1. A dog with a mass of 3.5 kg goes for a run at a pace of 1.4 m/s. What is the
dog's kinetic energy?

Answer:

200J or 196J

Explanation:

mass = 10kg

gravity = 9.8 or 10 ( the answer might vary if u take either 10 or 9.8 as the gravitational value )

height = 2 meter

p.e= mgh

option 1

10×10×2 = 200 Joules

option 2

10×9.8×2 = 196 Joules

better to take 9.8 as gravitational value

hope it helps

The AgCl would be least soluble in 0.1M NaCl

Strength of NaCl Solution = 0.1M

K(sp) AgCl = 1.20 * 10⁻¹⁰

Let solubility of AgCl be 'x'

∴ AgCl ⇔ Ag⁺ + Cl⁻

     x           x       x

∵ NaCl is strong electrolyte and is completely ionized. It shall provide Cl⁻ ion concentration = 0.1 M

[Ag⁺] = x                   [Cl⁻] = (x + 0.1) M

K(sp) = [Ag⁺] [Cl⁻] = x * (x + 0.1) = x² + 0.1x

∵ x << 1    ∴ x² is neglected

       ∴ 1.20 * 10⁻¹⁰ = 0.1x

         ⇒ [tex]x = \frac{1.20 * 10^{-10}}{0.1} = 1.20 * 10^{-9}[/tex]

Solubility of AgCl in 0.1M NaCl = 1.20 * 10⁻⁹M

By above method we find other four parts then we get,

Solubility of AgCl in 0.1M HCl = 1.8 * 10⁻¹⁰M

Solubility of AgCl in 0.1M NH₃ = 4.846 * 10⁻³M

Solubility of AgCl in 0.1M CaCl₂ = 1.8 * 10⁻¹⁰M

Solubility of AgCl in Pure Water = 1.5 * 10⁻¹⁰M

The order of solubility of AgCl in 1. 0.1M HNO₃. 2. 0.1M HCl. 3. 0.1M NH₃. 4. 0.1M CaCl₂. 5. Pure Water :-

0.1M NaCl < Pure Water < 0.1M HCl = 0.1M CaCl₂ < 0.1M NH₃

Thus, the AgCl would be least soluble in 0.1M NaCl

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The average atomic mass of Mg is 24.30505 u

The average atomic mass of Mg (0.7899 x 23.98504) + (0.100 x 24.98584) + (0.1101 x 25.98259)

= 24.30505 u

The total average mass of the atoms in a naturally occurring sample of an element is the average atomic mass, also known as atomic weight. The most common unit of measurement for average masses is the unified atomic mass unit (u), which is equal to precisely one-twelfth of the mass of a neutral carbon-12 atom.

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The object is not pure gold because it does not the same density as pure gold.

The density of an object is defined as the ratio of the mass to the volume of an object. We know that the density could be used to characterize an object. This is why we refer to the density of the object as an intrinsic property of the object. Since the density of the object does not change. We could use the density to determine if the second object is gold or not.

Mass of the object = 33.4.g

Inital volume of the water= 12.3mL

Final volume of the water = 21.7mL

Volume of the object = 21.7mL - 12.3mL = 9.4 mL

Density of the object = 33.4.g / 9.4 mL = 3.55 g/mL

The object is not pure gold because it does not the same density as pure gold.

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The molecules in a 5.15 g sample of dimethylmercury is 1.32 x 10²².

Dimethylmercury is a highly volatile, toxic liquid, which is reactive and colourless. Its molecular formula is C₂H₆Hg. It is used in NMR instruments for calibration. It has risks involved, so it has few applications and used in a lesser quantity.

Dimethylmercury C₂H₆Hg has a molecular mass of -

Molecular mass = (12x2) x (1x6) x 200.59 = 230.6 g/mol

Now,

number of moles in 5.15 g of dimethylmercury =  5.15   = 0.022 mol

                                                                                 230.6

1 mole of dimethylmercury contains 6.022 x 10²³ molecules

So, 0.022 moles of dimethylmercury contains 0.022 x (6.022 x 10²³) = 1.32 x 10²² molecules

Molecules contained in 5.15 g sample of dimethylmercury is

1.32 x 10²²

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The molar concentration of ethanol in this solution is 16.58 M.

percentage by weight of ethanol = 95%

M(CH₃CH₂OH) = 46.07 g/mol; molar mass od ethanol

d = 0.804 g/ml; density of the solution

Take 100 ml (0.1 l) of solution:

m(CH₃CH₂OH) = 100 ml × 0.804 g/ml × 0.95

m(CH₃CH₂OH) = 76.38 g; mass of pure ethanol

n(CH₃CH₂OH) = 76.38 ÷ 46.07 g/mol

n(CH₃CH₂OH) = 1.658 mol; amount of pure ethanol

c(CH₃CH₂OH) = 1.658 mol ÷ 0.1 l

c(CH₃CH₂OH) = 16.58 mol/l; the molar concentration of the solution

Molarity is defined as the number of moles of the solute per liter of the solution. It is denoted by M.

