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N76 [4]
3 years ago
6

The density of the hydrocarbon in part (a) is 2.0 g l-1 at 50°c and 0.948atm. (i) calculate the molar mass of the hydrocarbon. (

ii) determine the molecular formula of the hydrocarbon.
Chemistry
1 answer:
Vedmedyk [2.9K]3 years ago
7 0

1. Answer;

=56 g/mol

Explanation and solution;

PV = nRT

nRT= mass/molar mass (RT)

molar mass = (mass/V ) × (RT/P)

                   = Density × (RT/P)

Molar mass = 2.0 g/L × (0.0821 × 323 K)/0.948 atm

Molar mass = 56 g/mol


2. Answer;

Molecular mass is C4H8

Explanation;

Empirical mass × n = molar mass

Empirical mass for CH2 = 14 g/mol

Therefore;

56 g/mol = 14 g/mol × n

   n = 4

The molecular formula= 4(CH2)

    = C4H8


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How many grams of hydrochloric acid will react completely with a block of gold is 3.2 cm by 3.8 cm by 2.8 cm, if the density of
JulijaS [17]
Answer:
mass of HCl = 243.5426 grams

Explanation:
1- we will get the mass of the reacting gold:
volume of gold = length * width * height
volume of gold = 3.2 * 3.8 * 2.8 = 34.048 cm^3 = 34.048 ml<span>
density = mass / volume
Therefore:
mass = density * volume
mass of gold = </span>19.3 * 34.048 = 657.1264 grams

2- we will get the number of moles of the reacting gold:
number of moles = mass / molar mass
number of moles = 657.1264 / 196.96657 
number of moles = 3.3362 moles

3- we will get the number of moles of the HCl:
First, we will balanced the given equation. The balanced equation will be as follows:
Au + 2HCl ......> AuCl2 + H2
This means that one mole of Au reacts with 2 moles of HCl.
Therefore 3.3362 moles will react with 2*3.3362 = 6.6724 moles of HCL

4- we will get the mass of the HCl:
From the periodic table:
molar mass of H = 1 gram
molar mass of Cl = 35.5 grams
Therefore:
molar mass of HCl = 1 + 35.5 = 36.5 grams/mole
number of moles = mass / molar mass
Therefore:
mass = number of moles * molar mass
mass of HCl = 6.6724 * 36.5
mass of HCl = 243.5426 grams

Hope this helps :)
4 0
3 years ago
If you react 156.0 g of calcium chloride with an excess NaOH, how much sodium chloride should you get?
laiz [17]

Answer:

164.3g of NaCl

Explanation:

Based on the chemical equation:

CaCl2 + 2NaOH → 2NaCl + Ca(OH)2

<em>where 1 mole of CaCl2 reacts with 2 moles of NaOH</em>

To solve this question we must convert the mass of CaCl2 to moles. Using the chemical equation we can find the moles of NaCl and its mass:

<em>Moles CaCl2 -Molar mass: 110.98g/mol-</em>

156.0g CaCl₂ * (1mol / 110.98g) = 1.4057 moles CaCl2

<em>Moles NaCl:</em>

1.4057 moles CaCl2 * (2mol NaCl / 1mol CaCl2) = 2.811 moles NaCl

<em>Mass NaCl -Molar mass: 58.44g/mol-</em>

2.811 moles NaCl * (58.44g / mol) = 164.3g of NaCl

7 0
3 years ago
I need help solving this!
zmey [24]

Answer: Moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, CH_{4}.

Explanation:

Given: Mass of methane = 146.6 g

As moles is the mass of a substance divided by its molar mass. So, moles of methane (molar mass = 16.04 g/mol) are calculated as follows.

Moles = \frac{mass}{molar mass}\\= \frac{146.6 g}{16.04 g/mol}\\= 9.14 mol

The given reaction equation is as follows.

C + 2H_{2} \rightarrow CH_{4}

This shows that 2 moles of hydrogen gives 1 mole of methane. Hence, moles of hydrogen required to form 9.14 moles of methane is as follows.

Moles of H_{2} = \frac{9.14}{2}\\= 4.57 mol

Thus, we can conclude that moles of hydrogen required are 4.57 moles to make 146.6 grams of methane, CH_{4}.

5 0
3 years ago
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(IV) 2ZnS + 3O_{2} --> 2ZnO + 2SO_2

(V) Pb(NO_3)_2 + 2NaCl --> 2NaNO_3 + PbCl_2

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

7 0
3 years ago
The density of gas particles in a section of Earth’s atmosphere decreases. Which of the following is the most likely explanation
Bad White [126]
<span>This is due to the fact that the air pressure in that certain section of Earth’s atmosphere decreased. As density of gas particles decreases as air pressure decreases. Therefore, density of gas particles and air pressure have a direct relationship. An increase in air pressure would then effect to an increase in gas particles. </span>
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