All categories

4 years ago

Name four products of incomplete combustion Which four?

Chemistry

1 answer:

Y_Kistochka [10]4 years ago

8 0

Answer:

1. Carbon Monoxide

2.Water

3. sulfuric acid

4.sulfur dioxide

Explanation:

If my answer helped please rate

You might be interested in

Pam observed many small earthworms in the backyard of her house after a heavy rain. What most likely formed the earthworms?

AlekseyPX

Answer: Option (b) is the correct answer.

Explanation:

When it rains, then mud in the backyards become wet. It is known that earthworms live inside earth and diffusion of oxygen gets slower through water than through air.

As a result, supply of oxygen to the worms decreases in wet mud hence, they come to the surface in order to breathe.

Thus, we can conclude that wet mud in the backyard most likely formed the earthworms.

4 0

4 years ago

Na3po4 dissolves in water to produce an electrolytic solution. what is the osmolarity of a 2. 0 × 10-3 m na3po4 solution?

vichka [17]

When Na3po4 dissolves in water to produce an electrolytic solution. The osmolarity of a 2. 0 × 10-3 m Na3po4 solution is 0.008osmol/L.

Osmolarity is defined as the number of osmoles of solute particles per unit volume of the solution.

In other words osmolarity is the multiple if molarity

Osmolarity = i× molarity

Here i represents the van't Hoff factor,

$Na_{3}PO_{4}$ ⇒ $3Na^{+} + PO_{4} ^{-}$

3 Moles of $Na_{} ^{+}$ + 1 mole $PO_{4} ^{-}$ = 4

The number of moles of particles of solute produced in solution are actually called osmoles.

As a result, the van't Hoff factor will be equal to

i=4 Moles ions produced (osmoles) 1mole $Na_{3} PO_{4}$ .dissolved =4

Since we know that,

$Na_{3} PO_{4}$ = $2.0 * 10^{-3} M$

Osmolarity =

$4*2.0*10^{-3} M$ = $8.0 * 10^{-3} osmol L -10s$

Thus, the Osmolarity of given solution is 0.008 osmol/L.

learn more about Osmolarity:

brainly.com/question/13597129

#SPJ4

6 0

2 years ago

Calculate Δ H o for the reaction. CH3OH + HCl → CH3Cl + H2O answer is in kJ/mol .

Nimfa-mama [501]

Answer:

$\Delta H=-28.08 \frac{kJ}{mol}$

Explanation:

Hello,

This types of reactions are likely to be carried out in gaseous phase as it is easier to induce reactions, therefore, for us to compute the change in the enthalpy of this reaction we should write the formation enthalpy of gaseous methanol, hydrogen chloride, methyl chloride and water as -205.1, -92.3, -83.68 and -241.8 kJ/mol respectively. Then, the reaction enthalpy for this reaction is:

$\Delta H=\Delta _fH_{CH_3Cl}+\Delta _fH_{H_2O}-\Delta _fH_{CH_3OH}-\Delta _fH_{HCl}\\\\\Delta H=-83.68\frac{kJ}{mol}-241.8\frac{kJ}{mol}-(-205.1\frac{kJ}{mol})-(-92.3\frac{kJ}{mol} )\\\\\Delta H=-28.08 \frac{kJ}{mol}$