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Roman55 [17]
3 years ago
14

How mant grams are there in 7.15 x 10^24 molecules of water?

Chemistry
2 answers:
AysviL [449]3 years ago
8 0

Answer:

\boxed {\boxed {\sf B. \ 214 \ g }}

Explanation:

We are asked to find the mass of a given number of water (H₂O) molecules.

<h3>1. Convert Molecules to Moles</h3>

First, we convert molecules to moles. We will use Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are molecules of water. Therefore,<u> there are 6.022 ×10²³ molecules of water in 1 mole of water</u>.

We will convert using dimensional analysis, so we must set up a ratio using the underlined information.

\frac {6.022 \times 10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O}

We are converting 7.15 × 10²⁴ molecules of water to moles, so we multiply the ratio by that value.

7.15 \times 10^{24} \ molecules \ H_2O*\frac {6.022 \times 10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O}

Flip the ratio so the units of molecules of water cancel each other out.

7.15 \times 10^{24} \ molecules \ H_2O*\frac {1 \ mol \ H_2O} {6.022 \times 10^{23} \ molecules \ H_2O}

7.15 \times 10^{24} *\frac {1 \ mol \ H_2O} {6.022 \times 10^{23} }

\frac {7.15 \times 10^{24}} {6.022 \times 10^{23} } \ mol \ H_2O

11.87313185\ mol \ H_2O

<h3>2. Convert Moles to Grams </h3>

Next, we convert moles to grams. We will use the molar mass. This is the mass of 1 mole of a substance. These values are found on the Periodic Table. They are equivalent to the atomic masses, but the units are grams per mole instead of atomic mass units.

Look up the molar masses of the individual elements in water: hydrogen and oxygen.

  • H: 1.008 g/mol
  • O: 15.999 g/mol

There is a subscript of 2 after H, so there 2 moles of hydrogen in 1 mole of water. Multiply the molar mass by 2 before adding oxygen's molar mass.

  • H₂: 1.008 * 2 = 2.016 g/mol
  • H₂O: 2.016 + 15.999 = 18.015 g/mol

We convert using dimensional analysis again and set up another ratio.

\frac {18.015 \ g \ H_2O}{ 1 \ mol \ H_2O}

Multiply by the number of moles we just calculated.

11.87313185 \ mol \ H_2O *\frac {18.015 \ g \ H_2O}{ 1 \ mol \ H_2O}

The units of moles of water cancel.

11.87313185  *\frac {18.015 \ g \ H_2O}{ 1 }

213.8944703 \ g \ H_2O

The original measurement of molecules has 3 significant figures, so our answer must have the same. For the number we found, that is the one place. The 8 in the tenths place tells us to round the 3 up to a 4.

214 \ g \ H_2O

There are approximately <u>214 grams of water</u> and <u>choice B</u> is correct.

worty [1.4K]3 years ago
4 0

Answer:

B. 214.02

Explanation:

1 mol of water weighs 18.015 gm and contains 6.023 × 10²³ molecules

From question, We have 7.15 × 10²⁴ molecules

Dividing we get (7.15 × 10 ²⁴) ÷ ( 6.023 × 10²³) = 11.871 molecules

Now, Weight of water = 11.871 × 18.015 = 213.85 which is nearer to option B

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In which of the following types of solution is the rate of dissociation equal to the rate of crystallization? Concentrated Super
Butoxors [25]

In a <u>Saturated </u>solution, the rate of dissociation equal to the rate of crystallization

Explanation:

A saturated solution is one than cannot dissolve any more solute because the solutes inter-molecular spaces are filled with the solute molecules at that temperature. When an attempt is made to dissolve more solute into the solution, the rate at which the solute is dissolved into the solution is equal to the rate at which excess solute is precipitated and crystallized.

5 0
3 years ago
Answers to all of these
Paul [167]

Answer:

1. Percent composition of  Al = 13.423 %

2.

  • 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
3 years ago
A concentration cell is constructed using two Ni electrodes with Ni2+ concentrations of 1.0 M and 1.00 � 10�4 M in the two half-
Komok [63]

<u>Answer:</u> The cell potential of the cell is +0.118 V

<u>Explanation:</u>

The half reactions for the cell is:

<u>Oxidation half reaction (anode):</u>  Ni(s)\rightarrow Ni^{2+}+2e^-

<u>Reduction half reaction (cathode):</u>  Ni^{2+}+2e^-\rightarrow Ni(s)

In this case, the cathode and anode both are same. So, E^o_{cell} will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Ni^{2+}_{diluted}]}{[Ni^{2+}_{concentrated}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = ?

[Ni^{2+}_{diluted}] = 1.00\times 10^{-4}M

[Ni^{2+}_{concentrated}] = 1.0 M

Putting values in above equation, we get:

E_{cell}=0-\frac{0.0592}{2}\log \frac{1.00\times 10^{-4}M}{1.0M}

E_{cell}=0.118V

Hence, the cell potential of the cell is +0.118 V

5 0
3 years ago
How many mL are 2.3 mol of CO2 at STP?
kupik [55]

Answer:

0.052mL

Explanation:

1mole of a gas occupy 22.4L.

Therefore, 1 mole of CO2 will also occupy 22.4L.

If 1mole of CO2 occupies 22.4L,

Then 2.3moles of CO2 will occupy = 2.3 x 22.4 = 51.52L

coverting this volume to mL, we simply divide by 1000 as shown below:

51.52/1000 = 0.05152mL = 0.052mL

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A joule times a second :)
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