Classify each phrase according to whether it applies to photophosphorylation, oxidative phosphorylation, or both
Photophosphorylation Oxidative phosphorylation Both
1. occurs in plants produces ATP
2. occurs in chloroplasts
3. occurs in mitochondria
4. involves a larger electrical component
5. involves a smaller electrical component
6. involves a proton gradient

Answer:

Density = 19.3 g/cm³

Explanation:

In order to answer this question we need to keep in mind the following definition of density:

As both the mass and the volume are given by the problem, we can proceed to calculate the density of gold:

Answer:

potentiol or kenetic

Explanation:

The question is incomplete, the complete question is:

A certain substance X has a normal freezing point of [tex]-6.4^oC[/tex] and a molal freezing point depression constant [tex]K_f= 3.96^oC.kg/mol[/tex]. A solution is prepared by dissolving some glycine in 950. g of X. This solution freezes at [tex]-13.6^oC[/tex]. Calculate the mass of urea that was dissolved. Round your answer to 2 significant digits.

Answer: 129.66 g of glycine will be dissolved.

Explanation:

Depression in the freezing point is the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

[tex]\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m[/tex]

                                 OR

[tex]\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times \frac{\text{Given mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent (g)}}[/tex]        ....(1)

where,

i = Van't Hoff factor = 1 (for non-electrolytes)

Freezing point of pure solvent = [tex]-6.4^oC[/tex]

Freezing point of solution = [tex]-13.6^oC[/tex]

[tex]K_f[/tex] = freezing point depression constant  = [tex]3.96^oC/m[/tex]

[tex]M_{solute}[/tex] = Molar mass of solute (glycine) = 75.07 g/mol

[tex]w_{solvent}[/tex] = Mass of solvent = 950 g

Plugging values in equation 1:

[tex]-6.4-(-13.6)=1\times 3.96\times \frac{\text{Given mass of glycine}\times 1000}{75.07\times 950}\\\\\text{Given mass of glycine}=\frac{7.2\times 75.07\times 950}{1\times 3.96\times 1000}\\\\\text{Given mass of glycine}=129.66g[/tex]

Hence, 129.66 g of glycine will be dissolved.

Answer:

A 1.0 g sample of propane, C3H8, was burned in the calorimeter.

The temperature rose from 28.5 0C to 32.0 0C and the heat of combustion 10.5 kJ/g.

Calculate the heat capacity of the calorimeter apparatus in kJ/0C

Explanation:

[tex]Heat of combustion = heat capacity of calorimeter * deltaT\\[/tex]

Given,

The heat of combustion = 10.5kJ/g.

[tex]deltaT = (32.0-28.5)^oC\\deltaT = 3.5^oC[/tex]

Substitute these values in the above formula to get the value of heat capacity of the calorimeter.

[tex]deltaT =heat capacity of calorimeter * (change in temperature)\\10.5kJ/g = heat capacity of calorimeter * (3.5^oC)\\\\=>heat capacity of calorimeter = \frac{10.5kJ/g}{3.5^oC} \\=>heat capacity of calorimeter = 3.0 kJ/g.^oC[/tex]

Answer:

The heat capacity of the calorimeter is [tex]3.0kJ/g.^oC.[/tex]

Answer:

false all producers are at the top of food web

Answer: In the reaction:

a) The reactants are: P

b) The products are: M and N

Explanation:

The given reaction equation is as follows.

[tex]P \rightarrow M + N[/tex]

Reactants are the species present on the left side on an arrow in a chemical reaction equation.

On the other hand, products are the species which are present on the right side of an arrow in a chemical reaction equation.

Hence, in the given reaction equation P is the reactant. Whereas M and N are the products.

Thus, we can conclude that in the reaction:

a) The reactants are: P

b) The products are: M and N

Answer:

25%

Explanation:

Cystic fibrosis is a recessive allele, meaning that the child must be aa to suffer the symptoms. Only 1/4 of the paired alleles is aa, so the probability of getting cystic fibrosis would be 25%.

Answer:

d. 4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)

Explanation:

Aluminum metal reacts with oxygen gas in a combination reaction that forms a product that coats the metal preventing it from further oxidation: aluminum oxide. Aluminum is a cation with charge 3+ (Al³⁻) and oxide is an anion with charge 2- (O²⁻). Thus, the neutral compound aluminum oxide has the chemical formula Al₂O₃. The unbalanced chemical equation is:

Al(s) + O₂(g) → Al₂O₃(s)

We can balance using the trial and error method. First, we will balance O atoms by multiplying Al₂O₃ by 2 and O₂ by 3.

Al(s) + 3 O₂(g) → 2 Al₂O₃(s)

Finally, we get the balanced equation by multiplying Al by 4.

4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)

Answer:

1. 306,000: three significant figures because the last three zeros are not preceded by a decimal point.

2. 0.0073: two significant figures because the the leftmost zeros are not significant.

