# Lead nitrate can be decomposed by heating. what is the percent yield of the decomposition reaction if 9.9 g of pb(no3)2are heated to give 5.5 g of pbo? 2pb(no3)2( s) → 2pbo( s) + 4no2( g) + o2( g)

## Related Questions

It contains carbon atoms arranged in a ring structure.
B.
It has double bonds between all the carbon atoms.
C.
It contains molecules of many different types.
D.
It is made of large molecules with repeating units.

Explanation:

A polymer is a large molecule which contains repeating units.

A polymer forms long chain of atoms which are not necessarily arranged in a ring structure. There are not only double bonds between all the carbon atoms, as there can be single bonds or triple bonds between the atoms of a polymer.

A polymer does not contain many different type of molecules as there are many repeating units and not many molecules in a polymer.

A polymer is made of large molecules with repeating units.

Thus, we can conclude that option (D) is the correct answer.

Silicon has three naturally occurring isotopes (Si-28, Si-29, and Si-30). The atomic mass and natural abundance of Si-28 are 27.9769 amu and 92.2 %, respectively. The atomic mass and natural abundance of Si-29 are 28.9765 amu and 4.67 %, respectively. Find the atomic mass of Si-30.

The atomic mass of Si-30 is 29.955 amu

Explanation:

Step 1: Data given

Si-28 has an atomic mass of  27.9769 amu and a natural abundance of 92.2 %

Si-29 has an atomic mass of  28.9765 amu and a natural abundance of 4.67%

The atomic mass of Si is 28.0855 amu

Step 2: Calculate natural abundance of Si-30

100% - 92.2 % - 4.67% = 3.13 %

0.922 * 27.9769 + 0.0467 * 28.9765 + 0.0313* X = 28.0855

27.14790435 + 0.0313X = 28.0855

0.0313X = 0.93760

X = 29.955 amu

The atomic mass of Si-30 is 29.955 amu

Based upon the pH scale, a vegetable with a pH of 5.3 would be

It would not be a base, a base on the pH scale is from 8-14, an acid is from 1-6. There for it would be an acid.
A vegetable with a pH of 5.3 would be base, not acid. One exemple is celery

A student was given a solid containing a mixture of nitrate salts. The sample completely dissolved in water, and upon addition of dilute HCl, no precipitate formed. The pH was lowered to about 1 and H2S was bubbled through the solution. No precipitate formed. The pH was adjusted to 8 and H2S was again bubbled in. This time, a precipitate formed. Which compounds might have been present in the unknown?

• Ni(NO3)2
• Co(NO3)2
• Ca(NO3)2
• NaNO3

The last two require further qualitative analysis. So with only H2S tests the first two will appear.

Explanation:

The other ions would have turned solid before the second addition of H2S.