The energy required to run the Calvin cycle reactions of photosynthesis comes from which two substances produced during the light-dependent reactions?
O ATP and NADPH
O ADP and PO4
O H+ and PO2
O O2 and CO2

Answer:

HCO₂

Explanation:

From the information given:

The mass of the elements are:

Carbon C = 26.7 g;     Hydrogen H = 2.24 g     Oxygen O = 71.1 g

To determine the empirical formula;

First thing is to find the numbers of moles of each atom.

For Carbon:

[tex]=26.7 \ g\times \dfrac{1 \ mol }{12.01 \ g} \\ \\ =2.22 \ mol \ of \ Carbon[/tex]

For Hydrogen:

[tex]=2.24 \ g\times \dfrac{1 \ mol }{1.008 \ g} \\ \\ =2.22 \ mol \ of \ Hydrogen[/tex]

For Oxygen:

[tex]=71.1 \ g\times \dfrac{1 \ mol }{1.008 \ g} \\ \\ =4.44 \ mol \ of \ oxygen[/tex]

Now; we use the smallest no of moles to divide the respective moles from above.

For carbon:

[tex]\dfrac{2.22 \ mol \ of \ carbon}{2.22} =1 \ mol \ of \ carbon[/tex]

For Hydrogen:

[tex]\dfrac{2.22 \ mol \ of \ carbon}{2.22} =1 \ mol \ of \ hydrogen[/tex]

For Oxygen:

[tex]\dfrac{4.44 \ mol \ of \ Oxygen}{2.22} =2 \ mol \ of \ oxygen[/tex]

Thus, the empirical formula is HCO₂

Answer:

The atmosphere acts like a multi-layer shield that protects Earth from dangerous solar radiation. ... The stratospheric ozone layer absorbs ultraviolet (UV) radiation, preventing dangerous UV rays from hitting Earth's surface and harming living organisms. (Just look it up)

Explanation:

I hope this helps!

Answer:

D). NADH

Explanation:

Answer:

the correct answere to that is q

Answer:

D.

Explanation:

Answer:

There are 36 grams in 4.00 moles of Carbon.

Explanation:

In order to convert from moles of a substance into grams of the same substance, we need to use the substance's molar mass.

That means that in this case we use the molar mass of Carbon:

Thus, there are 36 grams in 4.00 moles of Carbon.

Answer:

False i think

Answer:

A. True

Explanation:

Answer:

Container 1 I think.

The containers have the same number of molecules and temperature. The container 1.

Collision is the encounter between two particles or object. There are three type of collision: perfectly elastic collision, inelastic collision, perfectly inelastic collision.

Container 1 is smallest, so there will be less distance between the molecules, so they will collide most.

Thus, the containers have the same number of molecules and temperature. The container 1.

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Complete Question

You and your lab partner are asked to determine the density of an aluminum bar. The mass is known accu­rately (to four significant figures). You use a simple metric ruler to measure its dimensions and obtain the results for Method A. Your partner uses a precision micrometer and obtains the results for Method B.

Method A (g/cm3) 2.2,2.3,2.7,2.4 Method B (g/cm3) 2.703 2.701 2.705 5.811

The accepted density of aluminum is 2.702 g/cm 3 .

(a) Calculate the average density for each method.

Should all the experimental results be included in your calculations? If not, justify any omissions.

(b) Calculate the percent error for each method’s average value.

(c) Calculate the standard deviation for each set of data.

(d) Which method’s average value is more precise? Which method is more accurate?

Answer:

a)[tex]Ma_{avg}= 2.4g/cm^3[/tex]

   [tex]Mb_{avg}=2701g/cm^3[/tex]

b)[tex]PE_a=7.5\%[/tex]

  [tex]P.E_b=28.8 \%[/tex]

c) [tex]\sigma_A=0.35[/tex]

   [tex]\sigma_B=0.0545[/tex]

d)Method A

Explanation:

From the question we are told that:

Method A Density values

2.7

Method B Density values

2.703,2.701,2.705,5.811

Density of aluminium [tex]p_d=2.702 g/cm^3[/tex].

a)  

Generally the average density for each method is mathematically given by

Method A

[tex]Ma_{avg}= \frac{2.2+2.3+2.7+2.4}{4}[/tex]

[tex]Ma_{avg}= 2.4g/cm^3[/tex]

Method B

[tex]Mb_{avg}=\frac{2.703+2.701+2.705}{3}[/tex]

[tex]Mb_{avg}=2701g/cm^3[/tex]

b)

Generally the equation for percentage error [tex]PE[/tex] is is mathematically given by

percent error = (measured- real)/real *100

Given real = 2.702 g/ cm​​​3

Method A =

[tex]PE_a=(2.5- 2.702)* \frac{100}{2.702}[/tex]

[tex]PE_a=7.5\%[/tex]

Method B =

[tex]PE_b=( 3.481- 2.702)*\frac{100}{2.702}[/tex]

[tex]P.E_b=28.8 \%[/tex]

c)