3. 39.9999: six significant figures because all these numbers are nonzero digits.

4. 25.00: four significant figures because right-handed zeros, after a decimal point, are significant.

5. 40,000,021: eight significant figures because intermediate zeros are significant.

6. 45,250.0: six significant figures because al the zeros are to the right of the nonzero digits.

7. 0.00011: two significant figures as well as #2.

8. 420.030700: nine significant figures because all the zeros are to the right of the first nonzero digits and after the decimal point.

Explanation:

Hello there!

In this case, by considering the given numbers, we can proceed as follows, by keeping in mind the rules for assigning significant figures:

1. 306,000: three significant figures because the last three zeros are not preceded by a decimal point.

2. 0.0073: two significant figures because the the leftmost zeros are not significant.

3. 39.9999: six significant figures because all these numbers are nonzero digits.

4. 25.00: four significant figures because right-handed zeros, after a decimal point, are significant.

5. 40,000,021: eight significant figures because intermediate zeros are significant.

6. 45,250.0: six significant figures because al the zeros are to the right of the nonzero digits.

7. 0.00011: two significant figures as well as #2.

8. 420.030700: nine significant figures because all the zeros are to the right of the first nonzero digits and after the decimal point.

Regards!

Answer:

When the negatively-charged rod is brought close to the electroscope, positive charges are attracted to it and negative charges are repelled away from it. The electroscope has a net neutral charge and the rubber rod has a net negative charge. If they are brought into contact, they will both take a net negative charge.

Explanation:

I looked it up

Answer:

40.4 kJ

Explanation:

Step 1: Given data

Step 2: Calculate the moles corresponding to 55.0 g of CO₂

The molar mass of CO₂ is 44.01 g/mol.

n = 55.0 g × 1 mol/44.01 g = 1.25 mol

Step 3: Calculate the heat (Q) required to sublimate 1.25 moles of CO₂

We will use the following expression.

Q = n × ΔH°sub

Q = 1.25 mol × 32.3 kJ/mol = 40.4 kJ

Answer: The value of i is 1.4 and 40% dissociation of 100 particles of zinc sulfate will yield 60 undissociated particles.

Explanation:

The equation used to calculate the Vant' Hoff factor in dissociation follows:

[tex]\alpha =\frac{i-1}{n-1}[/tex]

where,

[tex]\alpha [/tex] = degree of dissociation = 40% = 0.40

i = Vant' Hoff factor

n = number of ions dissociated = 2

Putting values in above equation, we get:

[tex]0.40=\frac{i-1}{2-1}\\\\0.40=i-1\\\\i=1.4[/tex]

The equation used to calculate the degee of dissociation follows:

[tex]\alpha =\frac{\text{Number of particles dissociated}}{\text{Total number of particles taken}}[/tex]

Total number of particles taken = 100

Degree of dissociation = 40% = 0.40

Putting values in above equation, we get:

[tex]0.40=\frac{\text{Number of particles dissociated}}{100}\\\\\text{Number of particles dissociated}=(0.40\times 100)=40[/tex]

This means that 40 particles are dissociated and 60 particles remain undissociated in the solution.

Hence, 40% dissociation of 100 particles of zinc sulfate will yield 60 undissociated particles.

Answer:

See explanation

Explanation:

The electrophilic substitution of aromatic compounds occurs faster in substituted aromatic compounds due to the fact that the ring becomes more or less susceptible to electrophilic attack depending on the nature of the substituent in the ring.

Electron pushing substituents such as alkyl groups stabilize the positive charge developed during electrophilic substitution hence they activate the ring towards electrophilic substitution.

The methyl group is an ortho - para directing substituent hence the product obtained by nitration of toluene is o-nitrotoluene and p-nitrotoluene.

The stepwise mechanism for obtaining these products is shown in the image attached to this answer.

Answer:

37.85 L

Explanation:

3.785 x 10.00 = 37.85 L

it would take 37.85 L  to fill a 10.00 tank

(sorry if im wrong pls dont report)

(hope this helps can i plz have brainlist :D hehe)

Answer:

C₁₅N₃H₅

Explanation:

Let's assume we have 240 g of the dye (1 mol), in that case we'd have:

Now we convert the masses of each element into moles, using their respective molar masses:

Thus the molecular formula is C₁₅N₃H₅.

Answer:

i didnt understand

Explanation:

Answer:

x = 2 (C₂H₆)

Explanation:

The general formula for alkanes is CₓH₂ₓ₊₂

2x + 2 = 6

Simply solve for n:

2x = 4

x = 2

Answer:

[tex]m_{H_2O}=73.0gH_2O[/tex]

Explanation:

Hello there!

