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3 years ago

Give the empirical formula for C↓8H↓8

Chemistry

1 answer:

mr_godi [17]3 years ago

4 0

Answer:

Empirical formula of C₈H₈ = CH

Explanation:

Data Given:

Molecular Formula = C₈H₈

Empirical Formula = ?

Solution

Empirical Formula:

Empirical formula is the simplest ration of atoms in the molecule but not all numbers of atoms in a compound.

So,

tha ration of the molecular formula should be divided by whole number to get the simplest ratio of molecule

C₈H₈ Consist of Carbon (C), and Hydrogen (H)

Now

Look at the ratio of these two atoms in the compound

C : H

8 : 8

Divide the ratio by two to get simplest ratio

C : H

8/8 : 8/8

1 : 1

So for the empirical formula is the simplest ratio of carbon to hydrogen 1 : 1

So the empirical formula will be

Empirical formula of C₈H₈ = CH

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A sample of flammable liquid is placed into an enclosed cylinder which is then fitted with a movable piston. Initially the cylin

polet [3.4K]

Answer:

12.09 L

Explanation:

Step 1: Convert 826.1 mmHg to atm

We will use the conversion factor 760 mmHg = 1 atm.

826.1 mmHg × 1 atm/760 mmHg = 1.087 atm

Step 2: Convert 427.8 J to L.atm

We will use the conversion factor 101.3 J = 1 L.atm.

427.8 J × 1 L.atm/101.3 J = 4.223 L.atm

Step 3: Calculate the change in the volume

Assuming the work done (w) is 4.223 L.atm against a pressure (P) of 1.087 atm, the change in the volume is:

w = P × ΔV

ΔV = w/P

ΔV = 4.223 L.atm/1.087 atm = 3.885 L

Step 4: Calculate the final volume

V₂ = V₁ + ΔV

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3 years ago

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aniked [119]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

How many moles of NH3 can be produced by the reaction of 2.00 g of N2 with 3.00 g H2?

yulyashka [42]

0.235 if I’m not mistaken!

8 0

3 years ago

How many grams are in 4.5 x 10^22 molecules of Ba(NO2)2

NNADVOKAT [17]

Answer:

17 g Ba(NO₂)₂

General Formulas and Concepts:

Chemistry

Stoichiometry
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

Step 1: Define

4.5 × 10²² molecules Ba(NO₂)₂

Step 2: Define conversion

Molar Mass of Ba - 137.33 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Ba(NO₂)₂ - 137.33 + 2(14.01) + 4(16.00) = 229.35 g/mol

Step 3: Dimensional Analysis

$4.5 \cdot 10^{22} \ mc \ Ba(NO_2)_2(\frac{1 \ mol \ Ba(NO_2)_2}{6.022 \cdot 10^{23} \ mc \ Ba(NO_2)_2} )(\frac{229.35 \ g \ Ba(NO_2)_2}{1 \ mol \ Ba(NO_2)_2} )$

= 17.1384 g Ba(NO₂)₂

Step 4: Check

We are given 2 sig figs. Follow sig fig rules.

17.1384 g Ba(NO₂)₂ ≈ 17 g Ba(NO₂)₂

7 0

3 years ago

Show the calculation of the molar mass (molecular weight) of a solute if a solution of 5.8 grams of the solute in 100 grams of w

Leni [432]

Answer : The molar mass of solute is, 89.9 g/mol

Explanation : Given,

Mass of solute = 5.8 g

Mass of solvent (water) = 100 g

Formula used :

$\Delta T_f=K_f\times m\\\\T_f^o-T_f=T_f\times\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of water}}$

where,

$\Delta T_f$ = change in freezing point

$T_f^o$ = temperature of pure solvent (water) = $0^oC$

$T_f$ = temperature of solution = $1.20^oC$

$K_f$ = freezing point constant of water = $1.86^oC/m$

m = molality

Now put all the given values in this formula, we get

$(0^oC)-(1.20^oC)=1.86^oC/m\times \frac{5.8g\times 1000}{\text{Molar mass of solute}\times 100g}$

$\text{Molar mass of solute}=89.9g/mol$

Therefore, the molar mass of solute is, 89.9 g/mol

5 0

3 years ago