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bazaltina [42]
3 years ago
7

What agent of weathering is acid precipitation?

Chemistry
1 answer:
Alona [7]3 years ago
3 0
It would be considered acid rain in the sense of dew fall
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List what you are thankful for! (It's Thanksgiving! Answer if you bake! Or if you like cakes pies cookies and treats!!! )
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I do not bake but my mom is a professional baker.  

Explanation:

A roof over my head and a warm home.

Plenty of drinkable water.

I don't have to go hungry.

I can enjoy the small and free pleasures of life.

Access to the internet.

My friends and family.

My health.

The kindness of people I have never met before.

8 0
2 years ago
Read 2 more answers
HELP PLEEAAASSE ITS NOT AS BAD AS IT LOOKS I PROM ISE ITS PHYSICAL AND CHEMICAL CHANGE
lions [1.4K]
I can’t really understand your question sorry
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2 years ago
A compound with molecular formula C5H10O2 has the following 1H NMR spectrum: 1.15 (t, 3H), 1.25 (t, 3H), 2.33 (q, 2H), 4.13 (q,
Vilka [71]

Answer:

Ethyl propionate

Explanation:

Calculating the double bond equivalence as:

DBE =  C - (H/2) - (X/2) + (N/2) +1

Where  

C is the number of carbon atoms

N is the number of nitrogen atoms

X is the number of halogens

H is the number of hydrogen atoms

So, according to the formula, C_5H_{10}O_2

DBE =  5 - (10/2) - (0/2) + (0/2) +1 = 1

It means there is one double bond or one ring.

from the NMR signal, it is clear that compound has two triplet and two quardrate group that means two -CH3 and two -CH2 groups are present.

From the splitting information, it is clear that each the -CH2- group is next to a -CH3, and vice-versa. In this case ring structure is not possible.

The chemical shift of two protons 4.13 ppm(q, 2H) is for protons next to an O atom, therefore, compound must have

-OCH_2CH_3.

Then the signal at 2.33 is for a -CH2-group next to a C=O. So compound must have a structure CH3-CH2-C=O. Now, there is only one way to complete the structure - by bonding the O to the C=O carbon.

So the compound is ethyl propionate

3 0
3 years ago
A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c
natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

  • Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

  • The mass of one mole of CaCO₃ is the same as the molar mass of this compound: \rm 100\; g.
  • The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: \rm 40.078\; g.

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}.

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be \rm 30 \% \times 100\; g = 30\; g of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio \displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}:

\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}.

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%.

3 0
2 years ago
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