Name four products of incomplete combustion.

In an ion with an unknown charge, the total mass of all the electrons was determined to be 2.19 ✕ 10−26 g, while the total mass

Ilya [14]

Answer:

$\large \boxed{\text{Fe$^{{2+}}$}}$

Explanation:

1. Number of electrons

$\text{Number of electrons} = 2.19 \times 10^{-26}\text{ g} \times \dfrac{\text{1 electron}}{9.109 \times 10^{-28}\text{ g}} = \text{24 electrons}$

2. Number of protons

$\text{Number of protons} = 4.34 \times 10^{-23}\text{ g} \times \dfrac{\text{1 proton}}{1.673 \times 10^{-24}\text{ g}} = \text{26 protons}$

3. Identify the ion

An atom with 26 protons is iron, Fe.

A neutral atom of iron would have 26 electrons.

The ion has only 24 electrons, so it has lost two. The ion must have a charge of +2.

$\text{The symbol for the ion is $\large \boxed{\textbf{Fe$^{\mathbf{2+}}$}}$}$

4 0

3 years ago

At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react

erastova [34]

Answer:

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

$N_{2 (g)}$ + $O_{2 (g)}$ ⇄ 2 $NO_{(g)}$

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no $N_{2 (g)}$ nor $O_{2 (g)}$ present. Immediately, $N_{2 (g)}$ and $O_{2 (g)}$ are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [ $N_{2 (g)}$ ]=0 ; [ $O_{2 (g)}$ ]= 0 ; [ $NO_{(g)}$ ]=0.60M

C: [ $N_{2 (g)}$ ]=+x ; [ $O_{2 (g)}$ ]= +x ; [ $NO_{(g)}$ ]=-2x

E: [ $N_{2 (g)}$ ]=0+x ; [ $O_{2 (g)}$ ]= 0+x ; [ $NO_{(g)}$ ]=0.60-2x

Now we can use the constant information:

$K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }$

$4.10* 10^{-4} =\frac{(0.60-2x)^{2}}{(x)*(x)}$

$4.10* 10^{-4}$ = $\frac{(0.60-2x)^{2}}{x^{2} }$

$4.10* 10^{-4} * x^{2}$ = $(0.60-2x)^{2}}$

$\sqrt{4.10* 10^{-4} * x^{2}}$ = $\sqrt{(0.60-2x)^{2}}}$

$0.0202 x =0.60 - 2x$

$2x+0.0202x=0.60$

$x=\frac{0.60}{2.0202}$ $= 0.30$

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M

3 0

3 years ago

The reaction shown below occurs in the blood between hemoglobin (Hb) and oxygen.

OlgaM077 [116]

Answer:

A, B, C

Explanation:

Notice that this reaction involves double arrows, meaning this represents an equilibrium reaction in which we observe a forward reaction (combination of hemoglobin and oxygen) and a reverse reaction (decomposition of the oxyhemoglobin complex).

Upon inhalation of oxygen, it accesses the blood of a person and binds to hemoglobin, so the following reaction proceeds to the right.

Similarly, the opposite process takes place in muscles, oxyhemoglobin is decomposed back into hemoglobin and oxygen.

The equilibrium constant reaction is relatively high, since at standard conditions, this is a spontaneous reaction, hemoglobin combines with oxygen without any additional external source of energy.

7 0

4 years ago

Convert 3.01 x 10^24 molecules of ammonium sulfate to mass

Dmitry [639]

Mass of ammonium sulfate = 660.7 g

<h3>Further explanation</h3>

Given

3.01 x 10²⁴ molecules of ammonium sulfate

Required

mass

Solution

The mole is the number of particles(molecules, atoms, ions) contained in a substance

1 mol = 6.02.10²³ particles

Can be formulated

N=n x No

N = number of particles

n = mol

No = Avogadro's = 6.02.10²³

mol ammonium sulfate (NH₄)₂SO₄ :

n = N : No

n = 3.01 x 10²⁴ : 6.02 x 10²³

n = 5

mass ammonium sulfate :

= mol x MW

= 5 x 132,14 g/mol

= 660.7 g

5 0

3 years ago

Consider the reaction of magnesium metal with hydrochloric acid to produce magnesium chloride and hydrogen gas. If 3.56 mol of m

pishuonlain [190]

<h3>Answer:</h3>

43.27 g Mg

<h3>Explanation:</h3>

The balanced equation for the reaction between magnesium metal and hydrochloric acid is;

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

From the equation;

1 mole of magnesium reacts with 2 moles of HCl

We are given;

3.56 moles of Mg and 3.56 moles of HCl

Using the mole ratio;

3.56 moles of Mg would react with 7.12 moles of HCl, and

3.56 moles of HCl would react with 1.78 moles of Mg

Therefore;

The amount of magnesium was in excess;

Moles of Mg left = 3.56 moles - 1.78 moles

= 1.78 moles

But; 1 mole of Mg = 24.305 g/mol

Therefore;

Mass of magnesium left = 1.78 moles × 24.305 g/mol

= 43.2629 g

= 43.27 g

Thus, the mass of magnesium that remained after the reaction is 43.27 g

4 0

3 years ago