Generally the standard deviation [tex]\sigma[/tex] for each method is mathematically given by

Method A

[tex]\sigma=\sqrt{\frac{\sum(x-a)^2}{N} }[/tex]

[tex]\sigma=\sqrt{\frac{0.25+0.16+0+0.09}{4} }[/tex]

[tex]\sigma_A=0.35[/tex]

Method B

[tex]\sigma=\sqrt{\frac{\sum(x-a)^2}{N} }[/tex]

[tex]\sigma=\sqrt{\frac{0.000001+0.00000+0.000009+0.011881}{4} }[/tex]

[tex]\sigma_B=0.0545[/tex]

d)

Therefore with lower percentage error the method who's average value is more precise is

Method A

Answer:

d.7 is the correct answer

Answer: The mass of sodium acetate anhydrous required is 0.820 g

The mass of sodium acetate trihydrate required is 1.36 g

The mass of Iron (III) chloride required is 2.70 g

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]

where,

n = moles of solute

= volume of solution in ml

1. moles of =

Now put all the given values in the formula of molarity, we get

[tex]0.100=\frac{x\times 1000}{82.03\times 100.0}[/tex]

[tex]x=0.820g[/tex]

Therefore, the mass of sodium acetate anhydrous required is 0.820 g

2.   moles of [tex]CH_3COONa.3H_2O[/tex] = [tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{xg}{136.08g/mol}[/tex]

Now put all the given values in the formula of molarity, we get

[tex]0.100=\frac{x\times 1000}{136.08\times 100.0}[/tex]

[tex]x=1.36[/tex]

Therefore, the mass of sodium acetate trihydrate required is 1.36 g

3. moles of [tex]FeCl_3.6H_2O[/tex] = [tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{xg}{g/mol}[/tex]

Now put all the given values in the formula of molarity, we get

[tex]0.100=\frac{x\times 1000}{270.33\times 100.0}[/tex]

[tex]x=2.70g[/tex]

Therefore, the mass of Iron (III) chloride required is 2.70 g

Answer: Thus hydrogen ion concentration is 0.031622777

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

[tex]pH=-\log [H^+][/tex]

Putting in the values:

[tex]1.5=-\log[H^+][/tex]

[tex][H^+]=10^{-1.5}=0.031622777[/tex]

Thus hydrogen ion concentration is 0.031622777

Answer:

neutralization reaction is when an acid and a base react to form water and salt and involves the combination of hydrogen ions and hydroxyl ions to generate water. The neutralization of a strong acid and strong base has a pH equal to 7.

Answer:

cu and Na

Explanation:

When an atom losses electrons this results in a positive charge. A positively charged ion is called a cation.

Sodium and copper will form positive ions. Sodium is a monovalent cation while copper is a divalent cation.

The ions which contain a positive charge it is called positive ions. The atomic number of sodium is 11. It contains (2, 8, 1) electrons. When it loses it's one electron it will be a positive ion.

Copper has valence electrons. When it loses its two valence electrons it will be a copper ion. There are two types of ions such as positive ions and negative ions.

Small molecules that have acquired a positive charge are known as positive ions. Positive ions are the predominant type of air pollution, including hazardous substances, pollen, mold, pet dander, and other dangerous chemicals.

Therefore, Sodium and copper will form positive ions. Sodium is a monovalent cation while copper is a divalent cation.

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Answer:

23,600 collisions.

Explanation:

Hello there!

In this case, according to the Avogadro's law, which help us to understand the molecules-volume relationship as directly proportional:

[tex]\frac{molecules_2}{V_2}=\frac{molecules_1}{V_1}[/tex]

It is possible for us to compute the collisions once the volume is increased as shown below:

[tex]molecules_2=\frac{molecules_1V_2}{V_1}\\\\molecules_2=\frac{2950*46.4L}{5.80L}\\\\molecules_2=23600[/tex]

Which means 23,600 collisions will occur per square centimeter.

Regards!

Answer:

Yes.

Explanation:

Yes, this method is accurate just as the slow titration because in both type of titrations we want to measure the point at which the indicator change the colour of the solution. Both fast and slow titrations are the same in their function i.e. both tell us the point or the amount at which the indicator change the colour of the solution so we can conclude from this discussion that fast titration gives the same result just as the slow titration.

Answer:

what I got was 0.8435160945347224 moles

Answer:

b

Explanation:

one clue is the double bond at 4 carbon c=c

The correct structure that corresponds to the name.

3-bromo-6-methyl-4-octene is   [tex]$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHBrCH}=\mathrm{CHCH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{CH}_{3}$[/tex] .

The molecule in which bromine is attached with C-3 carbon and double present at C-4 carbon containing total number of 8 carbon atom can be considered as 3-bromo-6-methyl-4-octene.

A chemist specifies the molecular geometry and, where possible and required, the electronic structure of such given molecule as well as other solid during a chemical structure determination.

