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3 years ago

What is the fourth most abundant element in the universe in terms of mass?

2 answers:

8 0

The fourth most abundant element in the universe in terms of mass would be Carbon.

The first is Hydrogen, the second is Helium, and the third is oxygen.

iren [92.7K]3 years ago

4 0

The fourth most abundant element in terms of mass is Carbon.

The first, second, and third most abundant element in terms of mass are Hydrogen, Helium, and Oxygen, respectively.

Hope this helps~

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Question: A boy swimming in a lake smacks the surface of the water with his hand, creating sound waves that travel through the w

rodikova [14]

<span>The speed of sound in air is about 340 m/s, while the speed of sound in water is about 1,500 m/s. In general, sound travels faster in denser media, so the fish hears the noise first.</span>

7 0

3 years ago

Which of these BEST describes a physical change?

OLEGan [10]

Answer:

C) a log in the campfire turns into ashes

8 0

3 years ago

9. Given the following reaction:CO (g) + 2 H2(g) CH3OH (g)In an experiment, 0.45 mol of CO and 0.57 mol of H2 were placed in a 1

WITCHER [35]

Answer:

$Keq=11.5$

Explanation:

Hello,

In this case, for the given reaction at equilibrium:

$CO (g) + 2 H_2(g) \rightleftharpoons CH_3OH (g)$

We can write the law of mass action as:

$Keq=\frac{[CH_3OH]}{[CO][H_2]^2}$

That in terms of the change $x$ due to the reaction extent we can write:

$Keq=\frac{x}{([CO]_0-x)([H_2]_0-2x)^2}$

Nevertheless, for the carbon monoxide, we can directly compute $x$ as shown below:

$[CO]_0=\frac{0.45mol}{1.00L}=0.45M\\$

$[H_2]_0=\frac{0.57mol}{1.00L}=0.57M\\$

$[CO]_{eq}=\frac{0.28mol}{1.00L}=0.28M\\$

$x=[CO]_0-[CO]_{eq}=0.45M-0.28M=0.17M$

Finally, we can compute the equilibrium constant:

$Keq=\frac{0.17M}{(0.45M-0.17M)(0.57M-2*0.17M)^2}\\\\Keq=11.5$

Best regards.

3 0

3 years ago

The reaction gives off heat energy, so it is an __________ reaction

Alex_Xolod [135]

Since the reaction gives off heat energy it is considered to be an exothermic reaction

3 0

3 years ago

Oxidation number of Al(OH)4

Kobotan [32]

Explanation:

The sum of all oxidation numbers in a neutral compound is zero. The sum of all oxidation numbers in a polyatomic (many-atom) ion is equal to the charge on the ion. The oxidation number of oxygen in a compound is usually –2. The oxidation state of hydrogen in a compound is usually +1.

The oxidation state of Al in Al(OH)

−

x+4(+1−2)=−1

∴x=+3

The oxidation state of Mn in MnO

y+2(−2)=0

∴y=+4

thank u

8 0

3 years ago