What is the name of the structures that help move substances across a tract surface?

Answers

Answer 1

Answer:

Answer:

cilia and flagella

Explanation:

In prokaryotic species , cilia are present , and in eukaryotic species , flagella is present .

Cilia and flagella both have same function , i.e. , to enable the movement of the cell , along with the movement of some substance and direct the flow of these substance along the tracts.

Cilia and flagella are composed of basal bodies.

Hence , from the given statement of the question,

The correct term is cilia and flagella .

Answer 2

Answer:

Final answer:

Cilia are the structures that help move substances across the surface of a tract. They are primarily found in certain types of cells such as those in the respiratory tract and the oviducts.

Explanation:

The structures that assist in moving substances across a tract surface are primarily called cilia. In biology, cilia are tiny hair-like structures that line certain types of cells, especially those in the respiratory tract and the oviducts. They work much like oars on a boat, moving in coordinated waves to propel substances (like mucus or egg cells) along the surface of the tract they occupy. They are responsible in movement of substances across tract.

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Please it's due today

Answers

Answer:

Explanation:

Newton's third law. states that:

Action and reaction are equal and opposite.

The destruction of the ozone layer by chlorofluorocarbons (CFC’s) can be described by the following reactions:ClO(g) + O3(g) ? Cl(g) + 2 O2(g) ?H°rxn = –29.90 kJ2 O3(g) ? 3 O2(g) ?H°rxn = 24.18 kJDetermine the value of heat of reaction for the following:Cl(g) + O3(g) ? ClO(g) + O2(g) ?H°=_____________?

Answers

Answer:

ΔH°rxn = 54.08 kJ

Explanation:

Let's consider the following equations.

a) ClO(g) + O₃(g) ⇄ Cl(g) + 2 O₂(g) ΔH°rxn = –29.90 kJ

b) 2 O₃(g) ⇄ 3 O₂(g) ΔH°rxn = 24.18 kJ

We have to determine the value of heat of reaction for the following reaction: Cl(g) + O₃(g) ⇄ ClO(g) + O₂(g)

According to Hess's law, the enthalpy change in a chemical reaction is the same whether the reaction takes place in one or in several steps. That means that we can find the enthalpy of a reaction by adding the corresponding steps and adding their enthalpies. According to Lavoisier-Laplace's law, if we reverse a reaction, we also have to reverse the sign of its enthalpy.

Let's reverse equation a) and add it to equation b).

-a) Cl(g) + 2 O₂(g) ⇄ ClO(g) + O₃(g) ΔH°rxn = 29.90 kJ

b) 2 O₃(g) ⇄ 3 O₂(g) ΔH°rxn = 24.18 kJ

-------------------------------------------------------------------------------------------------

Cl(g) + 2 O₂(g) + 2 O₃(g) ⇄ ClO(g) + O₃(g) + 3 O₂(g)

Cl(g) + O₃(g) ⇄ ClO(g) +O₂(g)

ΔH°rxn = 29.90 kJ + 24.18 kJ = 54.08 kJ

Final answer:

The heat of the reaction (ΔH°rxn) for the reaction Cl(g) + O3(g) ? ClO(g) + O2(g) is calculated using Hess's Law. The sum of the heat of reversed first reaction and the second reaction provided is 54.08 kJ.

Explanation:

The chemistry question asks to determine the heat of the reaction for the reaction Cl(g) + O3(g) ? ClO(g) + O2(g). In Hess's Law, the heat of the reaction or ΔH for a reaction can be calculated from the sum of the heats of other reactions that sum to the desired reaction. In this case, we want to reverse the first reaction provided (which changes the sign of ΔH) and add it to the second reaction provided.

So, reversing the first reaction we get: Cl(g) + 2 O2(g) ? ClO(g) + O3(g) ?H°rxn = 29.90 kJ

Adding this to the second reaction: 2 O3(g) ? 3 O2(g), ?H°rxn = 24.18 kJ, gives the reaction Cl(g) + O3(g) ? ClO(g) + O2(g). Adding the ΔH values gives the ΔH for this reaction: 29.90 kJ + 24.18 kJ = 54.08 kJ. So, ?H°rxn for the reaction Cl(g) + O3(g) ? ClO(g) + O2(g) is 54.08 kJ.

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What makes physical change a physical change and a chemical change a chemical change

Answers

A physical change is any change in a substances form that does not change its chemical makeup. Examples of physical changes are breaking a stick or melting ice. A chemical change occurs when atoms of a substance are rearranged, and the bonds between the atoms are broken or formed. HOPE THIS HELPS!!

