Which of these is a mixture? A. carbon B. carbon dioxide C. air D. oxygen

A compounds empirical formula is H.O. If the formula mass is 34 amu what is the molecular formula?

BigorU [14]

Answer:

C

Explanation:

1. The molecule has to have a ratio of 1 hydrogen to 1 oxygen.

2. Hydrogen = 1amu, Oxygen = 16amu. (2*2)+(16*2) = 34amu

7 0

2 years ago

Menthol, the substance we can smell in mentholated cough drops, is composed of c, h, and o. a 9.045×10−2 −mg sample of menthol i

Ket [755]

Answer:

Empirical Formula = C₁₀H₂₀O

Solution:

Data Given:

Mass of Menthol = 9.045 × 10⁻² mg = 9.045 × 10⁻⁵ g

Mass of CO₂ = 0.2546 mg = 0.0002546 g

Mass of H₂O = 0.1043 mg = 0.0001043 g

Step 1: Calculate %age of Elements as;

%C = (mass of CO₂ ÷ Mass of sample) × (12 ÷ 44) × 100

%C = (0.0002546 ÷ 9.045 × 10⁻⁵) × (12 ÷ 44) × 100

%C = (2.814) × (12 ÷ 44) × 100

%C = 2.814 × 0.2727 × 100

%C = 76.73 %

%H = (mass of H₂O ÷ Mass of sample) × (2.02 ÷ 18.02) × 100

%H = (0.0001043 ÷ 9.045 × 10⁻⁵) × (2.02 ÷ 18.02) × 100

%H = (1.153) × (2.02 ÷ 18.02) × 100

%H = 1.153 × 0.1120 × 100

%H = 12.91 %

%O = 100% - (%C + %H)

%O = 100% - (76.73% + 12.91%)

%O = 100% - 89.64%

%O = 10.36 %

Step 2: Calculate Moles of each Element;

Moles of C = %C ÷ At.Mass of C

Moles of C = 76.73 ÷ 12.01

Moles of C = 6.3888 mol

Moles of H = %H ÷ At.Mass of H

Moles of H = 12.91 ÷ 1.01

Moles of H = 12.7821 mol

Moles of O = %O ÷ At.Mass of O

Moles of O = 10.36 ÷ 16.0

Moles of O = 0.6475 mol

Step 3: Find out mole ratio and simplify it;

C H O

6.3888 12.7821 0.6475

6.3888/0.6475 12.7821/0.6475 0.6475/0.6475

9.86 19.74 1

≈ 10 ≈ 20 1

Result:

Empirical Formula = C₁₀H₂₀O₁

8 0

3 years ago

Why can't polyatomic ions ever stand alone?

Basile [38]

The prefix "poly-" means "many," and "atomic" refers to atoms, so a polyatomic ion is made of 2+ atoms covalently bonded together. These ions are charged, which is why they can never stand alone.

7 0

3 years ago

What volume does 18.0g of bromine occupy?

Varvara68 [4.7K]

Hey there!

Atomic mass Bromine ( Br ) = 79.9 u

Therefore:

79.9 g Br --------------- 22.4 L ( at STP )

18.0 g Br --------------- volume ??

Volume Br = 18.0 * 22.4 / 79.9

Volume Br = 403.2 / 79.9

Volume Br = 5.046 L

hope this helps!

4 0

3 years ago

How many grams of precipitate will be formed when 20.5 mL of 0.800 M

Anton [14]

Answer:

There will be formed 1.84 grams of precipitate (NaNO3)

Explanation:

Step 1: The balanced equation

CO(NO3)2 (aq) + 2 NaOH (aq) → CO(OH)2 (s) + 2 NaNO3 (aq)

Step 2: Data given

Volume of 0.800 M CO(NO3)2 = 20.5 mL = 0.0205 L

Volume of 0.800 M NaOH = 27.0 mL = 0.027 L

Molar mass of NaNO3 = 84.99 g/mol

Step 3: Calculate moles of CO(NO3)2

Moles CO(NO3)2 = Molarity * volume

Moles CO(NO3)2 = 0.800 M * 0.0205

Moles CO(NO3)2 = 0.0164 moles

Step 4: Calculate moles NaOH

moles of NaOH = 0.800 M * 0.027 L

moles NaOH = 0.0216 moles

Step 5: Calculate limiting reactant

For 1 mole CO(NO3)2 consumed, we need 2 moles of NaOH to produce 1 mole of CO(OH)2 and 2 moles of NaNO3

NaOH is the limiting reactant. It will completely be consumed.

CO(NO3)2 is in excess. There willbe 0.0216 / 2 = 0.0108 moles of CO(NO3)2 consumed. There will remain 0.0164 - 0.0108 = 0.0056 moles of CO(OH)2

Step 6: Calculate moles of NaNO3

For 2 moles of NaOH consumed, we have 2 moles of NaNO3

For 0.0216 moles of NaOH, we have 0.0216 moles of NaNO3

Step 7: Calculate mass of NaNO3

mass of NaNO3 = moles of NaNO3 * Molar mass of NaNO3

mass of NaNO3 = 0.0216 moles * 84.99 g/mol = 1.84 grams

There will be formed 1.84 grams of precipitate (NaNO3)

5 0

3 years ago