Antonym for temperature

Lauryl alcohol is a nonelectrolyte obtained from coconut oil and is used to make detergents. A solution of 8.80 g of lauryl alco

kipiarov [429]

Answer:

The molar mass of lauryl alcohol is approximately 180 g/mol

Explanation:

Step 1: Data given

Mass of lauryl alcohol = 8.80 grams

Mass of benzene = 0.100 kg = 100 grams

Freezing point of Benzene is 5.5 °C

Kf value for benzene = 5.12 °C/molal

Step 2: Calculate the freezing point depression

ΔTf = 5.5 - 3.0 °C = 2.5 °C

Step 3: Calculate molality

⇒ with ΔTf = the freezing point depression = 2.5 °C

⇒ with i = the van't Hoff factor = 1

⇒ with kf = the free point depression constat pf benzene = 5.12°C/m

⇒ with m =the molality = TO BE DETERMINED

molality = ΔTf / kf

molality = 2.5 °C / 5.12 °C /m

molality = 0.488 molal

Step 4: Calculate moles lauryl alcohol

Molality = moles lauryl alcohol / mass benzene

Moles lauryl alcohol = molality * mass benzene

Moles lauryl alcohol = 0.488 m * 0.100 kg

Moles lauryl alcohol = 0.0488 moles

Step 5: Calculate molar mass of lauryl alcohol

Molar mass lauryl alcohol = mass lauryl alcohol/ moles lauryl alcohol

Molar mass lauryl alcohol = 8.80 grams / 0.0488 moles

Molar mass lauryl alcohol = 180 g/mol

The molar mass of lauryl alcohol is approximately 180 g/mol

3 0

3 years ago

How many grams of H2O are produced when 35.0 g of NaOH reacts with 17.5 g of CO,?

zhenek [66]

Answer:

2NaOH + CO2 -> Na2CO3 + H2O

1) Find the moles of each substance

$\eq n(NaOH)=\frac{35.0}{22.99+16.00+1.008\\}\ =\frac{35.0}{39.998} \ = 0.8750437522 moles\\n(CO_{2} ) = \frac{17.5}{12.01+32.00} = \frac{17.5}{44.01} = 0.3976369007 moles\\$

2) Determine the limitting reagent

$\\NaOH = \frac{0.8750437522}{2} = 0.4375218761\\\\$

∴ Carbon dioxide is limitting as it has a smaller value.

3) multiply the limiting reagent by the mole ratio of unknown over known

n(H2O ) = 0.3976369007 × 1/2

= 0.1988184504 moles

4) Multiply the number of moles by the molar mass of the substance.

m = 0.1988184504 × (1.008 × 2 + 16.00)

= 0.1988184504 × 18.016

= 3.581913202 g

Explanation:

6 0

2 years ago

Name the process that changes matter into one or more new substances

Brums [2.3K]

Chemical change or process

6 0

3 years ago

Name the following compound with a polyatomic ion: Na₂SO₄ *

Helga [31]

Answer:

sodium sulfate

Explanation:

For naming an ionic compound with polyatomic anion, the metal is written first using its element name followed by name of the polyatomic anion. Therefore, the compound with Na+Na+ cation and SO2−4SO42− anion is named as sodium sulfate.

6 0

2 years ago

Read 2 more answers

When the following oxidation–reduction reaction in acidic solution is balanced, what is the lowest whole-number coefficient for

ruslelena [56]

Answer:

b. 16, reactant side

Explanation:

Let's consider the following redox reaction.

MnO₄⁻(aq) + I⁻(aq) → Mn²⁺(aq) + I₂(s)

We can balance it using the ion-electron method.

Step 1: Identify both half-reactions

Reduction: MnO₄⁻(aq) → Mn²⁺(aq)

Oxidation: I⁻(aq) → I₂(s)

Step 2: Perform the mass balance, adding H⁺(aq) and H₂O(l) where appropriate

MnO₄⁻(aq) + 8 H⁺(aq) → Mn²⁺(aq) + 4 H₂O(l)

2 I⁻(aq) → I₂(s)

Step 3: Perform the charge balance, adding electrons where appropriate

MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

2 I⁻(aq) → I₂(s) + 2 e⁻

Step 4: Multiply both half-reactions by numbers so that the number of electrons gained and lost are equal

2 × (MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))

5 × (2 I⁻(aq) → I₂(s) + 2 e⁻)

Step 5: Add both half-reactions and cancel what is repeated on both sides

2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 e⁻ + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s) + 10 e⁻

The balanced reaction is:

2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s)

5 0

2 years ago