In this case, since the formation of water from hydrogen and oxygen is:

[tex]2H_2+O_2\rightarrow 2H_2O[/tex]

Whereas we find a 2:2 mole ratio of hydrogen to water. In such a way, by using the Avogadro's number, the aforementioned mole ratio and the molar mass of water (18.02 g/mol), we obtain the following grams of water product:

[tex]m_{H_2O}=2.44x10^{24}molec*\frac{1molH_2}{6.022x10^{23}molec}*\frac{2molH_2O}{2molH_2}*\frac{18.02gH_2O}{1molH_2O}\\\\ m_{H_2O}=73.0gH_2O[/tex]

Regards!

Answer:

false

Explanation:

energy does not return to the sun, it returns to the plants or producers. if energy were to return to the sun, it would have to travel though space.

Answer:

7505.19 J

Explanation:

We'll begin by calculating the change in the temperature of copper. This can be obtained as follow:

Initial temperature (T₁) = 20 °C

Final temperature (T₂) = 875 °C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 875 – 20

ΔT = 855 °C

Finally, we shall determine the heat required. This can be obtained as follow:

Specific heat capacity (C) = 0.385 J/gºC

Change in temperature (ΔT) = 855 °C

Mass (M) = 22.8 g

Heat (Q) required =?

Q = MCΔT

Q = 22.8 × 0.385 × 855

Q = 7505.19 J

Thus, 7505.19 J of heat energy is required.

Answer:

25.46 mL

Explanation:

In a titration we use the volume and concentration of a solution to determine the previously unknown concentration of other solution. Let's consider the titration of an acid of unknown concentration with a standardized base (a base whose concentration we know). The generic neutralization reaction is:

HA + BOH ⇒ BA + H₂O

The base is in the buret and we will add it to the acid until the equivalence point is reached. The volume of base used is equal to the difference between the final reading of the buret and the initial reading of the buret.

V = 27.30 mL - 1.84 mL = 25.46 mL

Answer: hello your question is poorly written below is the complete question

answer:

For N1 :  sp³  orbital

For N2:  p orbital

For N3 : p orbital

For N4 : sp² orbital

For N5 : sp² orbital

Explanation:

Determining the type of orbital in which the lone pair on each N atom will reside.

From the configuration attached below we can determine the type of orbital and they are ;

For N1 :  sp³  orbital

For N2:  p orbital

For N3 : p orbital

For N4 : sp² orbital

For N5 : sp² orbital

Answer:

compound light microscope

Answer:

Identify each of the following as endothermic or exothermic.

a. Water in a pond evaporates.

b. Methane gas burns on a stovetop.

c. Water freezes to form ice.

d. Energy flows from the system to the surroundings.

e. Energy flows from the surroundings to the system.

Explanation:

An exothermic reaction is the one in which heat energy is released.

An endothermic reaction is one in which heat energy is absorbed.

a. Water in a pond evaporates.

This process absorbs heat energy.

Hence, this is an example of an endothermic process.

b. Methane gas burns on a stovetop and release heat energy and hence this is an example of an exothermic reaction.

c. Water freezes to form ice.

In this process heat energy is released.So this is an example of exothermic reaction.

d. Energy flows from the system to the surroundings.

That means heat energy is released into the surroundings.

So, this is an example of exothermic process.

e. Energy flows from the surroundings to the system.

That means energy is absorbed by the system.

So, it is an endothermic process.

Answer:

2.5 × 10² ppm

Explanation:

Step 1: Given data

Step 2: Convert 0.050 g to μg

We will use the conversion factor 1 g = 10⁶ μg.

0.050 g × 10⁶ μg/1 g = 5.0 × 10⁴ μg

Step 3: Calculate the concentration of NaCl in ppm

The concentration of NaCl in ppm is equal to the micrograms of NaCl per gram of the sample.

5.0 × 10⁴ μg NaCl/200. g = 2.5 × 10² ppm

Answer:250 ppm

Explanation:

Answer:

A) Electrons orbit the nucleus in shells at fixed distances

Answer:

the answer is A

Explanation:

Answer:

the value of H° is below -6535 kj. +6H2O

Explanation:

6H2O answer solved

For the given reaction, 2 moles of C₆H₆ the heat energy released is - 6535 KJ. Then, for 16 g of the compound or 0.205 moles needs 669.83 KJ of heat released in combustion.

Combustion is a chemical reaction that occurs between a fuel and an oxidizing agent, typically oxygen, resulting in the release of heat, light, and various combustion products, such as carbon dioxide and water vapor.

The process of combustion involves a rapid and exothermic (heat-releasing) oxidation reaction that produces a flame, which is visible in many cases.

Here, 2 moles of the hydrocarbon releases - 6535 KJ of energy.

molar mass of C₆H₆ = 78 g/mol

then no.of moles in 16 g = 16 /78 = 0.205 moles.

then energy released by 0.205 moles  = 0.205 moles × 6535 KJ /2 moles = 669.83 kJ

Therefore, the heat energy released by 16 g of the compound in combustion is 669.83 kJ.

Find more on combustion :

https://brainly.com/question/13153771

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The correct answer is mixture