How do matter cycles demonstrate the conservation of matter in a system?

Chemistry

1 answer:

damaskus [11]3 years ago

4 0

Answer:

These systems are defined both by the types of energy and matter they contain and by how that matter and energy move through and between systems. In natural systems, both energy and matter are conserved within a system. This means that energy and matter can change forms but cannot be created or destroyed.

Explanation:

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Fill in the blanks below.

Elenna [48]

Answer:

Explanation:

6CO₂ + 6 H₂O ⇄ C₆H₁₂0₆ + 6O₂

This is the chemical equation given .

1. The equation shows a __Chemical equation_______the breaking and forming of chemical bonds that leads to a change in the composition of matter.

2. In the equation, CO₂ is a___reactant_____.

3. In the equation, C₆H₁₂0₆ is a ___product________.

4. In O₂, the type of bond that holds the two oxygen atoms together is a_nonpolar_covalent bond_________.

5. In H₂O, the type of bond that holds one of the hydrogen atoms to the oxygen atom is a__polar_hydrogen bond____.

6. The number of oxygen atoms on the left side of the equation is__equal to_________ the number of oxygen atoms on the right side.

5 0

3 years ago

An unknown triprotic acid (H3A) is titrated with NaOH. After the titration Ka1 is determined to be 0.0013 and Ka2 is determined

e-lub [12.9K]

Answer:

6.68 X 10^-11

Explanation:

From the second Ka, you can calculate pKa = -log (Ka2) = 6.187

The pH at the second equivalence point (8.181) will be the average of pKa2 and pKa3. So,

8.181 = (6.187 + pKa3) / 2

Solving gives pKa3 = 10.175, and Ka3 = 10^-pKa3 = 6.68 X 10^-11

7 0

3 years ago

The half life for the decay of carbon-14 is 5.73 x 10 years. Suppose the activity due to the radioactive decay of the carbon-14

Elena-2011 [213]

Answer:

Age of the atifact is $4.2\times 10^{2}$ years

Explanation:

For first -order radioactive decay- $A_{t}=A_{0}(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}$
$A_{t}$ represents activity of radioactive nuclide after t time, $A_{0}$ represents initial activity of radioactive nuclide and $t_{\frac{1}{2}}$ represents half-life
Here, $A_{t}=19Bq$ , $A_{0}=20Bq$ and $t_{\frac{1}{2}}=5.73\times 10^{3}years$