Question 3 (2 points)
Using the following equation how many grams of water you would get from 846 g of
glucose:
C6H12O6 + 602 + 6CO2 + 6H20 + energy
Your Answer:
Answer
units g)

Answers

Answer 1

Answer: The mass of water produced is 507.6 g

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

Given mass of glucose = 846 g

Molar mass of glucose = 180 g/mol

Plugging values in equation 1:

[tex]\text{Moles of glucose}=\frac{846g}{180g/mol}=4.7 mol[/tex]

The given chemical equation follows:

[tex]C_6H_{12}O_6+6O_2\rightarrow 6CO_2+6H_2O+energy[/tex]

By the stoichiometry of the reaction:

If 1 mole of glucose produces 6 moles of water

So, 4.7 moles of glucose will produce = [tex]\frac{6}{1}\times 4.7=28.2mol[/tex] of water

Molar mass of water = 18 g/mol

Plugging values in equation 1:

[tex]\text{Mass of water}=(28.2mol\times 18g/mol)=507.6g[/tex]

Hence, the mass of water produced is 507.6 g


Related Questions

Suppose a piece of silver jewelry contains 7.49x10^22 atoms of silver (Ag). how many moles of silver are in the jewelry?​

Answers

Answer:

0.124 mol Ag

General Formulas and Concepts:

Atomic Structure

Moles

Stoichiometry

Using Dimensional Analysis

Explanation:

Step 1: Define

[Given] 7.49 × 10²² atoms Ag

[Solve] mol Ag

Step 2: Identify Conversions

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Step 3: Convert

[DA] Set up:                                                                                                    [tex]\displaystyle 7.49 \cdot 10^{22} \ atoms \ Ag(\frac{1 \ mol \ Ag}{6.022 \cdot 10^{23} \ atoms \ Ag})[/tex][DA] Divide [Cancel out units]:                                                                       [tex]\displaystyle 0.124377 \ mol \ Ag[/tex]

Step 4: Check

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

0.124377 mol Ag ≈ 0.124 mol Ag

Topic: AP Chemistry

Boiling point-methanol (65.0) 66.8c.Boiling point-unknown (record from video)——-c
Identify of unknown:
Possibilities are:Mathanol65.0c;Ethanol 78.5c; Acetone 56.0C

Answers

babshsuzikan s gif theme kaiauysvd dbdj

Describe why corrosion is a natural process

Answers

Answer :

Answer :because it happens due to moisture and oxygen

Soda contains phosphoric acid (H3PO4). To determine the concentration of phosphoric acid in 50.0 mL of soda, the available phosphate ions are precipitated with excess silver nitrate as silver phosphate (418.58 g/mol). The dry Ag3PO4 is found to have a mass of 0.0576 g. What is the concentration of phosphoric acid in the soda?

Answers

Answer:

0.0270w/v% H3PO4 in the soda

Explanation:

All phosphates reacts producing Ag3PO4. To solve this question we must convert the mass of Ag3PO4 to moles. These moles = moles of H3PO4. We can find, thus, the mass of H3PO4 and the w/v% as follows:

Moles Ag3PO4 -Molar mass: 418.58g/mol-

0.0576g * (1mol / 418.58g) = 1.376x10⁻⁴ moles Ag3PO4 = moles H3PO4

Mass H3PO4 -Molar mass: 97.994g/mol-

1.376x10⁻⁴ moles Ag3PO4 = moles H3PO4 * (97.994g/mol) = 0.0135g H3PO4

w/v%:

0.0135g H3PO4 / 50.0mL * 100 =

0.0270w/v% H3PO4 in the soda

When stirred in 30°C water, 5 g of powdered potassium bromide, KBr, dissolves faster than 5 g of large crystals of potassium bromide. Which of the following best explains why the powdered KBr dissolves faster?
A. Potassium ions and bromide ions in the powder are smaller than potassium ions and bromide ions in the large crystals.
B. Powdered potassium bromide exposes more surface area to water molecules than large crystals of potassium bromide.
C. Fewer potassium ions and bromide ions have been separated from each other in the powder than in the crystals.
D. Powdered potassium bromide is less dense than large crystals of potassium bromide.

Answers

Answer:

B

Explanation:

Do diện tích tiếp xúc ở dạng bột cao hơn dạng tinh thể

Suppose you are trying to separate three Proteins using Gel-Filtration chromatography. The sizes of each are given below:

Protein A: 1200 kDa
Protein B: 2000 kDa
Protein C: 800 kDa

Which protein will be the first to emerge from the column?

Answers

Answer: The correct answer is Protein B.

Explanation:

Gel-filtration chromatography is a separation technique that is based on the size of the molecules in a compound. It is also known as size-exclusion chromatography in which the eluent (carrier) used is an aqueous solution.

The matrix that is used is a porous material. When the sample is inserted in the column, the smaller particles interact strongly with the matrix than the large ones. Thus, as the eluent is passed through the matrix, larger molecules come out first, and the smallest molecule comes out last.

Given sizes of the proteins:

Protein A: 1200 kDa

Protein B: 2000 kDa

Protein C: 800 kDa

As protein B has the largest size of all the given proteins, it will emerge out first from the column.

Hence, the correct answer is Protein B.

What type of organic compound is a reactant in all substitution reactions?
1.alkyne
2.alkane
3.alkene

Answers

The organic compounds that are capable of being a reactant in all substitution reactions would belong to the alkane group.

Alkanes and substitution reactions

Alkanes are saturated compounds that ordinarily will not participate in addition reactions due to their saturated nature.

Thus, alkanes are only able to participate in substitution reactions involving the substitution of one or more of their component atoms for another atom.

This is unlike alkyne and alkenes which are naturally unsaturated. The unsaturation makes them a candidate for additional reactions.

More on saturation and substitution reactions can be found here: https://brainly.com/question/3735006

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The rate of the reaction is 1.6*10-2 M/s when the concentration of A is 0.15 M. Calculate the rate constant if the reaction is (a) first order in A and (b) second order in A.

Answers

Answer:

[tex]k_1=0.107s^{-1} \\\\k_2=0.711M^{-1}s^{-1}[/tex]

Explanation:

Hello there!

In this case, according to the given information and the attached picture in which we can see the units of the rate constant, it turns out possible for us to realize the two called rate laws are:

[tex]r=k[A]\\\\r=k[A]^2[/tex]

The former is first-order and the latter second-order; in such a way, we solve for the rate constant in both cases to obtain the following:

[tex]k=\frac{r}{[A]}=\frac{1.6x10^{-2}M/s}{0.15M}=0.107s^{-1} \\\\k=\frac{r}{[A]^2}=\frac{1.6x10^{-2}M/s}{(0.15M)^2}=0.711M^{-1}s^{-1}[/tex]

Regards!

Label each formula and name pair as correct or incorrect.
Formula Name Correct/Incorrect
Aluminum tribromide
Sulfur dioxide
Beryllium hydride
Magnesium(II) oxide
Copper(II) oxide
Calcium sulfate
Nitric acid

Answers

Answer:

Aluminum tribromide: AlBr₃, however, it should be just aluminum bromide.

Sulfur dioxide: SO₂.

Beryllium hydride: BeH₂

Magnesium(II) oxide: MgO; however the roman numeral is not used in Mg as it just has one oxidation number.

Copper(II) oxide: CuO.

Calcium sulfate: CaSO₄

Nitric acid: HNO₃.

Explanation:

Hello there!

In this case, it seems that the formulas were not given, however, we can write the correct one for each given compound according to the widely used nomenclature rules as shown below:

Aluminum tribromide: AlBr₃, however, it should be just aluminum bromide.

Sulfur dioxide: SO₂.

Beryllium hydride: BeH₂

Magnesium(II) oxide: MgO; however the roman numeral is not used in Mg as it just has one oxidation number.

Copper(II) oxide: CuO.

Calcium sulfate: CaSO₄

Nitric acid: HNO₃.

Regards!

Fe có tác dụng với HCL không

Answers

Fe có tác dụng với Hcl

Fe + 2hcl -> fecl2 +h2

Answer:

Explanatio:

Which of the following choices is a source of groundwater pollution?
O sewage
very warm water
O silt
O All of these choices are correct.

Answers

Answer:

very warm water consuving

The placement of carbonyl group of a kerose sugar is at second-carbon only (b) fir + carbon only () first and fast carbon (d) list carbon only Chirulit,​

Answers

Answer:

(d)

Explanation:

Carbonyl group can be the placement of kerosene sugar

separete the ALKALI from the following bases :

NH4OH(ammonium nitrate)

CuO(copper oxide)

Zn(OH)2 (zinc hydroxide)

MgO(magnesium oxide)

Na2O(sodium oxide)

NaOH(sodium hydroxide)

CoO(cobalt oxide)

Mg(OH)2(magnesium hydroxide)

LIOH(lithium hydroxide)

help me with this i will surely mark u as Brainliest

plss help!!!

Answers

Answer:

Ammonium hydroxide, NH₄OH

Magnesium hydroxide, Mg(OH)₂

Sodium hydroxide, NaOH

Lithium hydroxide, LiOH

Explanation:

A base is a substance which neutralizes acids to produce salt and water. Bases are hydroxide or oxides of metals. Bases may be soluble or insoluble in water. Bases generally have a bitter taste and turn red litmus paper or indicator red.

Alkalis are bases which are soluble in water. They form the hydroxide of the alkali metals or alkaline earth metals in solution and they ionize to produce hydroxide ions. They are slippery to touch and turn red litmus blue being bases.

Therefore, all alkalis are bases but not all bases are alkalis. Insoluble bases are not alkalis.

From the given chemical compounds the alkalis present in the list are:

Ammonium hydroxide, NH₄OH; since it is soluble in water and produces hydroxide ions

Magnesium hydroxide, Mg(OH)₂; since it is slightly soluble in water and produces hydroxide ions

Sodium hydroxide, NaOH; since it is soluble in water and produces hydroxide ions

Lithium hydroxide, LiOH; since it is soluble in water and produces hydroxide ions

CuO(copper oxide) is a base but not an alkali as it does not produce hydroxide ions.

Zn(OH)2 (zinc hydroxide) is amphoteric and is insoluble

MgO(magnesium oxide) is a base but not an alkali as it does not produce hydroxide ions.

Na2O(sodium oxide) is a base but not an alkali as it does not produce hydroxide ions.

CoO(cobalt oxide) is a base but not an alkali as it does not produce hydroxide ions.

All the properties listed below are characteristic of the transition elements except __. a) most are paramagnetic b) most are colored c) most have high electronegativities d) most have multiple oxidation states e) most form many different complexes

Answers

Answer:

c) most have high electronegativities

Explanation:

They tend to have high electric CONDUCTIVITY because of the free-flowing d-orbital electrons, but have low electron affinity, ionization energy, and electronegativities.

A scientist collects a sample that has 2.00 × 1014 molecules of carbon dioxide gas.How many grams is this, given that the molar mass of CO2 is 44.01 g/mol?

Answers

Answer:

1.46 × 10⁻⁸ g

Explanation:

Step 1: Given data

Molecules of CO₂: 2.00 × 10¹⁴ molecules

Step 2: Convert molecules to moles

We need a conversion factor: Avogadro's number. There are 6.02 × 10²³ molecules in 1 mole of molecules.

2.00 × 10¹⁴ molecules × 1 mol/6.02 × 10²³ = 3.32 × 10⁻¹⁰ mol

Step 3: Convert moles to mass

We need a conversion factor: the molar mass. The molar mass of CO₂is 44.01 g/mol.

3.32 × 10⁻¹⁰ mol × 44.01 g/mol = 1.46 × 10⁻⁸ g

A company manufacturing KMnO4 wants to obtain the highest yield possible. Two of their research scientists are working on a technique to increase the yield.

Both scientists started with 50.0 g of manganese oxide.

What is the theoretical yield of potassium permanganate when starting with 50.0 g MnO2?