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ΔSsys  and ΔSsurr both have values larger than 0.

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Answer:

\ \

=o + H+ ----> =o

/- /

conjugate acid is the product that is differ from the base by one proton

Answer: I don't know try your best

Explanation:

Structures that can be drawn to represent the covalent bonding in a molecule that has more than one position in the molecule for a multiple bond are known as Position Isomers.

Position isomers are isomers with same chemical formula but the functional group is attached to different position of carbon in different structures.

Example-

Hex1-ene , Hex2-ene, Hex3-ene are position isomers with double bond on first, second and third carbon respectively as shown below-

[tex]CH_2= CH-CH_2-CH_2-CH_2-CH_3\\CH_3-CH=CH-CH_2-CH_2-CH_3\\CH_3-CH_2-CH=CH-CH_2-CH_3[/tex]

Other examples are pent1-yne and pent2-yne where triple bond is on first and second carbon respectively.

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Answer:

Daffodil leaves always turn yellow a few weeks after the plant blooms. This is normal and indicates that their job is finished for the season. The leaves have absorbed sunlight, which creates energy for the production of sugar that replenishes the bulb for the coming growing season.

Explanation:

A pure substance has definite composition and constant properties in all of the sample

Characteristics and Properties Of Pure Substances

Thus we can conclude that  a pure substance has definite composition in all of the sample

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Oxalate concentrations ranged widely from 53.4 to 108.8 mg/g on a dry weight basis and 647.2 to 1286.9 mg/100 g on a fresh weight basis, exhibiting a variation that was close to normal.

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The maximum mass of carbon dioxide released in the atmosphere is 6072 g/mole.

Mass is defined as a dimensionless quantity representing the amount of matter in a particle or object.

1 kg = 1000gso 2kg = 2000g

To calculate mass of carbon dioxide released in atmosphere when butane gas is burned is:

Number of mole of butane = given mass/ molar mass

                                             = 2000g/58g/mole = 34.5 moles

1 mole of butane produce 4 mole of CO2

34.5 moles of butane = 4 x 34.5 = 138 mole

Maximum mass of CO2 released = number of mole x molar mass

                                                       = 138 mole x 44 g/ mole

                                                       = 6072 g/ mole

Thus, the maximum mass of carbon dioxide released in the atmosphere is 6072 g/mole.

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[tex]HClO_4 + NaOH[/tex]   →    [tex]NaClO_4 + H_2O[/tex]    Is the net ionic reaction when sodium hydroxide and hypochlorous acid is combined.

When the combine acid and base, neutralization reaction occurs which results in formation of salt and water because

          Acid + Base     →       Salt + Water

This reaction is known as neutralization reaction.

Here, Sodium hydroxide is base and hypochlorous acid is an acid .

So salt [tex]NaClO_4[/tex] and water is formed.

This reaction is also an example of double displacement reaction.

[tex]HClO_4 and NaOH[/tex]  breaks into [tex]H^{+} , ClO_4^{-} and Na^{+} , OH^{-}[/tex] ions respectively and displaces other ion from other compound.

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Answer:

When we dissolve sugar into water, the particles of sugar occupy the spaces between the molecules of water and eventually all the sugar disappears.

Explanation:

Answer:

you can see tiny bits of sugar become wet

Explanation:

then, you can see particles of sugar expand everywhere.

The significant digits are

4 significant digits (addition) = -3.220g

subtract: -5.820g, which has four significant digits.

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The equation below is photosynthesis .

6CO₂ + 6H₂O → [tex]C_{6} H_{12} O_{6}[/tex] + 6O₂

Water and carbon dioxide are converted into food by the process of photosynthesis , which occurs in the presence of sunlight in plants and photosynthetic bacteria . They create oxygen as a byproduct and glucose as the food . The photosynthesis equation is given as ,

6CO₂ + 6H₂O → [tex]C_{6} H_{12} O_{6}[/tex] + 6O₂

This indicates that the reactants, six carbon dioxide molecules and six water molecules, are transformed into a sugar molecule and six oxygen molecules, the products, by light energy being absorbed by chlorophyll (implied by the arrow).

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This style of boldface writing has a few authorized applications. In general, style manuals advise against using it to emphasize. Italics and sentence structure are used for this. It is preferable to utilize boldface text sparingly because it might be aesthetically obtrusive.

Often boldfaced and larger than the surrounding text, titles, headers, and subheadings are used. For instance, the heading "Titles and headings" in this section is bold. This is typical for web publishing.

Boldface is frequently used in textbooks, manuals, and various types of web writing to draw attention to the first occurrence of a crucial keyword that is crucial to the topic of a chapter or article. For instance, in our article on the distinction between program and programme, we boldfaced the text to draw attention to the first occurrence of each keyword (i.e., program and programme). Later, the words do not appear in boldface.

