Describe the role of chloroplasts in photosynthesis. (ASAP)

Answers

Answer 1

Answer:

Chloroplasts absorb sunlight and use it in conjunction with water and carbon dioxide gas to produce food for the plant. Chloroplasts capture light energy from the sun to produce the free energy stored in ATP and NADPH through a process called photosynthesis.

Explanation:

Answer 2
Sunlight is consumed by chloroplasts and used in combination with water and carbon dioxide gas to make food for the plant. In order to produce the free energy contained in ATP and NADPH via a process called photosynthesis, chloroplasts absorb light energy from the sun.

Related Questions

If you start with 34.0 grams of silver (I) nitrate, how many grams of solid silver would be
produced?

Answers

Answer:

6.36  g AgCl

Hope this helps!

the average kinetic energy of the gas molecule is greatest in which container

Answers

Answer:

Explanation:

average kinetic energy of a gas molecule really only depends on the temperature, so which ever container is at the highest temperature is the answer

In comparison to other fundamental states of matter, plasma has the highest kinetic energy. This is due to the fact that particles in a plasma travel more quickly than those in a corresponding solid, liquid, or gas.

What is kinetic energy ?

The energy that an object has as a result of motion is known as kinetic energy. It is described as the effort required to move a mass-determined body from rest to the indicated velocity. The body holds onto the kinetic energy it acquired during its acceleration until its speed changes.

The reason why molecules in gases have the largest kinetic energy is because they have more space between one another, experience less intermolecular force, and travel at a faster rate, which results in higher energy.

The temperature has a direct impact on the average kinetic energy of the particles inside a container.

Thus, In comparison to other fundamental states of matter, plasma has the highest kinetic energy.

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100. g of ice at 0 is added to 300.0 g of water at 60C. Assuming no transfer of heat to
the surroundings, what is the temperature of the liquid water after all the ice has
melted?

Answers

Heat of fusion of water = 333 J/g 100 g ice x 333 J/g = 33300 J (energy to melt 100 g ice at 0 deg C) 300

The final temperature of the liquid water after all the ice has melted is equal to 25°C.

What is the specific heat capacity?

The specific heat capacity can be described as the quantity of heat needed to raise the temperature in one unit (1kg) of substance by one-degree Celcius.

The formula to show the relationship between the specific heat capacity and heat absorbed or lost is:

Q = mCΔT

Given, the amount of ice, m = 100g

The amount of water, M = 300g

The specific heat capacity of the water, C = 1 cal /g°C

The heat of fusion of ice, C = 80 cal/g

Heat given by water = Heat taken by ice + Heat taken by water formed from ice

300 × 1 ×(60 - T) = 100 × 80  + 100 × 1 ×T

18000 - 300 T = 8000 + 100 T

400 T = 10000

T = 25°C

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Which compound most likely has Atoms Held together by ionic bonds?

A. Magnesium sulfide, MgS

B. Hydrogen peroxide, H2O2

C. Ammonia, NH3

D. Carbon monoxide, CO

Answers

Answer:

MgS Magnesium sulfide

according to the conveyor belt mode of ocean circulation, what happens when water reaches the poles

Answers

Answer: Some salt is trapped between ice crystals, but most are left behind in the unfrozen seawater.

Explanation:

Answer: The salinity of the water increases

Explanation: Hope this helps!

PLEASE HELP ME ASAP

N2 + 3H2 → 2NH3
How many grams of ammonia,
NH3, would be formed from the
complete reaction of 4.50 moles
hydrogen, H2?

Answers

Answer:

Mass =   51 g

Explanation:

Given data:

Mass of ammonia formed = ?

Number of moles of hydrogen = 4.50 mol

Solution:

Chemical equation:

N₂ + 3H₂     →     2NH₃

Now we will compare the moles of hydrogen and ammonia.

                H₂           :            NH₃

                  3           :              2

                 4.50       :           2/3×4.50 = 3 mol

Mass of ammonia formed:

Mass = number of moles × molar mass

Mass = 3 mol × 17 g/mol

Mass =   51 g

What is the mass of the oxygen in one mole of calcium phosphate

Answers

Answer:

16.00 g/mol

Explanation:

Answer:

16.00 g/mol

Explination:

1 mol of calcium phosphate contains 8 moles of oxygen. From this, we can compute for the amount of oxygen in grams using the molar mass of oxygen, which is 16.00 g/mol.