The equation for the production of potassium permanganate is as follows:

2 MnO2 + 4 KOH + O2 → 2 KMnO4 + 2 KOH + H2

Answers

Answer:

The theoretical yield potassium permanganate, KMnO₄ when starting with 50.0 g MnO₂ is 90.8 g

Explanation:

Molar mass of MnO₂ = (55 + 2 × 16) = 87.0 g/mol

Molar mass of KMnO₄ = (39 + 55 + 4 × 16) = 158 g/mol

Moles of MnO₂ in 50 g = reacting mass / molar mass

where reacting mass = 50 g

Moles of MnO₂ in 50 g = 50 g /87 g/mol = 0.575 moles

The equation for the production of potassium permanganate is as follows:

2 MnO2 + 4 KOH + O2 → 2 KMnO4 + 2 KOH + H2

From the equation of the reaction above, 2 moles of MnO₂ produces 2 moles of KmNO₄. The mole ratio of MnO₂ to KMnO₄ is 1 : 1

Therefore, 0.575 moles of MnO₂ will produce theoretically 0.575 moles of KMnO₄

Mass of 0.575 moles of KMnO₄ = number of moles × molar mass

Mass of 0.575 moles of KMnO₄ = 0.575 moles × 158 g/mol = 90.8 g of KMnO₄

Therefore, the theoretical yield potassium permanganate when starting with 50.0 g MnO₂ is 90.8 g

Balance each of the following equations. Then, drag and drop each equation to match the coefficient of H2O in the balanced chemical equation. A coefficient for water may be used once, more than once, or not at all. Drag and drop your selection from the following list to complete the answer:
C2H5OH + O2 + CO2 + H2O NH3 + O2 + NO2 + H20 C3H2 + O2 + CO2 + H2O H2SO4 + NaOH → Na2SO4 + H20 NO2 + H2O → HNO3 + NO

Answers

Answer:

C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O

2 NH₃ + 3.5 O₂ → 2 NO₂ + 3 H₂O

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O

3 NO₂ + H₂O → 2 HNO₃ + NO

Explanation:

We will balance the equation using the trial and error method.

C₂H₅OH + O₂ → CO₂ + H₂O

1) We balance C atoms by multiplying CO₂ by 2 and H atoms by multiplying H₂O by 3.

C₂H₅OH + O₂ → 2 CO₂ + 3 H₂O

2) We balance O atoms by multiplying O₂ by 3.

C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O

NH₃ + O₂ → NO₂ + H₂O

1) We balance H atoms by multiplying NH₃ by 2 and H₂O by 3.

2 NH₃ + O₂ → NO₂ + 3 H₂O

2) We balance N atoms by multiplying NO₂ by 2.

2 NH₃ + O₂ → 2 NO₂ + 3 H₂O

3) We balance O atoms by multiplying O₂ by 3.5

2 NH₃ + 3.5 O₂ → 2 NO₂ + 3 H₂O

C₃H₈ + O₂ → CO₂ + H₂O

1) We balance C atoms by multiplying CO₂ by 3 and H atoms by multiplying H₂O by 4.

C₃H₈ + O₂ → 3 CO₂ + 4 H₂O

2) We balance O atoms by multiplying O₂ by 5.

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

H₂SO₄ + NaOH → Na₂SO₄ + H₂O

1) We balance Na atoms by multiplying NaOH by 2.

H₂SO₄ + 2 NaOH → Na₂SO₄ + H₂O

2) We balance H and O atoms by multiplying H₂O by 2.

H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O

NO₂ + H₂O → HNO₃ + NO

1) We balance H atoms by multiplying HNO₃ by 2.

NO₂ + H₂O → 2 HNO₃ + NO

2) We balance N atoms by multiplying NO₂ by 3.

3 NO₂ + H₂O → 2 HNO₃ + NO

What is the concentration of s solution that contains 55 mL of alcohol per 145 mL solution?​

Answers

Answer:

37.9% v/v

Explanation:

Since both the alcohol and solution are presumed to be liquid, this concentration can be expressed as a volume concentration (or % v/v):

volume concentration = volume of solute / volume of solution

[tex]\% v/v = 55/145= 0.379[/tex]

5-Analysis of a 12.04-g sample of a liquid compound composed of carbon, hydrogen, and nitrogen showed it to contain 7.34 g C, 1.85 g H, and 2.85 g N. What is the percent composition of this compound

Answers

Answer:

%C = 60.9%; %H = 15.4%; %N = 23.7%

Explanation:

Step 1: Given data

Mass of the sample (m): 12.04 gMass of Carbon (mC): 7.34 gMass of Hydrogen (mH): 1.85 gMass of Nitrogen (mN): 2.85 g

Step 2: Calculate the percent composition of this compound

To calculate the percent by mass of any element (E), we will use the following expression.

