Answer:
a reaction in which electrons are transferred between different atoms
Explanation:
I took the test and got it right :)
c. a reaction in which electrons are released from the system
Oxidation:
Oxidation is a process which involves the addition of oxygen or any electronegative element or the removal of hydrogen or any electropositive element. In terms of electrons, oxidation is defined as the process in which an atom or ion loses one or more electrons.Oxidation occurs when the oxidation state of a molecule, atom or ion is increased. The opposite process is called reduction, which occurs when there is a gain of electrons or the oxidation state of an atom, molecule, or ion decreases.Learn more:
brainly.com/question/24333944
2. Alex pulls on the handle of a claw hammer with a force of 15 N. If
the hammer has a mechanical advantage of 5.2, how much force
is exerted on the nail in the claw?
Answer:
78n
Explanation:
The output force exerted on the nail in the claw is equal to 78 N which has a mechanical advantage of 5.2.
What is the mechanical advantage?The mechanical advantage can be demonstrated as the ratio of the output force to the Input force. The mechanical advantage of any machine can be expressed in the form of the ratio of the forces utilized to do the work.
The ratio of the resistance to the effort is said to be the actual mechanical advantage which will be less. The efficiency of a machine can be evaluated by equating the ratio of the output to its input.
Given, the input force = 15 N
The mechanical advantage of the hammer = 5.2
Mechanical advantage = Output force/ Input force
5.2 = Output/15
Output force = 15 ×5.2 = 78 N
Therefore, the force is exerted on the nail in the claw is equal to 78 N.
Learn more about the Mechanical advantage, here:
brainly.com/question/16617083
#SPJ2
a material that is not a mixture; has the same properties all the way through
Answer:
Explanation:
The material that is not a mixture; it has the same properties all the way through is called a substance. Thus the material that is not a mixture; it has the same properties all the way through is called a substance.
ALL THE BEST :)
which two types of food are homogeneous mixtures
A. mustard
B. mayonnaise
C. tossed salad
D. trail mix
the solubility of nitrogen gas is 1.90 mL/dL of blood at 1.00 atm. what is the solubility of nitrogen gas in a deepsea divers blood at a depth of 200 feet and pressure of 7.00 atm
The solubility of nitrogen gas in water is 1.90 mL/dL at 1.00 atm and 13.3 mL/dL at 7.00 atm.
We want to relate the solubility of a gas with its partial pressure.
We can do so using Henry's law.
What does Henry's law state?Henry's law states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid.
C = k × P
where,
C is the concentration of a dissolved gas. k is the Henry's Law constant. P partial pressure of the gas.The solubility of nitrogen gas is 1.90 mL/dL of blood at 1.00 atm.
Since the solvent is basically water, we can understand that the concentration of nitrogen gas is 1.90 mL/dL at 1.00 atm.
We can use this information to calculate Henry's Law constant.
k = C/P = (1.90 mL/dL)/1.00 atm = 1.90 mL/dL.atm
We want to calculate the solubility of nitrogen gas at a pressure of 7.00 atm.
We will use Henry's law.
C = k × P = (1.90 mL/dL.atm) × 7.00 atm = 13.3 mL/dL
The solubility of nitrogen gas in water is 1.90 mL/dL at 1.00 atm and 13.3 mL/dL at 7.00 atm.
Learn more about solubility here: https://brainly.com/question/11963573
There are two unknowns in this problem - the mass of potassium carbonates and the mass of sodium carbonate. Let's designate the grams of potassium carbonate as our first unknown (you may want to call it gKcarb, or x, some other variable name that makes sense to you) and the grams of sodium carbonate as our second unknown(you may want to call it gNacarb, or y, some other variable name that makes sense to you). Set up an equation for the sum of your two unknowns. Starting with 'unknown' grams of potassium carbonate, use stoichiometry to calculate the number of moles of nitric acid that would react with the potassium carbonate. Your answer will have a variable for your unknown grams of potassium carbonate in it. Starting with 'unknown' grams of sodium carbonate, use stoichiometry to calculate the number of moles of nitric acid that would react with the sodium carbonate. Your answer will have a variable for your unknown grams of sodium carbonate in it. Set up an equation for what you get if you add these two quantities.
This problem, is describing two scenarios, the first one, a reaction between potassium carbonate and nitric acid, and also this very same acid with sodium carbonate. In addition, it is asked to set up two equations whereby you can get the grams of nitric acid needed to react with the unknown grams of both carbonates.