Consider the reaction: 3Co2+(aq) + 6NO3¯(aq) + 6Na+(aq) + 2PO43¯(aq) â Co3(PO4)2(s) + 6Na+(aq) + 6NO3¯(aq) Identify the net ionic equation for this reaction.A. 3CO2(aq) + 6NO3(aq) + 6Na+(aq) + 2PO43-(aq) â Co3(PO4)2(s) + 6NaNO3(aq). B. Na+(aq) + NO3-(aq) â NaNO3(aq).C. 3CO2(aq) + NO3(aq) + Na*(aq) + 2PO43-(aq) â Co3(PO4)2(s) + NaNO3(aq).D. 3C02(aq) + 2PO43-(ag) â CO3(PO4)2(s).E. None of these choices are correct.

Answers

Answer:

D. 3 Co²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Co₃(PO₄)₂(s)

Explanation:

The complete ionic equation includes all the ions and the insoluble species. Let's consider the following complete ionic equation.

3 Co²⁺(aq) + 6 NO₃⁻(aq) + 6 Na⁺(aq) + 2 PO₄³⁻(aq) ⇒ Co₃(PO₄)₂(s) + 6 Na⁺(aq) + 6 NO₃⁻(aq)

The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and the insoluble species. The corresponding net ionic equation is:

3 Co²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Co₃(PO₄)₂(s)

(a) Given that the path length of the cuvette is 1 cm, what is the extinction coefficient of the 0.020 mM Yellow 5 dye at lambda max?(b) Given the data in the table below, what is the concentration of the diluted solution?Stock solution concentration 0.075 MStock solution absorbance 1.84Diluted solution absorbance 0.78

Answers

Answer:

(a). Kindly check the explanation section.

(b). 24.5 M^-1 cm^-

(c). 0.0318 M

Explanation:

So, we are given the following data parameters which is going to aid us in solving the above Question;

The path length = 1cm, extinction coefficient = unknown, absorbance is unknown and concentration= 0.02mM.

Using the formula below; we can determine the extinction coefficient as;

Absorbance = extinction coefficient × concentration × path length. -----------(1).

(a). Since, the absorbance is not given in the Question it won't be possible to determine the value of the extinction coefficient. Thus, say the value of absorbance = A.

Then, extinction coefficient = A/ 0.02 × 1 cm.

(b). Making use of the formula above, the extinction coefficient can be Determine as follows;

extinction coefficient = 1.84/0.075 × 1 = 24.5 M^-1 cm^-1.

(c). The concentration can be Determine by also making use of the formula above and making the concentration the subject of the formula;

Concentration = absorbance/ extinction coefficient × path length.

Concentration = 0.78/24.5 M^-1 cm^-1. × 1cm = 0.0318 M.

In an aqueous solution of a certain acid the acid is 0.094% dissociated and the pH is 4.55. Calculate the acid dissociation constant Ka of the acid. Round your answer to 2 significant digits

Answers

Answer:

Explanation:

pH = 4.55

[ H⁺ ] = 10⁻⁴°⁵⁵

= 2.82 x 10⁻⁵

Let the acid be HA and its molar concentration be a .

         HA         ⇄       H ⁺       +        A ⁻

a - .094x10⁻³a      .094x10⁻³a    .094x10⁻³a

.094x10⁻³a  = 2.82 x 10⁻⁵

a = 30 x 10⁻²

= .3

a - .094x10⁻³a   = .3 - .094 x 10⁻³ x .3

= .29997 approx

Ka =  2.82 x 10⁻⁵ x  2.82 x 10⁻⁵ /  .29997

= 7.95 x10⁻¹⁰ / .29997

= 26.5 x 10⁻¹⁰

= 27 x 10⁻¹⁰ ( rounding off to two digits )

At equilibrium, the value of [H ] in a 0.240M solution of an unknown acid is 0.00417M . Determine the degree of ionization and the Ka of this acid.

Answers

Answer:

[tex]ionization=1.74\%[/tex]

Explanation:

Hello!

In this case, since the degree of ionization of an acid is computed in terms of the concentration of hydrogen ions and the initial concentration of the acid:

[tex]ionization=\frac{[H^+]}{[HA]} *100\%[/tex]

Because the ionization reaction is represented by:

[tex]HA\rightleftharpoons H^++A^-[/tex]

Therefore, the degree or percent ionization turns out:

[tex]ionization=\frac{0.00417M}{0.240M} *100\%\\\\ionization=1.74\%[/tex]

Best regards!


Which lists the elements in order from least conductive to most conductive?

Answers

Answer:

Answer. Answer: Nitrogen (N), antimony (Sb), bismuth (Bi) is the order from least conductive to most conductive.