%E = mE/m × 100%

%C = 7.34 g/12.04 g × 100% = 60.9%

%H = 1.85 g/12.04 g × 100% = 15.4%

%N = 2.85 g/12.04 g × 100% = 23.7%

What are the characteristics of an acid-base neutralization reaction?​

Answers

You’re going to have to elaborate a little bit more, but basically the reaction (in most cases) usually goes acid + base = salt + water. If this isn’t the answer you’re looking for feel free to comment

For a particular chemical reaction the rate​ (g/hr) at which one of the reactants changes is proportional to the amount of that reactant present. If y represents the amount of that reactant at time​ t, StartFraction dy Over dt EndFraction equals minus0.7y. If there were 70 grams of the reactant when the process started ​(tequals ​0), how many grams will remain after 4 ​hours?

Answers

Answer:

Amount of reactant after four hours = 4,26 grams

Explanation:

Suppose y denotes the amount of reactant at the time (t)

The given function:

[tex]\dfrac{dy}{dt} = -0.7 y[/tex]

[tex]\dfrac{dy}{y} = -0.7 dt[/tex]

Taking integral on both sides

㏑(y) = -0.7t + c¹

[tex]e^{In(y)}= e^{-0.7t + c^1}[/tex]

[tex]y(t) = Ce ^{-0.7t}[/tex]

At t = 0 ; y (t) = 70

[tex]70 = Ce^{-0.7(0)}[/tex]

C = 70

As such; [tex]\mathtt{y(t) = 70 e^{-0.7*t}}[/tex]

After four hours, the amount of the reactant is:

[tex]\mathtt{y(t) = 70 e^{-0.7*4}}[/tex]

[tex]\mathtt{y(t) = 70 e^{-2.8}}[/tex]

[tex]\mathtt{y(t) = 4.26}[/tex]

Amount of reactant after four hours = 4,26 grams


Spell out the full name of the compounds
Help plz

Answers

Answer:

propanal

Explanation:

hope this helps :)

Perform the following
mathematical operation, and
report the answer to the correct
number of significant figures.
7.125 x 8.00 = [?]
Answer ?

Answers

57.0

Explanation:
It would be 3 sig figs b/c that is the least amount of figures in the operation; 8.00

Solution A has a pH of 7, and solution B has a pH of 14. Which statement
best describes these solutions?

Answers

Answer:

Option A. Solution B is basic, and solution A is neutral.

Explanation:

The pH of a solution is simply defined as the measure of acidity or alkalinity of a solution.

The pH scale ranges from 0 to 14 with the following readings:

0 to 6 => Acidic solution

7 => Neutral solution

8 to 14 => Alkaline / basic solution.

From the above, we understood that solutions with pH ranging from 0 to 6 are acidic solutions. Those with pH of 7 are neutral solutions while those with pH ranging from 8 to 14 are basic solutions.

With the above information in mind, let us answer the question given above. This is illustrated below:

pH of solution A = 7

pH of solution B = 14

Solution A has a pH of 7. This implies that solution A is a neutral solution

Solution B has a pH of 14. This implies that solution B is a basic solution.

Thus, option A gives the correct answer to the question.

why Mg(OH)2 is soluble in HCL​

Answers

Answer:

While Mg(OH)2 is practically insoluble, a certain amount of Mg(OH)2 dissociates into ions when put in water. ... As HCl is added to the beaker containing milk of magnesia, the H+ ions from the HCl react with the OH– ions (those that are actually in solution from the Mg(OH)2) according to Equation 3 below.

Select the net ionic equation for the reaction that occurs when magnesium sulfate and nickel(II) nitrate are mixed.
a. Ni2+(aq) + SO4^2- → NISO2 (s) + O2 (g).
b. Mg2+(aq) + 2NO3 (aq) → Mg(NO3)2(s).
c. Mg2+(aq) + NO3- → MgNO3 (s).
d. Mg2+(aq) + SO4^2- (aq) + Ni2+ (aq) + 2NO3- → Mg2+ (aq) + 2NO3 (aq) + NISO4 (s).
e. Ni2+(aq) + SO4^2- (aq) → NISO4 (s).
f. No reaction occurs.

Answers

Answer:

No reaction occurs.

