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2 years ago

If the reactants are 10 kg what is the mass of the products

Chemistry

1 answer:

nlexa [21]2 years ago

7 0

If the mass of both the reactants is 10kg then the mass of the products also equals 10kg.

It is due to the law of conservation of mass.

Mass can neither be created nor be destroyed.

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What will the pressure be if 89.9 moles of argon are contained in a 12.0 L cylinder that is pressurized at a temperature of 300

irinina [24]

P=nRT/V
p=89.9(8.314)(12)/300
P=8969.14/300
P=29.89atm
P=29.9atm

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2 years ago

Consider the solubilities of a particular solute at two different temperatures.

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2 years ago

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1 year ago

Pure chlorobenzene (C6H5Cl) has a normal boiling point of 131.00 °C. A solution of 32.5 g of 2,8-dibromodibenzofuran (C12H6Br2O)

vichka [17]

Answer:

Kb → 1.56 °C / m

Explanation:

This is all about boiling point elevation, the colligative property that shows that boiling point for a solution is higher than boiling point of pure solvent.

This is the formula: ΔT = Kb . m . i

where i is the Van't Hoff factor (ions dissolved in solution). As these are organic compounds, we assume they are non electrolytic,

m is molality (mol of solute / 1kg of solvent)

Kb is our unknown. The value for ebulloscopic constant, it is specific for each solvent.

ΔT = T° boiling from solution - T° boiling from solute

First of all, let's determine the moles of solute.

Mass / Molar mass → 32.5 g/ 113.45 g/mol = 0.286 mol

Molality is mol of solute/ 1 kg of solvent

We must convert the mass from g to kg

195g . 1kg /1000 = 0.195 kg

Molality = 0.286 mol / 0.195 kg = 1.47 m

Let's replace the values in the formula

133.30 °C - 131°C = Kb . 1.47m .1

2.30°C / 1.47 m = Kb → 1.56 °C / m

3 0

3 years ago

Mercury chloride is a commercial fungicide. If the molar mass is 470 g/mol and the percent composition is 85.0% Hg and 15.0% Cl,

ElenaW [278]

Data: molar mass 470 g/mol

Percent composition:

Hg = 85.0%
Cl = 15.0%

Solution:

1) Convert % to molar ratios

A. Base: 100 g

=> Hg = 85.0 g / 200.59 g/mol = 0.4235 mol

Cl = 15.0 g / 35.45 g/mol = 0.4231 mol

B. divide by the higher number and round to whole number

Hg = 0.4325 / 0.4231 = 1.00

Cl = 0.4231 / 0.4231 = 1.00

=> Empirical formula = Hg Cl

2) Find the mass of the empirical formula:

HgCl: 200.59 g/mol + 35.45 g/mol = 236.04

3) Determine how many times is the empirical mass contained in the molecular mass:

470 g/mol / 236.04 = 1.99 ≈ 2

=> Molecular formula = Hg2 Cl2.

Answers:

Empirical formula HgCl
Molecular Formula Hg2Cl2

6 0

3 years ago