Answer:

nitrogen (N), antimony (Sb), bismuth (Bi)

What is the mole fraction of water in a solution that contains 8.0 mol of ethanol (C2H5OH) and 1.6 mol of water?

Answers

Answer:

the mole fraction of water in a solution is 0.17

Explanation:

The computation of the mole fraction of water in a solution is shown below:

Given that

Ethanol be 8.0 mol

And, the water be 1.6 mol

Based on the above information, the mole fraction of water in a solution is

= Water ÷ (Water + ethanol)

= 1.6 mol ÷ (1.6 mol + 8.0 mol)

= 1.6 mol ÷ 9.6 mol

= 0.17

Hence, the mole fraction of water in a solution is 0.17

I have 2 samples of solid chalk (aka calcium carbonate). Sample A has a total mass of 4.12 g and Sample B has a total mass of 19.37 g. What is the difference between the samples?

A) Sample B has more calcium carbonate molecules
B) Sample B has a larger ratio of carbon, oxygen, and calcium atoms
C) Sample B has more calcium ion than carbonate ions
D) Sample B must have some impurity

Answers

Answer:

A) Sample B has more calcium carbonate molecules

Explanation:

M = Molar mass of calcium carbonate = 100.0869 g/mol

[tex]N_A[/tex] = Avogadro's number = [tex]6.022\times 10^{23}\ \text{mol}^{-1}[/tex]

For the 4.12 g sample

Moles of a substance is given by

[tex]n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{4.12}{100.0869}\\\Rightarrow n=0.0411\ \text{mol}[/tex]

Number of molecules is given by

[tex]nN_A=0.0411\times 6.022\times 10^{23}=2.48\times 10^{22}\ \text{molecules}[/tex]

For the 19.37 g sample

[tex]n=\dfrac{19.37}{100.0869}\\\Rightarrow n=0.193\ \text{mol}[/tex]

Number of molecules is given by

[tex]nN_A=0.193\times 6.022\times 10^{23}=1.16\times 10^{23}\ \text{molecules}[/tex]

[tex]1.16\times 10^{23}\ \text{molecules}>2.48\times 10^{22}\ \text{molecules}[/tex]

So, sample B has more calcium carbonate molecules.

The ratio of the elements of carbon, oxygen, calcium atoms, ions, has to be same in both the samples otherwise the samples cannot be considered as calcium carbonate. Same is applicable for impurities. If there are impurites then the sample cannot be considered as calcium carbonate.

Calculate the molarity of a solution that contains 183.51 grams of lead (II) bromide in 500.0 mL of the solution

Answers

Answer: 1M

Explanation:

Molarity = mols/L

moles of lead bromide: 183.51/ 367.0 = 0.5 mol

500 ml/ 1000 mL = .5L

.5 mol / .5 L = 1 mol/L = 1M

Which of the following is NOT a property of gases?
Your answer:
A.They are easy to compress.
B.They expand to fill their containers.
C.They have a definite shape and a definite volume.
D.They occupy far more space than the liquids or solids from which they form.

Answers

Answer: C

Explanation:

A. As they do not have a given shape and expand to fill the container, modifying the size of the container modifies the space the gas occupies

B. They are nearly free molecules, which means no bound between them, moving fast and freely around the container

C. NOT A PROPERTY. As said before, as there are no bindings between molecules and move freely, they cannot have a fixed size cause they're always on the move.

D. Because they expand to fill the container, it's easy to them to occupy more space than in any other state, as the container is the only thing that defines how much volume they occupy

given that the only known ionic charges of lead are pb(ii) and pb(iv), how can you explain the existence of the pb2o3 salt

Answers

Answer:

See Explanation

Explanation:

Pb2O3 is better formulated as PbO.PbO2. It is actually a mixture of the two oxides of lead, lead II oxide and lead IV oxide.

This implies that this compound Pb2O3  (sometimes called lead sesquioxide) is a mixture of the oxides of lead in its two known oxidation states +II and +IV.

Hence Pb2O3  contains PbO and PbO2 units.

Calculate the volume in milliliters of 1.57 M potassium hydroxide that contains 10.3 g of solute.

