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3 years ago

8

Create: Click Reset. Use the Gizmo to create a path in which the carbon atom goes from the atmosphere to the hydrosphere, biosph

ere and geosphere. Describe each transition briefly.

1 answer:

Solnce55 [7]3 years ago

7 0

Answer:

From the atmosphere to the hydrosphere by diffusion.

From the atmosphere to the biosphere by photosynthesis.

From the atmosphere to the geosphere by rainfall.

Explanation:

Carbon atom goes from the atmosphere to the hydrosphere by the process of diffusion because there is high concentration of carbondioxide present in the atmosphere. The carbon atom goes from the atmosphere to the biosphere by the process of photosynthesis in plants which uses carbondioxide gas as a raw material in the process for the preparation of organic compounds such as glucose. The carbon atom goes from the atmosphere to the geosphere with the help of rain. When carbondioxide gas react with water in the atmosphere, carbonic acid is formed and comes to the ground through rainfall.

You might be interested in

Barium-142 is used as a GI radiocontrast agent. After 1.25 hours, 9.25 μg remains in the patient. Determine the original dose gi

Trava [24]

Answer:

The answer to your question is the original dose of Ba-142 was 1184 μg

Explanation:

Data

Total time = 1.25 hours

The Final amount of Ba = 9.25 μg

The Half-life of Ba = 10.6 minutes

Process

1.- Convert total time to minutes

1 h ----------------- 60 min

1.25 h ------------ x

x = 75 min

2.- Draw a table of this process

Final amount of Ba Time

9.25 75 min

18.5 64.4 min

37 53.8 min

74 43.2 min

148 32.6 min

296 22 min

592 11.4 min

1184 0.8 min

4 0

4 years ago

What properties you can observe or measure?

Advocard [28]

We can observe physical properties of elements and compounds without changing the substance.

Examples of physical properties: Density, color, boiling point, state of matter, appearance: dull or shiny, etc.

But we can also observe and measure chemical properties by reacting a substance with something else. For example, like mixing baking soda and vinegar together. The vinegar reacts with the baking soda and produces carbon dioxide: a new substance.

Some examples of chemical properties: Flammability, amount of heat that is released during combustion, toxicity (how much damage it causes to other organisms), radioactivity, and ability to oxidize (when you have metal that becomes rusty looking).

4 0

4 years ago

Glycolysis is the process by which energy is harvested from glucose by living things. Several of the reactions of glycolysis are

nika2105 [10]

Answer: Reaction A: pi + glucose ⇒ glucose-6-phosphate + H2O ΔG = 13.8 kJ/mol

Reaction B: pi + frutose-6-phosphate ⇒fructose-1,6-biphosphate + H2O ΔG = 16.3kJ/mol

Explanation: ΔG is the representation of the change in Gibbs Free Energy and relates enthalpy and entropy in a single value, which is:

ΔG = ΔH - TΔS

where:

ΔH is enthalpy

T is temperature

ΔS is entropy (measure of the )

It can also predict the direction of the reaction with the conditions of temperature and pressure being constant.

When the change is positive, the reaction is non-spontaneous, which means the reaction needs external energy to occur. If the change is negative, it is spontaneous, i.e., happens without external help.

Analyzing the reaction, we see that reaction A and B have a positive ΔG, while reaction C is negative, so the reaction that are unfavorable or nonspontaneous are <u>reactions A and B</u>.

6 0

3 years ago

Diamonds are measured in carats, and 1 carat=0.200 g. The density of diamond is 3.51 g/cm^3. What is the volume of a 5.0 - carat

Blizzard [7]

Answer:- Volume of the 5.0 carat diamond is $0.28cm^3$ .

Solution:- It is a unit conversion problem. Need to convert carat to grams and then using grams and density the volume could easily be calculated.

We have 5.0 carat diamond and asked to calculate it's volume. Density of diamond is given as $3.51\frac{g}{cm^3}$ and 1 carat = 0.200 g. Let's do this step by step:

First step is the conversion of carat to grams:

$5.0carat(\frac{0.200g}{1carat})$ = 1.0 g

In second step, grams are converted to volume using the given density as:

$volume=\frac{mass}{density}$

$volume=1.0g(\frac{1cm^3}{3.51g})$

$volume=0.28cm^3$

So, the volume of the 5.0 carat diamond is $0.28cm^3$ .

6 0

3 years ago

Iron and vanadium both have the BCC crystal structure and V forms a substitutional solid solution in Fe for concentrations up to

Bess [88]

Answer:

Explanation:

To find the concentration; let's first compute the average density and the average atomic weight.

For the average density $\rho_{avg}$ ; we have:

$\rho_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }$

The average atomic weight is:

$A_{avg} = \dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }$

So; in terms of vanadium, the Concentration of iron is:

$C_{Fe} = 100 - C_v$

From a unit cell volume $V_c$

$V_c = \dfrac{n A_{avc}}{\rho_{avc} N_A}$

where;

$N_A$ = number of Avogadro constant.

SO; replacing $V_c$ with $a^3$ ; $\rho_{avg}$ with $\dfrac{100}{ \dfrac{C_{Fe} }{\rho_{Fe}} + \dfrac{C_v}{\rho_v} }$ ; $A_{avg}$ with $\dfrac{100}{ \dfrac{C_{Fe} }{A_{Fe}} + \dfrac{C_v}{A_v} }$ and

$C_{Fe}$ with $100-C_v$

Then:

$a^3 = \dfrac { n \Big (\dfrac{100}{[(100-C_v)/A_{Fe} ] + [C_v/A_v]} \Big) } {N_A\Big (\dfrac{100}{[(100-C_v)/\rho_{Fe} ] + [C_v/\rho_v]} \Big) }$

$a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100-C_v)A_{v} ] + [C_v/A_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100-C_v)/\rho_{v} ] + [C_v \rho_{Fe}]} \Big) }$

$a^3 = \dfrac { n \Big (\dfrac{100 \times A_{Fe} \times A_v}{[(100A_{v}-C_vA_{v}) ] + [C_vA_Fe]} \Big) } {N_A \Big (\dfrac{100 \times \rho_{Fe} \times \rho_v }{[(100\rho_{v} - C_v \rho_{v}) ] + [C_v \rho_{Fe}]} \Big) }$

Replacing the values; we have:

$(0.289 \times 10^{-7} \ cm)^3 = \dfrac{2 \ atoms/unit \ cell}{6.023 \times 10^{23}} \dfrac{ \dfrac{100 (50.94 \g/mol) (55.84(g/mol)} { 100(50.94 \ g/mol) - C_v(50.94 \ g/mol) + C_v (55.84 \ g/mol) } }{ \dfrac{100 (7.84 \ g/cm^3) (6.0 \ g/cm^3 } { 100(6.0 \ g/cm^3) - C_v(6.0 \ g/cm^3) + C_v (7.84 \ g/cm^3) } }$

$2.41 \times 10^{-23} = \dfrac{2}{6.023 \times 10^{23} } \dfrac{ \dfrac{100 *50*55.84}{100*50.94 -50.94 C_v +55.84 C_v} }{\dfrac{100 * 7.84 *6}{600-6C_v +7.84 C_v} }$

$2.41 \times 10^{-23} (\dfrac{4704}{600+1.84 C_v})=3.2 \times 10^{-24} ( \dfrac{284448.96}{5094 +4.9 C_v})$

$\mathbf{C_v = 9.1 \ wt\%}$

4 0

3 years ago