This is the process of separating a solid from a liquid by means of a porous material.

The concentration of Urea in final solution will be [tex].00725[/tex] M.

We know that

[tex]M_1/M_2 = V_1/V_2[/tex]

where

[tex]M_1[/tex] is the concentration of solute in first solution

[tex]M_2[/tex]  is the concentration of solute in second solution

[tex]V_1[/tex] is the volume of first solution

[tex]V_2[/tex] is the volume of second solution

According to question,

Solute is Urea whose molecular weight is 60gm

1,75 gram of urea is used

Therefore moles of urea used =  [tex]1.75/60[/tex]

                                                  =   .029

Molarity of urea =  [tex].029/1L[/tex]

                           = .029

Volume of first solution = 5ml

Volume of second solution = 20ml   ( [tex]25 - 5 = 20[/tex])

Therefore,

             [tex].029/M_2 = 5/20[/tex]

[tex]M_2 = .00725[/tex]

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Answer:

Explanation:

Scientists use the periodic table to quickly refer to information about an element, like atomic mass and chemical symbol. The periodic table's arrangement also allows scientists to discern trends in element properties, including electronegativity, ionization energy, and atomic radius.

Answer:

Scientists use the periodic table to quickly refer to information about an element, like atomic mass and chemical symbol. The periodic table's arrangement also allows scientists to discern trends in element properties, including electronegativity, ionization energy, and atomic radius.

Explanation:

The experiments of wöhler, kolbe, miller, and other chemists invalidated the concept of vitalism by demonstrating that B) organic molecules could be produced apart from living organisms using inorganic materials.

Concept of vitalism propose that organic molecules could only be synthesised by living organism. Modern science said that vitalism has no credibility.

Adolph Wilhelm Hermann Kolbe (1818 - 1884) contributed to invalidate the concept of vitalism. He made the organic substance acetic acid from inorganic carbon disulfide.

Friedrich Wöhler (1800 - 1882) synthesis of the organic compound urea from inorganic substances.

Abiotic factors are non-living chemical and physical things of the environment.

Abiotic factors can include water, light, radiation, temperature, humidity, atmosphere, and soil.

On the contrary, biotic factors are all of the living organisms within an ecosystem.

Missing options:

A) carbon atoms were common in both living organisms and nonliving substances such as rocks and soil

B) organic molecules could be produced apart from living organisms using inorganic materials

C) tissues extracted from living organisms could still promote organic reactions

D) one organic substance could be transformed into another organic substance

E) complex organic substances are the result of a series of sequential synthetic reactions occurring in both living organisms and nonliving systems.

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The following elements are given in the table:

For the complete table, refer attachment.

Simple rules can be used to calculate the number of protons, neutrons, electrons, mass number and atomic number in an atom:

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We could use a large size sieve to separate large stones from pear sized gravel. Option A

A sieve is a device that could be used for separation. When we are carrying out separation using a sieve, the separation is done according to the sizes of the materials. We know that the sieve allows a given size of the material to pass through while holding back the other sizes.

In this case, we can see a sieve that has been pictured in the question. The size of the sieve is what determines the kind of things that it could separate. As such, we could use a large size sieve to separate large stones from pear sized gravel. Option A

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Answer:

variables because they can change or be changed

Answer:

independent variable is what you change in a science experiment, dependent variable is what's being measured in a science experiment. any characteristics, number, or quantity that can be measured or counted. A variable may also be called a data item.

A constant is a quantity that does not change. Although you can measure a constant, you either cannot alter it during an experiment or else you choose not to change it. Contrast this with an experimental variable, which is the part of an experiment that is affected by the experiment.

for materials and equipment, can be a ruler, and a data chart to put information, and a timer if needed,

you will record the amount of baking powder you have, and make observations of making sure you have the right amount of baking powder.

also make sure your experiment is safe.

I guess you can use a data chart to keep track of your information, and you can make a hypothesis, conlusion, BUT NO OPINION, make sure to know that before starting a experiment

Ionization energy tends to drop from the top to the bottom of a group as electrons move away from the nucleus.

All elements in the periodic table possess ionisation energy, also known as ionisation potential.

It is the energy needed to remove one electron from a neutral atom, which results in the formation of an ion.

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the question you are looking for is

Which best explains why ionization energy tends to decrease from the top to the bottom of a group?

A.The number of orbitals decreases.

B.The number of neutrons decreases.

C.Electrons get closer to the nucleus.

D.Electrons get farther from the nucleus.

The average atomic mass of strontium reported to two decimal places is 87.62 amu

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KMnO4; NaMnO4; LiMnO4

FeO ; FeCl2; FeS ; FeSO4

HBrO3; KBrO3; NaBrO3; LiBrO3

PbO2; PbS2; Pb(SO4)2; Pb(CO3)2

Ionic compounds are made up of charged particles called ions, which are created when an atom (or group of atoms) gains or loses an electron. An ion is classified as either positively or negatively charged depending on its charge.