It can be seen that in molecule [tex]$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHBrCH}=\mathrm{CHCH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{CH}_{3}$[/tex] bromine is attached with C-3 carbona and double bond exist at C-4 carbon. the total number of carbon atom is 8.

therefore, the structure of the compound will be [tex]$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHBrCH}=\mathrm{CHCH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{CH}_{3}$[/tex].

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Answer:

Water acts as a base in the presence of a strong acid

Explanation:

Water,being an amphoteric compound, can act both as an acid and as a base.

In the presence of an acid , water acts as a base but in the presence of a base, water acts as an acid.

Answer:TRUE

Explanation:

A hydrogen bond is the electromagnetic attraction created between a partially positively charged hydrogen atom attached to a highly electronegative atom and another nearby electronegative atom. A hydrogen bond is a type of dipole-dipole interaction; it is not a true chemical bond.

Answer:

Explanation:

A1. Chemical indicator, any substance that gives a visible sign, usually by a colour change, of the presence or absence of a threshold concentration of a chemical species, such as an acid or an alkali in a solution. An example is the substance called methyl yellow, which imparts a yellow colour to an alkaline solution.

A2. The reaction of an acid with a base is called a neutralization reaction. The products of this reaction are a salt and water. ... For example, the reaction of hydrochloric acid, HCl, with sodium hydroxide, NaOH, solutions produces a solution of sodium chloride, NaCl, and some additional water molecules

A3. Methyl orange has the property to color alkaline and neutral water yellow. If the water becomes acidic, it turns red immediately.

Answer:

B. 4 K + O₂ → 2 K₂O

Explanation:

Which of the following correctly shows a synthesis reaction involving potassium?

A. Br₂ + 2 KI → I₂ + 2 KBr

No, this is a single displacement reaction, in which Br displaces I from its salt.

B. 4 K + O₂ → 2 K₂O

Yes, this is a synthesis reaction, in which 2 elements combine to form a compound.

C. 2 KClO₃ → 2 KCl + 3 O₂

No, this is a decomposition reaction, in which a big substance decomposes into smaller ones.

D. 2 K₂O → 4K + O₂

No, this is a decomposition reaction, in which a big substance decomposes into smaller ones.

Answer:

After checking many websites as well

I haven't seen any answer for this reaction

While I found answers or other reactions

Therefore I think this reaction cannot occur

Answer:

8.5 mol H₂SO₄

Explanation:

It seems the balanced reaction the problem is referring to is absent, however the description matches the following balanced reaction:

Now we can convert 8.5 moles of SO₂ into moles of H₂SO₄, using the stoichiometric coefficients of the balanced reaction:

Answer:

170.38g is the mass of 1 mole of C₁₂H₂₆

Explanation:

To solve this question we must find the molar mass of the alkane C₁₂H₂₆. The molar mass is defined as the mass of one mole of a substance.

In 1 mole of C₁₂H₂₆ we have 12  moles of Carbon and 26 moles of Hydrogen. With the periodic table of elements we can know the molar mass of 1 mole of Carbon and 1 mole of hydrogen, and, thus, the molar mass of the alkane (Molar mass C = 12.01g/mol, H = 1.01g/mol)

Molar mass C₁₂H₂₆:

12C = 12*12.01g/mol = 144.12g/mol

26H = 26*1.01g/mol = 26.26g/mol

144.12g/mol + 26.26g/mol =

Answer:

1. The product has a higher Rf value on a silica gel TLC plate because it is more polar than the starting methyl benzoate.

2. False

3. True

Explanation:

In chromatography, there is a stationary phase and a mobile phase. The ratio of the distance moved by a component and the distance moved by the solvent gives the retention factor (Rf).

Since silica gel is a polar solvent, it will retain the more polar product methyl m-nitrobenzoate compared to the methyl benzoate starting material.

In comparing the electrophillic aromatic substitution of m-nitrobenzoate  and methyl benzoate, we must remember that the presence of electron withdrawing groups (such as -NO2 and -CHO) on the aromatic compound deactivates the compound towards electrophillic aromatic substitution hence, methyl m-nitrobenzoate is less reactive than methyl benzoate in Electrophilic Aromatic Substition and Methyl benzoate is less reactive than benzene in Electrophilic Aromatic Substition

The final result has a higher Rf value because the TLC plate has more polarity than the other compound.

The answer to the following question is as follows:-

In chromatography, it is a process that is used to separate the different components on the basis of their votality. The ratio of the distance moved by a component and the distance moved by the solvent gives the retention factor (Rf).

In comparing the electrophilic aromatic substitution of m-nitrobenzoate and methyl benzoate, we must remember that the presence of electron-withdrawing groups on the aromatic compound deactivates.

The compound towards electrophilic aromatic substitution hence, methyl m-nitrobenzoate is less reactive than methyl benzoate in Electrophilic Aromatic Substitution and Methyl benzoate is less reactive than benzene in Electrophilic Aromatic Substitution.  

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Answer:

Do you mean Esterification experiment?

Explanation: