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What is parasitism?

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Answer 1

Answer:

Answer:

the practice of living as a parasite in or on another organism.

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A chemist has two solutions of H2SO4. One has a 40% concentration and the other has a 25% concentration.How many liters of each solution must be mixed to obtain 78 liters of a 28% solution?

liters of the 40% solution and

liters of the 25% solution must be mixed to obtain a 28% solution of H2SO4.

(Round to the nearest tenth, if necessary.)

Answers

Answer:

16 liters of the solution with 40% concentration must be mixed with 62 liters of the solution with 25% concentration in order to obtain 78 liters, 25% concentration solution.

Explanation:

Let the required volume of solution 1 be represented by x.

The required volume of solution 2 would then be 78-x.

The number of moles of solution 1 that would be required = 0.4x

The number of moles of solution 2 that would be required = 0.25(78-x)

The number of moles of the final mixture = 78 x 0.28 = 21.84

moles of solution 1 + moles of solution 2 = moles of final mixture

0.4x + 0.25(78 - x) = 21.84

0.4x + 19.5 - 0.25x = 21.84

0.4x - 0.25x = 21.84 - 19.5

0.15x = 2.34

x = 15.6 liters

To the nearest tenth = 16 liters

Liters of 40% solution needed = 16 liters

Liters of 25% solution needed = 78 - 16 = 62 liters.

Hence, 16 liters of the solution with 40% concentration must be mixed with 62 liters of the solution with 25% concentration in order to obtain 78 liters, 25% concentration solution.

Develop expressions for the mole fraction of reacting species functions of the reaction coordinate for: A system initially containing 2 mol NH3 and 5 mol O2 and undergoing the reaction: 4NH3 (g) + 5O2 (g) ® 4NO (g) + 6 H20 (g) A system initially containing 3 mol NO2, 4 mol NH3, and 1 mol N2 and undergoing the reaction: 6NO2 (g) + 8NH3 (g) ® 7N2 (g) +12H2O (g)

Answers

Answer:

Individual mole fractions of all the species of the all reaction is as follows.

(a)

$y_{NH_(3)}=(2+(-4)\epsilon)/(7+\epsilon)$

$y_{O_(2)}=(5+(-5)\epsilon)/(7+\epsilon)$

$y_(NO)=(0+(4)\epsilon)/(7+\epsilon)=(4\epsilon)/(7+\epsilon)$

$y_{H_(2)O}=(0+6\epsilon)/(7+\epsilon)$

(b)

$y_{H_(2)S}=(3+(-2)\epsilon)/(8- \epsilon)$

$y_{O_(2)}=(5+(-3)\epsilon)/(8- \epsilon)$

$y_{H_(2)O}=(2\epsilon)/(8- \epsilon)$

$y_{SO_(2)}=(2\epsilon)/(8- \epsilon)$

(c)

$y_{NO_(2)}=(3+(-6)\epsilon)/(8+5\epsilon)$

$y_{NH_(3)}=(4+(-8)\epsilon)/(8+5\epsilon)$

$y_{N-{2}}=(1+7\epsilon)/(8+5 \epsilon)$

$y_{H_(2)O}=(12\epsilon)/(8+5\epsilon)$

Explanation:

(a)

Initial number of moles of $NH_(3)$ and $O_(2)$ are 2 mol and 5 mol respectively.

The given chemical reaction is as follows.

$4NH_(3)(g)+5O_(2)(g)\rightarrow 4NO(g)+6H_(2)O$

The stoichiometric numbers are as follows.

$v_{NH_(3)}=-4$

$v_{O_(2)}=-5$

$v_(NO)=4$

$v_{H_(2)O}=6$

The total number of moles initially present -7

$\Sigma v_(i)\epsilon = (-4-5+4+6)= 1\epsilon$

The expression for the mole fraction of species"i" is as follows.

$y_(i)=\frac{(n_{i_(o)})+(v_(i)\epsilon) }{n_(o)+v\epsilon}$

The individual mole fractions of all the species are as follows.

$y_{NH_(3)}=(2+(-4)\epsilon)/(7+\epsilon)$

$y_{O_(2)}=(5+(-5)\epsilon)/(7+\epsilon)$

$y_(NO)=(0+(4)\epsilon)/(7+\epsilon)=(4\epsilon)/(7+\epsilon)$

$y_{H_(2)O}=(0+6\epsilon)/(7+\epsilon)$

(b)

Initial number of moles of $H_(2)S$ and $O_(2)$ are 3 mol and 5 mol respectively.

The given chemical reaction is as follows.

$2H_(2)S(g)+3O_(2)(g)\rightarrow 2H_(2)O(g)+2SO_(2)$

The stoichiometric numbers are as follows.

$v_{H_(2)S}=-2$

$v_{O_(2)}=-3$

$v_{H_(2)O}=2$

$v_{SO_(2)=2$

The total number of moles initially present -8

$\Sigma v_(i)\epsilon = (-2-3+2+2)= - \epsilon$

The expression for the mole fraction of species"i" is as follows.

