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1 year ago

2. A quantity of 1.922g of methanol (CH3OH) was burned in a constant-volume

1 answer:

5 0

Mass of methanol (CH3OH) = 1.922 g
Change in Temperature (t) = 4.20°C
Heat capacity of the bomb plus water = 10.4 KJ/oC
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
Let’s assume that no heat is lost to the surroundings. First, let’s calculate the heat changes in the calorimeter. This is calculated using the formula shown below:
qcal = Ccalt
Where, qcal = heat of reaction
Ccal = heat capacity of calorimeter
t = change in temperature of the sample
Now, let’s calculate qcal:
qcal = (10.4 kJ/°C)(4.20°C)
= 43.68 kJ
Always qsys = qcal + qrxn = 0,
qrxn = -43.68 kJ
The heat change of the reaction is - 43.68 kJ which is the heat released by the combustion of 1.922 g of CH3OH. Therefore, the conversion factor is:

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Most Bic lighters hold 5.0ml of liquified butane (density = 0.60 g/ml). Calculate the minimum size container you would need to "

Hatshy [7]

Answer:

Volume of container = 0.0012 m³ or 1.2 L or 1200 ml

Explanation:

Volume of butane = 5.0 ml

density = 0.60 g/ml

Room temperature (T) = 293.15 K

Normal pressure (P) = 1 atm = 101,325 pa

Ideal gas constant (R) = 8.3145 J/mole.K)

volume of container V = ?

Solution

To find out the volume of container we use ideal gas equation

PV = nRT

P = pressure

V = volume

n = number of moles

R = gas constant

T = temperature

First we find out number of moles

<em>As Mass = density × volume</em>

mass of butane = 0.60 g/ml ×5.0 ml

mass of butane = 3 g

now find out number of moles (n)

n = mass / molar mass

n = 3 g / 58.12 g/mol

n = 0.05 mol

Now put all values in ideal gas equation

V = (0.05 mol × 8.3145 J/mol.K × 293.15 K) ÷ 101,325 pa

V = 121.87 ÷ 101,325 pa

V = 0.0012 m³ OR 1.2 L OR 1200 ml

8 0

2 years ago

How many moles are in 19.6 g of Sodium (Na)? And the conversion factor?

Snezhnost [94]

Answer:

n=0.852 moles

Explanation:

Given mass is, m = 19.6 g

The molar mass of sodium is, M = 22.99 u

We need to find the no of moles in 19.6 of Sodium. We know that, no of moles is equal to given mass divided by molar mass.

$n=\dfrac{m}{M}\\\\n=\dfrac{19.6}{22.99}\\\\n=0.852$

So, there are 0.852 moles in 19.6 g of Sodium.

6 0

2 years ago

What is the amount of Al2S3 remains when 20.00 g of Al2S3 and 2.00 g of H2O are reacted? A few of the molar masses are as follow

Irina18 [472]

From the equation, we can tell that 1 mol of Al₂S₃ requires 6 moles of water.
The molar ratio is 1/6
Moles of Al₂S₃ present = 20/150.17
= 0.133
Moles of water present = 2/18.02
= 0.111
The moles of Al₂S₃ that will react are:
0.111/6
= 0.0185
The remaining amount:
0.133 - 0.0185
= 0.1145 mol
Or
0.1145 * 150.17
= 17.19 grams

6 0

2 years ago

Read 2 more answers

How many moles are in 6.5e23 atoms of Ne?

Savatey [412]

<h3>Answer:</h3>

1.1 mol Ne

<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>

Moles

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

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 6.5 × 10²³ atoms Ne

[Solve] moles Ne

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

Avogadro's Number

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 6.5 \cdot 10^{23} \ atoms \ Ne(\frac{1 \ mol \ Ne}{6.022 \cdot 10^{23} \ atoms \ Ne})$
[DA] Divide [Cancel out units]: $\displaystyle 1.07938 \ mol \ Ne$

<u>Step 4: Check</u>

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

1.07938 mol Ne ≈ 1.1 mol Ne

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1 year ago

Using what you know of map making terms and concepts and word parts, define hemisphere in your own words.

Lady_Fox [76]

Answer:

The Earth is divided into two halves by the prime meridian, forming two half spheres divided vertically. The Earth is also divided into two halves by the equator, creating two half spheres divided horizontally.

Explanation:

I took it on edge ( :

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2 years ago