Jupiter's atmosphere is composed predominantly of hydrogen and helium, but if you have to select any one option then we can look at the percentage of existence of these elements that would be
<span>90 percent hydrogen.
remaining 10 percent is helium
so choose Hydrogen.</span>

3 0

3 years ago

Which statement best describes how today’s scientific community views light?

arlik [135]

ITS A SIMPLE.............................
FuFu Boys

5 0

3 years ago

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Here is a more complex redox reaction involving the dichromate ion in acidic solution: 3N O 2 − + 8H + + C r 2 O 7 2− → 3N O 3 −

Fantom [35]

Answer:

NO2- is the reducing agent.

Cr2O7_2- is the oxidizing agent.

H+ is neither

Explanation:

Reduction is the gain in electron. A chemical specie that undergoes reduction is called the oxidizing agent.

Oxidation is simply the loss in electrons. A chemical specie that undergoes oxidation is called the reducing agent.

Let us look at the species.

The first specie is the NO2-. In this specie, the oxidation number of nitrogen changed from +3 to +5 in NO3-. Thus we can see that there is more loss of electron to have caused an increase in the oxidation number positively. This shows an oxidation. Hence, NO2- is the reducing agent.

Let us look at the chromium. We can see that the oxidation number of chromium changed from +7 to +3.

Now we can see that it is a decrease and hence, it is a gain of electron and thus it is reduction. This means the first chromium specie is the oxidizing agent.

The hydrogen ion is simply placed there to balance the ions and hence it is neither the oxidizing nor the reducing agent.

4 0

3 years ago

A slender uniform rod 100.00 cm long is used as a meter stick. two parallel axes that are perpendicular to the rod are considere

kvv77 [185]

Refer to the diagram shown below.

The second axis is at the centroid of the rod.
The length of the rod is L = 100 cm = 1 m

The first axis is located at 20 cm = 0.2 m from the centroid.
Let m = the mass of the rod.

The moment of inertia about the centroid (the 2nd axis) is
$I_{g} = \frac{mL^{2}}{12} = (m \, kg) \frac{(1 \, m)^{2}}{12} = \frac{m}{12} \, kg-m^{2}$

According to the parallel axis theorem, the moment of inertia about the first axis is
$I_{1} = I_{g} + (m \, kg)(0.2 m)^{2} \\ I_{1} = \frac{m}{12}+ 0.04m = 0.1233m \, kg-m^{2}$

The ratio of the moment of inertia through the 2nd axis (centroid) to that through the 1st axis is
$\frac{I_{g}}{I_{1}} = \frac{0.0833m}{0.1233m} =0.6756$

Answer: 0.676

6 0

3 years ago

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