$y_(i)=\frac{(n_{i_(o)})+(v_(i)\epsilon) }{n_(o)+v\epsilon}$

The individual mole fractions of all the species are as follows.

$y_{H_(2)S}=(3+(-2)\epsilon)/(8- \epsilon)$

$y_{O_(2)}=(5+(-3)\epsilon)/(8- \epsilon)$

$y_{H_(2)O}=(2\epsilon)/(8- \epsilon)$

$y_{SO_(2)}=(2\epsilon)/(8- \epsilon)$

(c)

Initial number of moles of $NO_(2)$ , $NH_(3)$ and $N_(2)$ are 3 mol,4 mol and 1 mol respectively.

The given chemical reaction is as follows.

$6NO_(2)(g)+8NH_(3)(g)\rightarrow 7N_(2)(g)+12H_(2)O$

The stoichiometric numbers are as follows.

$v_{NO_(2)}=-6$

$v_{NH_(3)}=-8$

$v_{N_(2)}=7$

$v_{H_(2)O}=12$

The total number of moles initially present -8

$\Sigma v_(i)\epsilon = (-6-8+7+12)= 5 \epsilon$

The expression for the mole fraction of species"i" is as follows.

$y_(i)=\frac{(n_{i_(o)})+(v_(i)\epsilon) }{n_(o)+v\epsilon}$

The individual mole fractions of all the species are as follows.

$y_{NO_(2)}=(3+(-6)\epsilon)/(8+5\epsilon)$

$y_{NH_(3)}=(4+(-8)\epsilon)/(8+5\epsilon)$

$y_{N-{2}}=(1+7\epsilon)/(8+5 \epsilon)$

$y_{H_(2)O}=(12\epsilon)/(8+5\epsilon)$

Final answer:

Expressions for the mole fractions of reacting species are determined using stoichiometry and the initial molar amounts, taking into account the stoichiometric coefficients of the chemical reactions.

Explanation:

To develop expressions for the mole fraction of reacting species as functions of the reaction coordinate for the given systems, we will examine each reaction individually. For the reaction 4NH3 (g) + 5O2 (g) ® 4NO (g) + 6 H2O (g), we can use stoichiometry to correlate the molar amounts of each species with reaction progress. Given the initial amounts, we will track how the molar amount changes for each mole of NH3 reacted.

Starting with 2 mol NH3 and 5 mol O2, the mole ratio from NH3 to NO and H2O is 1:1 and 1:1.5, respectively. The mole ratio from NH3 to O2 is 4:5. If x moles of NH3 react, the mole fractions for each species at any point in the reaction can be expressed as follows:

Mole fraction of NH3: (2 - x)/(Total moles)
Mole fraction of O2: (5 - 5x/4)/(Total moles)
Mole fraction of NO: (4x)/(Total moles)
Mole fraction of H2O: (6x)/(Total moles)

Note that 'Total moles' is the sum of the ongoing moles of all species. The mole fractions must always add up to 1 at any point during the reaction.

For the second reaction 6NO2 (g) + 8NH3 (g) ® 7N2 (g) +12H2O (g), with initial amounts of 3 mol NO2, 4 mol NH3, and 1 mol N2, similar steps are taken. For every mole of NH3 reacted, the corresponding changes in molar amounts can be calculated from the stoichiometry of the balanced equation.

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Carbon and oxygen combine to form the molecular compound CO2, while silicon and oxygen combine to form a covalent network solid with the formula unit SiO2. Explain the difference in bonding between the two group 4A elements and oxygen. g

Answers

Answer:

See explanation below.

Explanation:

Both carbon and silicon are members of group 4A(now group 14) i n the periodic table. Carbon is the first member of the group. CO2 is a gas while SiO2 is a solid. In SiO2, there are single bonds between silicon and oxygen and the geometry around the central atom is tetrahedral while in CO2, there are double carbon-oxygen bonds and the geometry around the central atom is linear. CO2 molecules are discrete and contain only weak vanderwaals forces.

Again, silicon bonds to oxygen via its 3p orbital while carbon bonds to oxygen via a 2p orbital. As a result of this, there will be less overlap between the pi orbitals of silicon and that of oxygen. This is why tetrahedral bonds are formed with oxygen leading to a covalent network solid rather than the formation of a silicon-oxygen pi bond. A covalent network solid is known to be made up of a network of atoms of the same or different elements connected to each other continuously throughout the structure by covalent bonds.

In SiO2, each silicon atom is surrounded by four oxygen atoms. Each corner is shared with another tetrahedron. SiO2 forms an infinite three dimensional structure and melts at a very high temperature.

Final answer:

Carbon and oxygen form a molecular compound CO2 with weaker covalent bonds, while silicon and oxygen form a covalent network solid SiO2 with stronger, three-dimensional covalent bonds.

Explanation:

The difference in bonding between carbon and oxygen compared to silicon and oxygen is due to the different nature of their chemical bonds. In the case of carbon and oxygen, they form a molecular compound CO2, where carbon and oxygen atoms share electrons to form covalent bonds. This is because carbon and oxygen have similar electronegativities, so they can share electrons equally. The covalent bonds in CO2 are relatively weak, allowing the compound to exist as a gas at room temperature and pressure.