Answer:

A physical change is reversible and the changes made are only physical (smell, physical state, volume, etc) and most importantly no new substance is formed. In a chemical change, a new substance is formed, the chemical properties are changed, and it's permanent.

Under what set of conditions does HCl(g) deviate the most from ideal behavior?a) high temperature and low pressure.b) low temperature and high pressure.c) high temperature and high pressure.d) low temperature and low pressure. e) standard temperatore and pressure.

Answers

Answer:

b) low temperature and high pressure

Explanation:

Deviation of gases from their ideal behavior could be as a result of two things which include a very small volume for the gases and the collisions not being elastic enough.

Small volume will decrease the frequency of elastic collisions a gas will experience as a result of the tiny space.

Low temperature will decrease the amount of elastic collisions and energy of the gas and high pressure will decrease the volume which is why there will be a deviation in the ideal behavior of Hcl(g)

A chemist titrates 160.0mL of a 0.3403M aniline C6H5NH2 solution with 0.0501M HNO3 solution at 25°C . Calculate the pH at equivalence. The pKb of aniline is 4.87 . Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HNO3 solution added.

Answers

Answer : The pH of the solution is, 5.24

Explanation :

First we have to calculate the volume of $HNO_3$

Formula used :

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the initial molarity and volume of $C_6H_5NH_2$ .

$M_2\text{ and }V_2$ are the final molarity and volume of $HNO_3$ .

We are given:

$M_1=0.3403M\nV_1=160.0mL\nM_2=0.0501M\nV_2=?$

Putting values in above equation, we get:

$0.3403M* 160.0mL=0.0501M* V_2\n\nV_2=1086.79mL$

Now we have to calculate the total volume of solution.

Total volume of solution = Volume of $C_6H_5NH_2$ + Volume of $HNO_3$

Total volume of solution = 160.0 mL + 1086.79 mL

Total volume of solution = 1246.79 mL

Now we have to calculate the Concentration of salt.

$\text{Concentration of salt}=(0.3403M)/(1246.79mL)* 160.0mL=0.0437M$

Now we have to calculate the pH of the solution.

At equivalence point,

$pOH=(1)/(2)[pK_w+pK_b+\log C]$

$pOH=(1)/(2)[14+4.87+\log (0.0437)]$

$pOH=8.76$

$pH+pOH=14\n\npH=14-pOH\n\npH=14-8.76\n\npH=5.24$

Thus, the pH of the solution is, 5.24

Calculate your experimentally determined percent mass of water in Manganese(II) sulfate monohydrate. Report your result to 2 or 3 significant figures, e. g. 9.8% or 10.2%.

Answers

The mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.

What is percentage mass?

The percentage mass is the ratio of the mass of the element or molecule in the given compound.

The percentage can be given as:

$\text{Percent Mass} = \frac{\text{Mass of molecule}}{\text{total mass of compound}} * 100 \%$

The mass of the water is 18.02 g/mol and the molar mass of hydrated magnesium sulfate (MnSO4 . H2O) is 169.03 g/mol.

Thus,

$\text{Percent Mass} = \frac{\text{18.02}}{\text{169.03 }} * 100 \%\n\n\text{Percent Mass} = 10.6 \%}$

Therefore, the mass percentage of the water in hydrated magnesium sulfate (MnSO4 . H2O) is 10.6%.

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Answer:

10.6%

Explanation:

The determined percent mass of water can be calculated from the formula of the hydrate by

dividing the mass of water in one mole of the hydrate by the molar mass of the hydrate and

multiplying this fraction by 100.

Manganese(ii) sulphate monohydrate is MnSO4 . H2O

1. Calculate the formula mass. When determining the formula mass for a hydrate, the waters of

hydration must be included.

1 Manganes 52.94 g = 63.55 g

1 Sulphur 32.07 g =

32.07 g 2 Hydrogen is = 2.02 g

4 Oygen =

64.00 g 1 Oxygen 16.00 = 16.00 g

151.01 g/mol 18.02 g/mol

Formula Mass = 151.01 + (18.02) = 169.03 g/mol

2. Divide the mass of water in one mole of the hydrate by the molar mass of the hydrate and

multiply this fraction by 100.

Percent hydration = (18.02 g /169.03 g) x (100) = 10.6%

The final result is 10.6% after the two steps calculations

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