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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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What is the simplest formula of a compound if a sample of the compound contains 0.221 mol X, 0.442 mol Y, and 0.884 mol Z? chemP

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Answer:

XY₂Z₄

2.35 mol Z

Explanation:

A sample of the compound contains 0.221 mol X, 0.442 mol Y, and 0.884 mol Z. We can find the simplest formula (empirical formula) by <em>dividing all the numbers of moles by the smallest one</em>.

X: 0.221/0.221 = 1

Y: 0.442/0.221 = 2

Z: 0.884/0.221 = 4

The simplest formula is XY₂Z₄.

The molar ratio of X to Z is 1:4. The moles of Z in a sample that contained 0.588 moles of X is:

0.588 mol X × (4 mol Z/1 mol X) = 2.35 mol Z

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2 years ago

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Give a possible explanation for the relative amounts of the isometric methyl nitrobenzoates formed in the nitration reaction. Co

vovikov84 [41]

Answer:

The electrophilic aromatic substitution reaction nitration is used to nitrate methyl benzoate and acetanilide with a nitronium ion. Crystallization was used to purify the product. The melting point was used to determine its purity and the regiochemistry of the products.

Explanation:

Methyl m-Nitrobenzoate is formed in this

reaction rather that ortho/para isomers

because of the ester group of your starting

product of methylbenzoate. The functional

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causing nitrobenzene (N02) to become in the

meta position. Thus N02 is a deactivating

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Basically you must look at the substituents

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in order to figure out whether your reaction

with be ortho/para directors or meta

directors. If the substituents are electron

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1 year ago

A student has a 1g sample of each of the following compounds: NaCl, KBr, and KCL Which of the following lists the samples in ord

krek1111 [17]

Given :

A student has a 1 g sample of each of the following compounds: NaCl, KBr, and KCl.

To Find :

The samples in order of increasing number of moles in the sample.

Solution :

Molecular mass of NaCl, KBr, and KCl is 58.5 g/mol , 119 g/mol and

74.5 g/mol respectively .

Moles of NaCl , $n_1=\dfrac{1}{58.5}=0.017\ mol$ .

Moles of KBr , $n_2=\dfrac{1}{119}=0.008\ mol$ .

Moles of KCl , $n_2=\dfrac{1}{74.5}=0.013\ mol$ .

The order of moles in increasing order is :

KBr , KCl and NaCl .

Hence , this is the required solution .

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2 years ago

In order to reduce the exposure to organic solvents like turpentine, some art instructors recommend the students clean brushes a

kykrilka [37]

Generally speaking, organic molecules tend to dissolve in solvents that have similar physical properties. A good rule of thumb is that "like dissolves like". Meaning, polar compounds can dissolve polar compounds and nonpolar compounds can dissolve nonpolar compounds.

To apply this to the current problem, we are told that the brushes are being cleaned with vegetable oil or mineral oil. In this case, the oils are used as solvents. In order for these solvents to be effective, the compounds they are trying to dissolve must be similar in structure and properties to other oils. Therefore, vegetable oil or mineral oil will be most effective in removing oil-based paints, as these will have the similar properties needed to dissolve in the oil solvents.

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2 years ago

1. A gas has a pressure of 799.0 mm Hg at 50.0 degrees C. What is the temperature at standard Pressure?

larisa [96]

Answer:

A = 674.33mmHg

B = 0.385atm

Explanation:

Both question A and B requires the application of pressure law which states that the pressure of a fixed mass of gas is directly proportional to its temperature provided that volume is kept constant.

Mathematically,

P = kT, k = P / T

P1 / T1 = P2 / T2 = P3 / T3 =.......= Pn/Tn

Data:

P1 = 799mmHg

T1 = 50°C = (50 + 273.15) = 323.15K

P2 = ?

T2 = 273.15K

P1 / T1 = P2 / T2

Solve for P2

P2 = (P1 × T2) / T1

P2 = (799 × 273.15) / 323.15

P2 = 674.37mmHg

The final pressure is 674.37mmHg

P1 = 0.470atm

T1 = 60°C = (60 + 273.15)K = 333.15K

P2 = ?

T2 = 273.15K

P1 / T1 = P2 / T2

Solve for P2,

P2 = (P1 × T2) / T1

P2 = (0.470 × 273.15) / 333.15

P2 = 0.385atm

The final pressure is 0.385atm

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2 years ago

Read 2 more answers