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Gnom [1K]
3 years ago
9

Which of the following is an example of land used as a protected area

Chemistry
1 answer:
LenaWriter [7]3 years ago
8 0
I think the answer is A. Grand Canyon
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Is C2H6 empirical molecular or both
marshall27 [118]

Answer:

molecular

Explanation:

7 0
3 years ago
In the EXPLORE section of your lesson 4.08 on Potential energy there were several animations to watch that provided a graphic il
Lunna [17]

Answer:

This is because no energy is being created or destroyed in this system

Explanation:

I think this is correct? I hope it helps.        

7 0
3 years ago
Read 2 more answers
Unit: Chemical Quantities
Vaselesa [24]

Answer:

(See explanation for further details)

Explanation:

1) The quantity of moles of sulfur is:

n = \frac{1.20\times 10^{24}\,atoms}{6.022\times 10^{23}\,\frac{atoms}{mol} }

n = 1.993\,moles

2) The number of atoms of helium is:

x = (1.5\,moles)\cdot \left(6.022\times 10^{23}\,\frac{atoms}{mole} \right)

x = 9.033\times 10^{23}\,atoms

3) The quantity of moles of carbon monoxide is:

n = \frac{4.15\times 10^{23}\,molecules}{6.022\times 10^{23}\,\frac{molecules}{mol} }

n = 0.689\,moles

4) The number of molecules of sulfur dioxide is:

x = (2.25\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)

x = 1.355\times 10^{24}\,molecules

5) The quantity of moles of sodium chloride is:

n = \frac{2.4\times 10^{23}\,molecules}{6.022\times 10^{23}\,\frac{molecules}{mol} }

n = 0.399\,moles

6) The number of formula units of magnesium iodide is:

x = (1.8\,moles)\cdot \left(6.022\times 10^{23}\,\frac{f.u.}{mole} \right)

x = 1.084\times 10^{24}\,f.u.

7) The quantity of moles of potassium permanganate is:

n = \frac{3.67\times 10^{23}\,f.u.}{6.022\times 10^{23}\,\frac{f.u.}{mol} }

n = 1.214\,moles

8) The number of molecules of carbon tetrachloride is:

x = (0.25\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)

x = 1.506\times 10^{23}\,molecules

9) The quantity of moles of aluminium is:

n = \frac{3.67\times 10^{23}\,atoms}{6.022\times 10^{23}\,\frac{atoms}{mol} }

n = 0.609\,moles

10) The number of molecules of oxygen difluoride is:

x = (3.52\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)

x = 2.120\times 10^{24}\,molecules

3 0
3 years ago
Calculate each of the following quantities.<br> (a) mass in kilograms of 3.7 x 1020 molecules of NO2
HACTEHA [7]

Answer:

The answer to your question is:  0.028 kg of NO2

Explanation:

Data

3.7 x 10²⁰ molecules of NO2 in kg

MW of NO2 = 14 + (16 x 2) = 14 + 32 = 46 kg

                   1 mol of NO2 ---------------------  6.023 x 10 ²³ molecules

                   x                     --------------------- 3.7 x 10²⁰ molecules

                   x = 3.7 x 10²⁰ x 1 / 6.023 x 10 ²³

                   x = 0.00061 mol

         

                 1 mol of NO2 ---------------------  46 kg of NO2

                 0.00061 mol     ------------------    x

                 x = 0.00061 x 46/1

                x = 0.028 kg of NO2

7 0
3 years ago
Use the following equation to answer the questions below.
kaheart [24]

Answer:

1-Double replacement

Explanation:

3 0
3 years ago
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