Chemical formulas for compounds that simply specify the proportions of the elements they contain rather than the exact number or arrangement of atoms are known as empirical formulas. The element in the compound with the lowest whole number ratio would be this one.

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The heat change for vaporization of 6.00 kg of water at 100°C is 13490.584 kJ.

The heat or enthalpy for vaporization is the amount of heat energy which is required to bring change in the state of a substance, such as from a liquid into a vapor or gas. It is also known as enthalpy of vaporization. The unit of enthalpy for vaporization is given in joules (J) or calories (cal).

Given data :

mass of water = 6.00 kg

Hvap = 40.7 kJ/mol

To calculate the amount of heat -

we know,

the heat required for 1 mol of water to vaporize at 100°C is 40.7 kg

and

we know molar weight of water is  18.015 gram

So,

moles of 6.00 kg water is  = 2600  

                                               18.1015

mole of water = 331.464 mol

Hence,

Heat change for vaporization is = 331.464 × 40.7

                                                     = 13490.584 kJ

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Based on the Law of conservation of mass, the product form when baking soda decomposes is also 25 g.

NaHCO₃ => Na₂CO₃ + H₂O + CO₂

reactant mass = product mass  (law of conservation)

The law of conservation of mass or principle of mass conservation states that for any machine close to all transfers of rely and energy, the mass of the gadget has to remain consistent over the years because the gadget's mass cannot alternate, so the amount can neither be added nor be removed.

The law of conservation of mass states that during a chemical reaction mass is neither created nor destroyed. as example, the carbon atom in coal turns into carbon dioxide while it is burned. The carbon atom modifications from a strong shape to a gas however its mass does not trade.

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The liquid with a total mass of 88.30 grams and a volume of 165.0 milliliters has a density of: 0.5351 g/ml

To solve this problem the formula and the procedure that we have to use  is:

d = m/v

Where:

Information about the problem:

Applying the density  formula we get:

d = m/v

d = 88.30 g/165.0 ml

d = 0.5351 g/ml

It is a physical quantity that expresses the ratio of the body mass to the volume it occupies.

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The ground state electron configuration of the element is; 1s2 2s2 2p6 3s2.

The mass spectrum shows the identity of an element. We can use it to identify the number of isotopes in the elements as well as the abundance of each of the isotopes.

Now, the element that is here has three isotopes, the element is magnesium. We can obtain the relative atomic mass as the sum of the products of the mass of each isotope and its relative abundance. The ground state electron configuration of the element is; 1s2 2s2 2p6 3s2.

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The large piece of jewelry  that has a mass of 132.6 g and when is submerged in a graduated cylinder that initially contains 48.6 ml water and the volume increases to 61.2 ml once the piece of jewelry is submerged, has a density of: 10.523 g/ml

To solve this problem the formulas and the procedures that we have to use  are:

Where:

Information about the problem:

Applying the volume formula we get:

v = v(f)-v(i)

v = 61.2 ml - 48.6 ml

v = 12.6 ml

Applying the density  formula we get:

d = m/v

d = 132.6 g/12.6 ml

d = 10.523 g/ml

It is a physical quantity that expresses the ratio of the body mass to the volume it occupies.

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The article aims to highlight the most recent improvements in the PCET procedures for electrocatalysis at clearly characterized electrode–electrolyte interfaces.

Aim : Here, we focus on the most recent developments in the PCET methods for electrocatalysis at clearly defined electrode–electrolyte interfaces.

Method: Since the groundbreaking and ongoing studies made by Frumkin and his colleagues initiated in the 1930s, non-covalent interaction, which is controlled by the cation/anion features, is known to play a crucial role in the interfacial water and adsorbate environments correlated to electrode–electrolyte catalytic properties.

The microscopic perspective of this interaction is still a significant scientific mystery, nevertheless. Recent research based on the well-defined surface and solid-liquid interfacial structure of single-crystal model electrodes revealed novel characteristics of the cation effect.

Practical usage: Designing high-performance electrode and electrolyte materials requires an understanding of the roles played by the electrical double layer (EDL) structure in proton-coupled electron transfer (PCET) electrode processes.

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percentage error in the student’s experiment is 0.8%

To find percentage error we use the formula I E -T I / I T I X100

so by the given data experimental value is 99.2ºC and actual theoretical value is 100.0ºC by putting the values we get 0.8% error.

To determine how closely the measured number resembles the real value, the percentage error is calculated. It is a component of a thorough error analysis. In most domains, % error is always reported as a positive figure, although in some, either a positive or negative value is acceptable.