In this case, we can start off by designating the unknown mass of potassium carbonate as X and that of sodium carbonate as Y, so that we will be able to provide a reliable answer. Next, we write the corresponding chemical equations that take place:
K2CO3 + 2HNO3 --> 2KNO3 + H2O + CO2
Na2CO3 + 2HNO3 --> 2NaNO3 + H2O + CO2
After that, we can set up the conversion by considering the following calcultion track:
g Carbonate --> mol Carbonate --> mol HNO3 --> g HNO3
The conversion from grams to moles involve the carbonates' molar mass and the conversion from moles of nitric acid to grams, its molar mass well. In addition, we need the 1:2 mole ratio of the carbonates to nitric acid that it is evidenced in the reaction.
Therefore, the resulting equations that can be set up are shown as follows:
[tex]X g K_2CO_3 *\frac{1mol K_2CO_3 }{138.2 gK_2CO_3 } \frac{2molHNO_3}{1molK_2CO_3 } *\frac{63.1gHNO_3}{1molHNO_3} \\\\Y g Na_2CO_3 *\frac{1mol K_2CO_3 }{105.99 gK_2CO_3 } \frac{2molHNO_3}{1molNa_2CO_3 } *\frac{63.1gHNO_3}{1molHNO_3}[/tex]
Learn more:
https://brainly.com/question/22889208https://brainly.com/question/24384921An example of kinetic energy being converted into heat energy
Answer:
if you drop a water balloon onto the ground, its kinetic energy is converted mostly to thermal energy. If the balloon weighs 1 kilogram and you drop it from about 2 meters, it will heat up by less.
Explanation:
As you say, kinetic energy of large objects can be converted into this thermal energy. For example, if you drop a water balloon onto the ground, its kinetic energy is converted mostly to thermal energy. If the balloon weighs 1 kilogram and you drop it from about 2 meters, it will heat up by less than.
Tristan wraps some gifts and then brings them to the post office where they are delivered to people in different parts of the country. Which organelle is Tristan most like?
Answer:
Tristan wraps some gifts and then brings them to the post office where they are delivered to people in different parts of the country. Which organelle is Tristan most like?
Answer: Tristan is most like the Golgi Body
Explanation:
Find the mass of 1.220 moles of PH5.
Answer:
43.93642
Explanation:
Phosphorus P 30.973762 1 86.0061
Hydrogen H 1.00794 5 13.9939
I don't know physics, but I hope this helps :)
(I'm a seventh grader so don't judge if this is wrong please)
You have three gases in a mixture where P1= 100 kPa, P2 = 50 kPa, and P3 = 75
kPa. What is the total pressure of the gas mixture?
A. 225 kPa
B. 25 kPa
C. 75 kPa
D. None of the above
Answer:
Ptotal=P1+P2+… +Pn. + P nExplanation:
its c
PLEASE HELP ME!!! When two atoms combine to form a compound, one atom pulls electrons from the other atom towards itself. The atom that pulls electrons is (reduced or oxidized). The atom whose electrons are being pulled is (reduced or oxidized)?
Answer:
The pulling atom is oxidized while the pulled atom is reduced. Grade 9 Chemistry
Explanation:
Determine the reducing agent in the following reaction. Explain your answer. 2 Li(s) + Fe(C2H3O2)2(aq) → 2 LiC2H3O2(aq) + Fe(s)
The reducing agent in the reaction 2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) is lithium (Li).
The general reaction is:
2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) (1)
We can write the above reaction in two reactions, one for oxidation and the other for reduction:
Oxidation reactionLi⁰(s) → Li⁺(aq) + e⁻ (2)
Reduction reactionFe²⁺(aq) + 2e⁻ → Fe⁰(s) (3)
We can see that Li⁰ is oxidizing to Li⁺ (by losing one electron) in the lithium acetate (reaction 2) and that Fe²⁺ in iron(II) acetate is reducing to Fe⁰ (by gaining two electrons) (reaction 3).
We must remember that the reducing agent is the one that will be oxidized by reducing another element and that the oxidizing agent is the one that will be reduced by oxidizing another species.
In reaction (1), the reducing agent is Li (it is oxidizing to Li⁺), and the oxidizing agent is Fe(CH₃COO)₂ (it is reducing to Fe⁰).
Therefore, the reducing agent in reaction (1) is lithium (Li).
Learn more here:
https://brainly.com/question/10547418?referrer=searchResultshttps://brainly.com/question/14096111?referrer=searchResults
I hope it helps you!
