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3 years ago

Why is boiling point for carbon tertrachloride igher than methanol?

1 answer:

8 0

Carbon tetrachloride looks like this

Cl
Cl-C-Cl
Cl

Whereas methane looks as so,

H
H-C-H
H

——————————————————————
In structure/geometry, they’re the exact same. They have the same surface area, and neither one is polar.

The only attribute to the difference in boiling points is the total molecular weights of the atoms

CH4 is MUCH lighter than CCl4, making the molecules easier to float around.

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Is 3Al + 3FeO → 2Al2O3 + 3Fe a balenced or unbalenced equation?

Murrr4er [49]

3f+5=eo got it right

8 0

2 years ago

You determine the volume of your plastic bag (simulated human stomach) is 1.08 L. How many grams of NaHCO3 (s) are required to f

dsp73

Answer:

3.636 grams of sodium bicarbonate is required.

Explanation:

Using ideal gas equation:

PV = nRT

where,

P = Pressure of gas = 753.5 mmHg = 0.9914 atm

( $1 atm = 760 mmHg$ )

V = Volume of gas = 1.08 L

n = number of moles of gas = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = 24.5 °C= 297.65 K

Putting values in above equation, we get:

$(0.9914 atm)\times 1.08 L=n\times (0.0821L.atm/mol.K)\times 297.65K\\\\n=0.0438 mole$

Percentage recovery of carbon dioxide gas = 49.4%

Actual moles of carbon dioxide formed: 49.4% of 0.0438 mole

$\frac{49.4}{100}\times 0.0438 mol=0.02164 mol$

$2NaHCO_3\righarrow Na_2CO_3+H_2O+CO_2$

According to reaction ,1 mol is obtained from 2 moles of sodium bicarbonate.

Then 0.02164 moles f carbon dioxide will be obtained from:

$\frac{2}{1}\times 0.02164 mol=0.04328 mol$

Mass of 0.04328 moles pf sodium bicarbonate:

0.04328 mol × 84 g/mol = 3.636 g

3.636 grams of sodium bicarbonate is required.

5 0

3 years ago

How many moles of S would I have if I had 11 grams? (Stoichiometry) HELP

zepelin [54]

<h3>Answer:</h3>

0.34 mol S

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

11 g S

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of S - 32.07 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 11 \ g \ S(\frac{1 \ mol \ S}{32.07 \ g \ S})$
Multiply/Divide: $\displaystyle 0.343 \ mol \ S$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

0.343 mol S ≈ 0.34 mol S

4 0

2 years ago

How many litres (L) of oxygen at STP can be obtained from 110 g of potassium chlorate? [R=0.0821 L.atm/K.mol]

lisabon 2012 [21]

Balance the equation first:

2 KClO3 (s) ---> 2 KCl (s) + 3 O2 (g)

Moles of KClO3 = 110 / 122.5 = 0.89

Following the balanced chemical equation:
We can say moles of O2 produce = $\frac{3}{2}$ x moles of KClO3

So, O2 = (3 / 2) x 0.89

= 1.34 moles

So, Volume at STP = nRT / P

T = <span>273.15 K
P = 1 atm

So, V = (1.34 x 0.0821 x 273.15) / 1 = 30.2 L</span>

3 0

3 years ago

PLEASE HELP ASAP I NEED IT

Lapatulllka [165]

the answer is true the first one

3 0

3 years ago