Explanation:

The molecular reaction is as follows;

MgSO4(aq) + Ni(NO3)2(aq) ----> Mg(NO3)2(aq) + NiSO4(aq)

We can see from the reaction above that the both products of the reaction are soluble. Recall that a double replacement reaction often yields one insoluble product which separates as a precipitate.

This reaction does not occur since the two products that ought to be obtained are soluble in water.

Compare the solubility of silver iodide in each of the following aqueous solutions:

a. 0.10 M AgCH3COO
b. 0.10 M NaI
c. 0.10 M KCH3COO
d. 0.10 M NH4NO3

1. More soluble than in pure water.
2. Similar solubility as in pure water.
3. Less soluble than in pure water.

Answers

Answer:

Compare the solubility of silver iodide in each of the following aqueous solutions:

a. 0.10 M AgCH3COO

b. 0.10 M NaI

c. 0.10 M KCH3COO

d. 0.10 M NH4NO3

1. More soluble than in pure water.

2. Similar solubility as in pure water.

3. Less soluble than in pure water.

Explanation:

This can be explained based on common ion effect.

According to common ion effect the solubility of a sparingly soluble salt decreases further in a solution which has a common ion to it.

The solubility of AgI(s) silver iodide in water is shown below:

[tex]AgI(s) <=> Ag^{+}(aq)+I^{-}(aq)\\[/tex]

a. a. 0.10 M AgCH3COO has a common ion Ag+ with AgI.

So, AgI is less soluble than in pure water in this solution.

b. 0.10 M NaI has a common ion I- with AgI.

So, AgI is less soluble than in pure water in this solution.

c. 0.10 M KCH3COO:

This solution has no common ion with AgI.

So, AgI has similar solubility as in pure water.

d. 0.10 M NH4NO3:

In this solution, AgI can be more soluble than in pure water.

When zinc nitrate is heated, zinc oxide, nitrogen dioxide(NO2) and oxygen gas are
produced.
i.
Calculate the mass of Zinc oxide produced if 38.5 g of zinc nitrate is heated.
ii.
Determine the volume of Nitrogen dioxide gas evolved at rtp

Answers

Answer:

i) 16.5g of ZnO

ii) 9.8 dm³ of NO2

Explanation:

The working is shown in the photo so kindly refer to it

You have run an experiment studying the effects of the molecular weight of a compound on the rate of diffusion in agar. Compound X has a molecular weight of 25.3 g/mol and compound Y has a molecular mass of 156.2 g/mol. On two separate agar plates, 0.1 g of each substance were transferred and allowed to diffuse for 2 hours. The results below were obtained. Use this information about this situation to answer the following three questions.
a. What was the independent variable in this experiment?
b. What was the independent variable in this experiment?
c. List two controls that were held constant for this experiment or that you would hold constant for this experiment

Answers

Answer:

A) Molecular weight

B) Rate of diffusion

C) Agar plates

Time of diffusion

Explanation:

A) The independent variable is the molecular weight because it is the variable that the researcher changes at will to aid his research. It doesn't depend on other variables.

B) The dependent variable is "Rate of diffusion because it depends on the molecular weight of the compound.

C) A control variable is one that would be held constant throughout the research.

In this case, the agar plates and the time of 2 hours diffusion remain the same throughout.

Which best describes how the total mass of the substances that go into
photosynthesis compares to the mass of substances that are present
afterward?
O A. The mass increases because the molecules that are produced are
larger than those that are used.
B. The mass increases because some light energy changes into
mass.
O C. The mass stays the same because the total number of atoms
does not change
O D. The mass decreases because plants destroy some of the atoms
during photosynthesis.

Answers

Answer:

C. The mass stays the same because the total number of atoms does not change

Explanation:

According to the law of conservation of matter/mass, matter cannot be created nor destroyed, hence, the amount of matter in the reactants must be the same amount in the products.

Using the photosynthetic reaction as a case study, carbon dioxide (CO2) and water (H2O) are the compounds that go into the reaction (reactants) while glucose and oxygen (O2) are the products of the reaction.

Using the law of conservation of matter to explain, the total mass of both the reactants and products stays the same because the total number of atoms does not change i.e. if 6 atoms of Carbon starts the reaction, 6 atoms of carbon will end it.

Other Questions
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