Learning Task 2: Read each statement or question below carefully and fill in the blank(s) with the best answer by choosing the words inside the box. Write your answers in a separate sheet of paper. cation 1 -ide -ine nonmetals O ion ionic compound anion metals root name 1. Any atom or molecule with a net charge, either positive or negative, is known as en 2. An atom that gains one extra electron forms an with a 1- charge. 3. A positive ion, called a is produced when one or more electrons are lost from a neutral atom. 4. Unlike a cation, which is named for the parent atom, an anion is named by taking the of the atom and changing the ending. 5. The name of each anions is obtained by adding the suffix to the root of the atom name. 6. The always form positive ions. 7. on the other hand, form negative ions by gaining electrons. 8. It is very important to remember that a chemical compound must have a net charge of
The table below shows the dimensions of two colored cubes.
Dimensions of Cubes
Cube Side (cm) Mass
(g)
Yellow 3 135
Black 2 48
Answer: The correct answer is black because the product of its side and mass is lower.
Explanation: The density of a substance is defined as the amount of matter that can be stored in a given volume.
Hence, the black cube will be denser because the product of its side and mass is lower.
Answer: it was wrong on my test
Explanation:
literally dont believe them
How can you include osmosis in animal cell
Calculate the number of molecules present in 11 moles of H2O.
Answer:
[tex]11 \times 6.022 \times {10}^{23} \\ = 66.242\times {10}^{23} \: of \: \\ water \: molecules[/tex]
A chemist heats the block of copper as shown in the interactive, then places the metal sample in a cup of oil at 25.00 °C instead of a cup of water. The temperature of the oil increases to 27.33 °C . Calculate the mass of oil in the cup. The specific heat of copper is 0.387 J/g⋅°C and the specific heat of oil is 1.74 J/g⋅°C .
When the oil is added to the heated copper, the energy in the system is
conserved.
The mass of the oil in the cup, is approximately 64.73 grams.Reasons:
The question parameters are;
Temperature of the oil in the cup = 25.00°C
Final temperature of the oil and copper, T₂ = 27.33 °C
Specific heat of copper, c₂ = 0.387 J/(g·°C)
Specific heat capacity of oil, c₁ = 1.74 J/(g·°C)
Required:
The mass of oil in the cup.
Solution:
The mass of the copper, m₂ = 17.920 g
Temperature of copper after heating, T₂ = 65.17°C
Temperature of the copper after being placed in the cup of oil, T₂ = 27.33°C
Heat lost by copper = Heat gained by the oil
m₂·c₂·(T₂ - T₃) = m₁·c₁·(T₃ - T₁)Therefore, we get;
17.920 × 0.387 × (65.17 - 27.33) = m₁ × 1.74 × (27.33 - 25)
262.4219136 = 4.0542·m₁
m₁ ≈ 64.73
The mass of the oil in the cup, m₁ ≈ 64.73 gLearn more here:
https://brainly.com/question/21406849
Possible part of the question obtained from a similar question online, are;
The mass of the copper, m₂ = 17.920 g
Temperature of copper after heating = 65.17°C
what is the common name of hydrated calcium sulphate?
Answer:
Gypsum
Explanation:
Calcium sulphate, is a naturally occurring calcium salt. It is commonly known in its dihydrate form, a white or colourless powder called gypsum.
HELP!! what are the usual products of combustion reactions?
Explanation:
Carbon dioxide and water
I hope it helps
Answer:
The usual products of combustion reactions are carbon dioxide and water.
Explanation:
Combustion reaction is when a substance reacts with oxygen gas, resulting in a release of energy in the form of light and heat. Combustion reactions must have oxygen (O2) as one of the reactants.
How many moles of potassium iodide, KI, are required to precipitate all of the lead (II) ion from 25.0 mL of a 1.6 M Pb(NO3)2 solution? (First, write a balanced equation for the reaction.)
0.04 moles of iodide is required to precipitate all the lead II ions from 25.0 mL of a 1.6 M Pb(NO3)2 solution.
The reaction equation is;
2KI(aq) + Pb(NO3)2(aq) -------> PbI2(s) + 2KNO3(aq)
The net ionic equation is;
Pb^2+(aq) + 2I^-(aq) ------> PbI2(s)
Number of moles of Pb(NO3)2 = 25/1000 L × 1.6 M = 0.04 moles
Number of moles of Pb^2+ = 0.04 moles /2 = 0.02 moles
Since 2 moles of iodide reacts with 1 mole of Pb^2+
x moles of iodide reacts with 0.02 moles of Pb^2+
x = 2 moles × 0.02 moles/ 1 mole
x = 0.04 moles of iodide
Hence, 0.04 moles of iodide is required to precipitate all the lead II ions from 25.0 mL of a 1.6 M Pb(NO3)2 solution.