On the other hand, silicon and oxygen form a covalent network solid with the formula unit SiO2, known as quartz. In this case, silicon and oxygen atoms are covalently bonded in a three-dimensional network structure, where each silicon atom is bonded to four oxygen atoms and each oxygen atom is bonded to two silicon atoms. This network structure gives SiO2 its high melting point and hardness, making it a solid at room temperature and pressure.

In summary, the difference in bonding between carbon and oxygen compared to silicon and oxygen is that carbon and oxygen form a molecular compound with weaker covalent bonds, while silicon and oxygen form a covalent network solid with stronger, three-dimensional covalent bonds.

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To plan ahead using REPS and Checks, choose the minimum space cushion needed to maintain folloying distance traveling at 25 MPH (for a cargo van)?

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The minimum space cushion defines the required amount of space which vehicles should maintain in other to afford them the time and space to gain control in emergency scenarios. Hence, the minimum space cushion required in the scenario is 4 seconds.

In cases of mishaps or accidents, the space cushion might just afford other cars the space to maneuver their way to safety rather than being caught up in the collison or accident.

The required space cushion in most scenario is usually between 2 - 5 seconds, with additional space afforded depending on the length and type of the vehicle.

Therefore, to ensure safety, the required minimum spacecushion to be left when driving being a cargo van traveling at a speed of 25mph is 4 seconds.

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Final answer:

In order to ensure safety while driving a cargo van at 25 MPH, the driver should maintain a space cushion of about 3-4 van lengths, which accounts for speed, reaction time, and distance needed to apply brakes and avert a collision.

Explanation:

The subject of your question revolves around optimal space cushion required for safety while driving a cargo van at the speed of 25 MPH, adhering to REPS (Reference point, Eye lead time, Posting and Scanning) and Checks (Check side mirrors and Rearview mirror every 5-8 seconds). This question falls under the domain of physics, as it involves velocity (speed of the vehicle), distance (space cushion), and time.

As a general rule of thumb, for every 10 miles per hour, a driver should ideally stay approximately one car length away from the car in front of them. Therefore, at 25 MPH, the driver should maintain a distance of at least 2.5 car lengths. In the case of a cargo van, which is typically larger than a regular car, this distance should ideally be increased to 3-4 van lengths to ensure safe stopping distance and reaction time in case of any sudden stoppage by the vehicle ahead.

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Does dissolved potassium chloride affect the surface tension between water molecules?

Answers

Answer:

Inorganic impurities present in the bulk of a liquid such as KCl tend to increase the surface tension of water.

Explanation:

As potassium chloride (KCl) dissolves in water, the ions are hydrated. ... When ionic compounds dissolve in water, the ions in the solid separate and disperse uniformly throughout the solution because water molecules surround and solvate the ions, reducing the strong electrostatic forces between them.

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200.0 grams of an organic compounds known to contain 98.061 grams of carbon, 10.381 grams of hydrogen, 32.956 grams of oxygen and the rest is nitrogen. what is the empirical formula of the compound? what is the molecular formula of the compound if its molar mass is 194.101?

Answers

Answer:

1. The empirical formula is C₄H₅N₂O

2. The molecular formula is C₈H₁₀N₄O₂

Explanation:

The following data were obtained from the question:

Mass of compound = 200 g

Carbon (C) = 98.061 g

Hydrogen (H) = 10.381 g

Oxygen (O) = 32.956 g

Empirical formula =?

Molecular formula =?

Next, we shall determine the mass of nitrogen in the compound. This can be obtained as follow:

Nitrogen (N) = 200 – (98.061 + 10.381 + 32.956)

Nitrogen (N) = 200 – 141.398

Nitrogen (N) = 58.602 g

1. Determination of the empirical formula of the compound.

C = 98.061 g

H = 10.381 g

O = 32.956 g

N = 58.602 g

Divide by their molar masses

C = 98.061 /12 = 8.172

H = 10.381 /1 = 10.381

O = 32.956 /16 = 2.060

N = 58.602 /14 = 4.186

Divide by the smallest

C = 8.172 /2.060 = 4

H = 10.381 / 2.060 = 5

O = 2.060 / 2.060 = 1

N = 4.186 / 2.060 = 2

Thus, the empirical formula of the compound is C₄H₅N₂O

2. Determination of the molecular formula of the compound.

Empirical formula of the compound => C₄H₅N₂O

Molar mass of compound = 194.101 g/mol

Molecular formula =.?

[C₄H₅N₂O]n = 194.101

[(12×4) + (1×5) + (14×2) + 16]n = 194.101

[48 + 5 + 28 + 16]n = 194.101

97n = 194.101

Divide both side by 97

n = 194.101 /97

n = 2

Molecular formula => [C₄H₅N₂O]n

=> [C₄H₅N₂O]2

=> C₈H₁₀N₄O₂

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I need soon this is due at 10:30 pm its curently 6:00 pmWhat is parasitism?

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Explanation:

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Final answer:

Explanation:

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Final answer:

Explanation:

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I need soon this is due at 10:30 pm its curently 6:00 pm
What is parasitism?