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Answer: B

Explanation: :)

Half-reaction for the cell's anode is given below:

Anode : [tex]\\Sn(s) \rightarrow Sn^{2+} (aq) +2e^{-}[/tex]

The anode is defined as the electrode at which electrons leave the cell and oxidation occurs, and the cathode as the electrode at which electrons enter the cell and reduction occurs. The anode is usually the positive side.

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Your given question is quite incomplete here is complete question.

A voltaic cell is based on the reduction of _ Agt(aq) to Ag(s) and the oxidation of Sn(s) to Sn2+(aql) : Part 1  Include the phases of all species in the chemical equation: (aqh Anode: Sn(s) Sn?+ (aq)

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Using an empirical approach when attempting to answer mean your answer should be based on data that is backed by experiments and evidence.

It is an experiment-based approach based on real-world data and results. The empirical approach-based answer should not be made up of theories and concepts. Your calculations and results should be based on experiences in the real world. Empirical research can be Quantitative research (information based on numerical data) or Qualitative (non-numerical).

On the other hand, non-empirical or theoretical answers are based on theories and concepts that have not been experimented as of now.

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The largest pH lowering agent is HNO3 . We need to add more H3O+ to the solution in order to reduce the pH.

There are two ways to measure pH: colorimetrically with indicator fluids or sheets and electrochemically with electrodes and a millivoltmeter for greater accuracy (pH meter).

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Answer:

310 u

Explanation:

Hence, the formula unit mass for Ca 3 ( PO 4 ) 2 is 310 u .

Taking into account the rule of three, you eat 480 kcal.

The rule of three is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them.

That is, what is intended with it is to find the fourth term of a proportion knowing the other three.  

If the relationship between the magnitudes is direct, the direct rule of three must be applied as follow, being a, b and c known data and x the variable to be calculated:

a ⇒ b

c ⇒ x

So: x= (c×b)÷ a

To know the mass of eaten gummy bears, the following rule of three applies: if 1 gummy bear weighs 3 g, 40 gummy bears weigh how much mass?

1 gummy bear ⇒ 3 g

40 gummy bears ⇒ x

So: x= (40 gummy bears× 3 g)÷ 1 gummy bear

Solving:

x= 120 g

If gummy bears are 100% carbohydrates and carbohydrates are 4 kcal/g, the following rule of three applies: If 1 g has 4 kcal, 120 g has how many kcal?

1 g ⇒ 4 kcal

120 g⇒ x

So: x= (120 g× 4 kcal)÷ 1 g

Solving:

x= 480 kcal

In summary, you eat 480 kcal.

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The histone octamer has a total charge of +146 at pH 7 if histidine residues are assumed to be uncharged.

There are 9 negatively charged (Asp + Glu) and 26 positively charged (Arg + Lys) residues in total on histone H2B.

The net charge on H2A is 26 - 9 = +17.

There are 10 negatively charged (Asp + Glu) and 28 positively charged (Arg + Lys) residues in total on histone H2B.

The net charge on H2B is 28 - 10 = +18.

There are 11 negatively charged (Asp + Glu) and 31 positively charged (Arg + Lys) residues in total on histone H3.

The net charge on H3 is 31 - 11 = +20.

There are 7 negatively charged (Asp + Glu) and 25 positively charged (Arg + Lys) residues in total on histone H4.

The net charge on H4 is 25 - 7 = +18.

2 x (17 + 18 + 20 + 18) = +146 is the total charge of the histone octamer.

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If the bears don't pack on enough pounds before they hibernate, they won't enough energy that will sustain them during the hibernation period.

Hibernation is a condition when an animal slows its heart rate to save energy and survive the winter without eating much.

Some animals just slow down and move less frequently during hibernation, but others go into a deep sleep and don't wake up till spring.

If the bears don't pack on enough pounds before they hibernate, they won't return to the mountains to den for the winter.

Packing on enough pounds will provide them with enough energy that will sustain them during the hibernation period.

Thus, If the bears don't pack on enough pounds before they hibernate, they won't enough energy that will sustain them during the hibernation period.

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Answer:

3HCl + Cr (HCO3)3 --> CrCl3 + 3CO2 + 3H2O

The reason why this small change between Neon and sodium cause a very big effect is simply because neon is a noble gas with an octet structure in its electronic configuration while sodium is an alkali metal which is univalent electropositive element.

However, despite the fact that these two different elements has a difference of one electron in their outermost shell, yet they individually carry and posses very powerful electronic structure due to their position in the periodic table which makes them react differently

An element can simply be a chemical substance which cannot be split into simpler forms by an ordinary chemical process.

Consider the following differences between Neon and sodium in the periodic table:

So therefore, the reason why this small change between Neon and sodium cause a very big effect is simply because neon is a noble gas with an octet structure in its electronic configuration while sodium is an alkali metal which is univalent electropositive element.

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