Question convert 70mi/hr into ft/s

The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different react

murzikaleks [220]

Answer: The value of $K_{eq}$ is $4.84\times 10^{-5}$

Explanation:

We are given:

Initial moles of ammonia = 0.0280 moles

Initial moles of oxygen gas = 0.0120 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of ammonia = $\frac{0.0280}{1.00}=0.0280M$

Concentration of oxygen gas = $\frac{0.0120}{1.00}=0.0120M$

The given chemical equation follows:

$4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)$

Initial: 0.0280 0.0120

At eqllm: 0.0280-4x 0.0120-3x 2x 6x

We are given:

Equilibrium concentration of nitrogen gas = $3.00\times 10^{-3}M=0.003$

Evaluating the value of 'x', we get:

$\Rightarrow 2x=0.003\\\\\Rightarrow x=0.0015M$

Now, equilibrium concentration of ammonia = $0.0280-4x=[0.0280-(4\times 0.0015)]=0.022M$

Equilibrium concentration of oxygen gas = $0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M$

Equilibrium concentration of water = $6x=(6\times 0.0015)]=0.009M$

The expression of $K_{eq}$ for the above reaction follows:

$K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.009)^6\times (0.003)^2}{(0.022)^4\times (0.0075)^3}\\\\K_{eq}=4.84\times 10^{-5}$

Hence, the value of $K_{eq}$ is $4.84\times 10^{-5}$

4 0

3 years ago

Correct name for #10 please

ira [324]

Answer:

CH₃CH(CH₃)CH(C₃H₇)CH₂CH(CH₃)₂:

4-isopropyl-2-methylpentane.

Explanation:

Step One: Draw the structure formula of this compound. Parentheses in the formula indicate substitute groups that are connected to the carbon atom to the left.

For example, the first (CH₃) indicates that the second carbon atom from the left is connected to:

the CH₃- on the left-hand side,
the -CH(C₃H₇)CH₂CH(CH₃)₂ on the right-hand side,
a hydrogen atom, and
an additional CH₃- group that replaced one hydrogen atom.

Each carbon atom in this compound is connected to four other atoms. All bonds between carbon atoms are single bonds.

The C₃H₇ in the second pair of parentheses is the condensed form of CH₃CH₂CH₂-. See the first sketch attached. Groups in parentheses are highlighted.

Step Two: Find the carbon backbone. The backbone of a hydrocarbon is the longest chain of carbon atoms that runs through the compound. See the second sketch attached. The backbone of this compound consists of seven carbon atoms and is highlighted in green. The name for this backbone shall be heptane.

Step Three: Identify and name the substitute groups.

The two substitute groups are circled in blue in the second sketch.

The one on the right -CH₃ is a methyl group.
The one on the left is branched. $\begin{aligned}\text{CH}_3-&\text{CH}-\text{CH}_3\\[-0.5em]&\;|\end{aligned}$ This group can be formed by removing one hydrogen from the central carbon atom in propane. The name for this group is isopropyl.

Step Four: Number the atoms.

Isopropyl shall be placed before methyl. Start from the right end to minimize the index number on all substitute groups. The methyl group is on carbon number two and the isopropyl group on carbon number four. Hence the name:

4-isopropyl-2-methylheptane.