Answers

Answer:

[tex]V = 0.117 \ L[/tex]

General Formulas and Concepts:

Chem

Reading a Periodic TableWriting compounds and polyatomic ionsMolarity = moles of solute / liters of solution

Explanation:

Step 1: Define

1.57 M KOH (potassium hydroxide)

10.3 g KOH

Step 2: Define conversions

Molar Mass of K - 39.10 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of H - 1.01 g/mol

Molar Mass of KOH - 39.10 + 16.00 + 1.01 = 56.11 g/mol

Step 3: Convert

[tex]10.3 \ g \ KOH(\frac{1 \ mol \ KOH}{56.11 \ g \ KOH} )[/tex] = 0.183568 mol KOH

Step 4: Solve for Volume

Substitute:                    [tex]1.57 \ M=\frac{0.183568 \ mol}{x \ L}[/tex]Move x:                         [tex]x1.57 \ M=0.183568 \ mol[/tex]Isolate x:                       [tex]x=\frac{0.183568 \ mol}{1.57 \ M}[/tex]Evaluate:                      [tex]x=0.116922 \ L[/tex]

Step 5: Check

We are given 3 sig figs. Follow sig fig rules.

[tex]0.116922 \ L \approx 0.117 \ L[/tex]

Metamorphic rocks with a non-foliated texture show metamorphic change that involves ____.
a.
mineral grains arranging into layers
b.
growth in the size of the mineral grains
c.
mineral grains flattening under pressure
d.
mineral grain melting

Answers

Answer:

b.  growth in the size of the mineral grains

Explanation:

Non-foliated texture shown by a metamorphic change is depicted by growth in the size of the mineral grains.

Examples of non-foliated metamorphic rocks are quartzite and marble. In these metamorphic rocks, mineral grains are not aligned with their long axis. Non-foliated texture occurs under high temperature and low pressure conditions.As minerals are able to grow, the size can be used to show a metamorphic change.

Answer:

its b home slice

Explanation:

How do living systems follow the laws of conservation of mass?

Answers

Answer:

they use it when a living organism dies.

Explanation:

when d organism decomposes carbon is released that enters back into d earth. this follows d law of mass

Suppose the Cu ions are produced by the reaction of 0.94 g of copper turnings with excess nitric acid. How many moles of Cu are produced?

Answers

Answer:

0.0148 moles

Explanation:

Solid copper is oxidized with HNO₃:

Cu + 4HNO₃ → Cu²⁺ + 2 NO₃⁻ + 2H₂O + 2NO₂

Where 1 mole of Cu produce 1 mole of Cu²⁺ when nitric acid is in excess.

Moles of Cu that react (Molar mass Cu = 63.546g/mol):

0.94g * (1mol / 63.546g) = 0.0148 moles of Cu

And moles of Cu²⁺ produced are also:

0.0148 moles

Washing machines use a large amount of water. A student suggested that old pairs of stained jeans which have to be washed more frequently should be replaced by new pairs of jeans to conserve water. Which of these statements best describes the suggestion made by the student? Question 9 options: It is not practical because an old pair of jeans needs less water to be washed than a new pair of jeans. It is practical because a huge amount of water can be conserved by this method. It is practical because it is easy to implement. It is not practical because it takes a huge amount of water to produce a new pair of jeans.

Answers

Answer:

C |||| It is not practical because it takes a huge amount of water to produce a new pair of jeans

Explanation:

If you're doing flvs then it's C.

Answer:

C!

Explanation:

i got it right on the test UwU

How many grams of CO2 are produced by the combustion of 344 g of a mixture that is 33.6% CH4 and 66.4% C3H8 by mass

Answers

Answer:

1,002.936 g

Explanation:

The combustion equation of each will be;

CH4 + 2O2 = CO2 + H2O

C3H8 + 5O2 = 3CO2 + 4H2O

We are told the mixture is 344g.

Thus;

For CH4 combustion;

Amount of CH4 = 0.336 × 344 = 115.584g

Molar mass of CH4 is 16 g/mol

Number of moles of CH4 is;

n = 115.584g/(16 g/mol)

n = 7.224 moles

n(CO2) = n(CH4) = 7.224 moles

For C3H8 combustion;

Amount of C3H8 = 0.664 × 344 = 228.416 g

Molar mass of C3H8 = 44 g/mol

Thus;

Number of moles of C3H8 = 228.416 g/(44 g/mol) = 5.19 mol

n(CO2) = 3n(C3H8) = 3 × 5.19 = 15.57 moles

Total moles of CO2 = 7.224 moles + 15.57 moles = 22.794 mol

Molar mass of CO2 = 44 g/mol

Thus amount of CO2 by mass = 22.794 mol × 44 g/mol = 1,002.936 g

Which element would most likely have a chemical reactivity similar to bromine?

Answers

Answer:

Bromine is a very reactive element. While it is less reactive than fluorine or chlorine, it is more reactive than iodine. It reacts with many metals, sometimes very vigorously. For instance, with potassium, it reacts explosively.

