Two comets are 300 m apart. One

A 0.1014 g sample of a purified CHO compound was burned in a combustion apparatus and produced 0.1486 g CO2 and 0.0609 g of H2O.

kirza4 [7]

Answer: the empirical formula is CH2O

Explanation:Please see attachment for explanation

4 0

3 years ago

Consider a hydrogen atom in the ground state. What is the energy of its electron?

mr Goodwill [35]

The energy of an electron in a hydrogen atom is:
E= (-2.179 x 10 ^-18 J)(1/n^2)

where n is the principle energy level of the electron.

E= (-2.179 x 10 ^-18 J)(1/4)
E= -2.179 x 10 ^-18 J
E = -5.4 x 10 ^-19 J

4 0

3 years ago

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In this graph, what are the independent and dependent variables?

NeTakaya

Answer:

mass of hummingbirds is dependent and whatever the other one is the independent

Explanation:

4 0

3 years ago

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QUESTION 10

worty [1.4K]

I have no idea honestly I don’t remember I had it and I forgot it

6 0

3 years ago

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