Learn more: https://brainly.com/question/9743981
At 298 K, the reaction 2 HF (g) ⇌ H2 (g) + F2 (g) has an equilibrium constant Kc of 8.70x10-3. If the equlibrium concentrations of H2 and F2 gas are both 1.33x10-3 M, determine the initial concentration of HF gas assuming you only started with HF gas and no products initially.
This problem is describing the equilibrium whereby hydrofluoric acid decomposes to hydrogen and fluorine gases at 298 K whose equilibrium constant is 8.70x10⁻³, the equilibrium concentrations of all the reactants are both 1.33x10⁻³ M and asks for the initial concentration of hydrofluoric acid which turns out to be 2.86x10⁻³ M.
Then, we can write the following equilibrium expression for hydrofluoric acid once the change, [tex]x[/tex], has taken place:
[tex][HF]=[HF]_0-2x[/tex]
Now, since both products are 1.33x10⁻³ M we infer the reaction extent is also 1.33x10⁻³ M, and thus, we can calculate the equilibrium concentration of HF via the law of mass action (equilibrium expression):
[tex]8.70x10^{-3}=\frac{(1.33x10^{-3} M)^2}{[HF]} }[/tex]
[tex][HF]=\frac{(1.33x10^{-3} M)^2}{8.70x10^{-3}} }=2.03x10^{-4}M[/tex]
Finally, the initial concentration of HF is calculated as follows:
[tex][HF]_0=[HF]+2x=2.033x10^{-4}+2*(1.33x10^{-3})=2.86x10^{-3}M[/tex]
Learn more:
https://brainly.com/question/13043707https://brainly.com/question/16645766PLEASE HELPPPPPPPPPPPPPPPP
What instruments are used to measure moleculars vibration , rotation and translation?
Answer:
you hear
Explanation:
beautiful hear can be wonderful day in your life so to be rotations is not a big deal
Convert the heat of neutralization of acetic acid from -49,8 kj/mmol to calories per
millimole and ROUND TO ONE DECIMAL PLACE (1 cal = 4.184 J)
DO NOT INCLUDE UNITS
This question is providing the exothermic heat of neutralization of acetic acid in units of kilojoules per mollimole (-49,8 kj/mmol) and asks for the same value but in calories per millimole which results -11,902.5 cal/mmol.
In this case, according to the given problem, it turns out necessary to solve a two-factor conversion in order to convert the kilojoules to joules and finally to calories as shown below:
[tex]-49.8\frac{kJ}{mmol}*\frac{1000J}{1kJ}*\frac{1cal}{4.184J}[/tex]
Thus, we cancel out the kJ and J, to obtain the following result, rounded to one decimal place:
[tex]-11,902.5\frac{cal}{mmol}[/tex]
Learn more:
https://brainly.com/question/2731380https://brainly.com/question/2921187The Difference between the number of protons and electrons give an atom it’s _________.
Answer:
it gives an atom its charge
Using chemical equations, show how the triprotic acid H3PO4 ionizes in water. Phases are optional.
Ka1:
Ka2:
Ka3:
Explanation:
H3PO4 is a weak acid so it partially dissociates in water
Ka1
H3PO4 (aq) + H2O(l) <----> H2PO4-(aq) + H3O+ (aq)
Ka2
H2PO4- (aq) + H2O(l) <----> HPO4 2- (aq) + H3O+ (aq)
Ka3
HPO4 2- (aq) + H2O(l) < ---> PO4 3- (aq) + H3O+ (aq)
What would the empirical formula be for the molecular compound CoH904?
What is it called when you have hydrogen peroxide that just eventually turns into water
Answer:
chlorine
Explanation:
Answer:chlorine reacts with hydrogen peroxide
Explanation:
If such an ion is negatively charged and includes one or more oxygen atoms
Answer:
atom
Explanation:
The sodium atom has a single valence electron that it can easily lose. (If the sodium atom loses its valence electron, it achieves the stable electron configuration of neon.) The chlorine atom has seven valence electrons and can easily gain one electron.
Why is the first one (A) correct?
Answer: yes it is correct
Explanation: the higher it is the cooler.
Which is true of protons and neutrons?
1. They have approximately the same mass and the same charge.
2) They have approximately the same mass but different charge.
by The have different mass and different charge.
O sette
4) They have different mass but the same charge.
Answer:
[tex]\blue{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}[/tex]
[tex]\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}\red{\rule{40pt}{999999pt}}[/tex]