Explanation:

Answer: Bromine is a high reactive element. While it is not as reactive as fluorine or chlorine, it is still more reactive than iodine. It reacts with many metals, and even sometimes very aggressively. For example, when mixed with potassium, it will explode.  

Explanation:

An igneous intrusion show that a magma body emerged in a stratigraphic section. This magma body is _____ than the rocks around it.

A) older
B) the same age
C) newer

Answers

Answer:

Younger/newer

Explanation:

Stratagraphic means an intrusion or fault is younger than the layer it affects.

How many molecules are in 2.0 grams of
Na2SO4?

Answers

Answer:

450 grams of Na2SO4

Explanation:

hope that helps

Answer

450 grams of Na2SO4.

Hope this helps!

How many valence electrons are found in P3- ?

Answers

Answer:

5 valence elctrons

Explanation:

the outer most orbitals, 3s2 and 3p3 contains 5 electrons, thus valences electrons for P is 5

Compound Y has a distribution coefficient of 4.0 when extracted from water with chloroform, with Y being more soluble in chloroform. How many 10.0 mL chloroform extractions would be required to extract at least 95% of Y from a 50.0 mL aqueous solution of water?

Answers

For the extraction of at least 95% of Y in water, 3 extractions are to be performed.

Distribution coefficient can be defined as the ratio of the concentration of solute in an organic solvent to water.

Distribution coefficient = [tex]\rm \dfrac{concentration\;in\;solvent}{concentration\;in\;water}[/tex]

For the first extraction, the concentration in organic solvent = X per 10 ml

Concentration in water = 100-X per 50 ml

Distribution coefficient = [tex]\rm \dfrac{\frac{X}{10} }{\frac{100-X}{50} }[/tex]

4 = [tex]\rm \dfrac{\frac{X}{10} }{\frac{100-X}{50} }[/tex]

4 = [tex]\rm \dfrac{50X}{1000-X}[/tex]

4000 - 4X = 50X

X = 74.1 %

Thus, after the first extraction, the amount of Y extracted is 74.1%.

We have to extract at least 95% of Y. Thus, the second extraction is performed.

Remaining y = 100 - 74.1

Remaining y = 25.9%

The concentration in organic solvent = X per 10 ml

Concentration in water = 25.9 -X

Distribution coefficient = [tex]\rm \dfrac{\frac{X}{10} }{\frac{25.9-X}{50} }[/tex]

4 = [tex]\rm \dfrac{\frac{X}{10} }{\frac{25.9-X}{50} }[/tex]

4 = [tex]\rm \dfrac{50X}{259-X}[/tex]

1036 - 4X = 50X

X = 19.2%

Thus, after the second extraction the amount of Y extarcted = first extraction + second extraction

The amount of Y extracted = 74.1 + 19.2 %

The amount of Y extracted = 93.3%

To reach at least 95% extraction, the third extraction has to be performed.

The remaining Y for third extraction = 100 - 93.3

The remaining Y for the third extraction = 6.7%

Concentration in water = 100 - 6.7

Distribution coefficient = [tex]\rm \dfrac{\frac{X}{10} }{\frac{6.7-X}{50} }[/tex]

4 = [tex]\rm \dfrac{\frac{X}{10} }{\frac{6.7-X}{50} }[/tex]

4 = [tex]\rm \dfrac{50X}{67-X}[/tex]

268 - 4X = 50X

X = 5.0%

The total extraction after third extraction = 93.3 + 5%

The total extraction after third extraction = 98.3%.

Thus for the extraction of at least 95% of Y in water, 3 extractions are to be performed.

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Calculate the maximum mass of a metal, with equivalent mass of 20 g, that should be used in an equivalent mass determination with a 100-mL eudometer tube for collection. For the purposes of calculation, assume 80 mL of hydrogen gas produced at standard conditions. (Round to 2 sig figs)

Answers

Answer:

Explanation:

In the equivalent mass measurement , one gram equivalent of metal produces 11200 mL of hydrogen gas .

11200 mL hydrogen gas is produced by 1 gram equivalent of gas at STP

11200 mL hydrogen gas is produced by 20 g  of gas at STP

80 mL hydrogen gas is produced by 20 x 80 / 11200  g  of gas at STP

= 142.85 mg .

= 140 mg . ( rounded to 2 sig figures )

how do you determine number of valence electrons an element has

Answers

Answer:the first column has one valance electron the second has 2 and so on.

The only exception is helium that has 2.

Explanation:

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