 min^{-1}$

To determine how long can the food be heated at this temperature and retain 74% of its buttery flavor; we use the formula:

$\frac{N_t}{N_0}= -kt$

$t = - (\frac{1}{k}*In\frac{N_t}{N_0} )$

Substituting our values; we have:

$t = - (\frac{1}{0.077}*In\frac{74}{100} )$

t = 3.91 minutes

∵ The time needed for the food to be heated at this temperature and retain 74% of its buttery flavor is 3.91 minutes

4 0

3 years ago

The compound HNO3 can be described as an

xz_007 [3.2K]

The answer is (1) Arrhenius acid and an electrolyte. The HNO3 is a H+ ion donor and OH- receptor. Electrolyte means the compound will release ion when dissolving in suitable ionized solution.

6 0

2 years ago

Which molecule is nonpolar and has a symmetrical shape hcl?

Alchen [17]

HCl is considered as polar molecule and that is incorrect answer, The choices provided are A) HCl , B) CH₄ , C) H₂O, D) NH₃

And the correct answer is : B) CH₄

In case of H₂O and NH₃ the high difference in electronegativity between nitrogen or oxygen and hydrogen make the bonds acquire partial charges and become polar while in case of CH₄ the difference in electronegativity between carbon and hydrogen is only 0.4 so it considered as non-polar bond
also carbon connected from all directions with Hydrogens

4 0

2 years ago

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