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The reported value of adding 4.56 g to 0.01860 g to the correct significant figures will be 4.58 g.

According to one of the rules guiding addition operation in significant figures., the number with the lowest significant figure to the right of the decimal place determines the number of significant figures in the final answer.

Thus, in operation: 4.56 + 0.01860, the number with the lowest significant figures to the right of the decimal place is 4.56 with 2 significant figures after the decimal place.

In other words, the final answer of the operation will also have two significant figures after the decimal.

Hence;

                     4.56 + 0.01860 = 4.58 g

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The correct dimensional analysis for the conversion of 14,612 mL to daL  is shown in option A.

Many times, it important to determine the value of measurement across different units. This is why we need to convert from on unit to another. Such a conversion is made possible by the use of a conversion factor.

Now we know that there is a conversion factor that could be used to convert mL to daL. This conversion factor has to be laid out in a way that is dimensionally correct so that the units could cancel out.

The correct dimensional analysis for the conversion of 14,612 mL to daL  is shown in option A.

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0.0105 mol of non-electrolyte solute, with 297.14 g/mol molar mass, is dissolved in water to make a 0.0488 M solution.

Given:

Osmotic pressure is expressed mathematically as π=M*R*T. The concentration of the solution "M" should be expressed in molarity, mol/L. Through the use of the formula, let us solve for the concentration of the solution. The osmotic pressure must be converted to atm.

Conversion factor: 1 atm has 760 torr.

913 torr * (1 atm/760 torr) = 1.20131 atm

π = 1.20131 atm = M*(0.08206 L-atm/mol-K)*(300 K)

M = 0.0488 M

From the molar concentration of the solution, we can solve for the number of moles of the solute.

M = mol solute / volume solution

mol solute = M * volume solution

mol solute = 0.0488 M * (215 mL) * (1 L/1000 mL)

mol solute = 0.0105 mol

Then, we can solve for the molar mass of the solute.

molar mass solute = mass solute / mol solute

molar mass solute = 3.12 g / 0.0105 mol

molar mass solute = 297.14 g/mol

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Answer:

5 g

Explanation:

50/10 = 5,

25/5 = 5grams

hope this works: )

The answer would be

A. The sample was green, the color of chloroplast

Hope this helps

The atomic number, which is equal to the number of protons in the nucleus, controls the electrons in the atoms, which in turn control the chemistry.

The periodic table is all about chemistry, for one simple reason. The atomic number, which is equal to the number of protons in the nucleus, controls the electrons in the atoms, which in turn control the chemistry.

The atomic number is what distinguishes an element.

Additionally, if we have two distinct isotopes of the same element, their atom types will differ.

As a result, there is significant power in using isotopic systems that only use one element, as demonstrated by the analysis of the three oxygen isotopes (16O, 17O, 18O), which have different properties, should be treated as different types of atoms in an atoms table.

As each isotope of an element is listed as a different species, the chart of the nuclides can be thought of as a "chart of atoms."

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According to Avogadro’s law, the number of moles of each gas is same in the two different containers of same size at same temperature and pressure conditions.

According to Avogadro’s law, equal volume of different gases at same temperature and pressure conditions have equal number of molecules. At standard temperature and pressure conditions, Avogadro's law states that 6.022 x [tex]10^2^3[/tex] molecules of gas is present in 22.4L.

1 mole of gas is equal to the number of atoms of the elements that make up the gas. According to Avogadro’s law, 1 mole of gas contains 6.022 x [tex]10^2^3[/tex] molecules. Therefore, the number of moles of the different gases at same temperature and pressure conditions is same.

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Through unit conversion, it has been determined that a crystal of sugar contains 45,000 atoms.

The problem can be solved through unit conversion. The atoms of the table sugar, with a molecular formula of C12H22O11, are 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms. As stated in the problem, a molecule of sugar has 45 atoms and a crystal of sugar has 1000 molecules.

In solving for the number of atoms in a sugar crystal, we will be getting the product between the number of molecules per crystal of sugar to the number of atoms per molecule.

(1000 molecules of sugar/crystal of sugar) * (45 atoms of sugar/molecule of sugar) = 45,000 atoms of sugar/crystal of sugar.

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Answer:

A for number 3 and C for number 4

Explanation:

I hope you pass this homework

The diameter of the nucleus in angstrom, given the data is 1×10⁻⁵ angstrom

1 nm = 10 angstrom

The following data were obtained from the question:

1 m = 1×10⁹ nm

Therefore,

1×10⁻¹⁵ m = (1×10⁻¹⁵ m × 1×10⁹ nm) / 1 m

1×10⁻¹⁵ m = 1×10⁻⁶ nm

1 nm = 10 angstrom

Therefore,

1×10⁻⁶ nm = (1×10⁻⁶ nm × 10 angstrom) / 1 nm

1×10⁻⁶ nm = 1×10⁻⁵ angstrom

Thus, the diameter of the nucleus in angstrom is 1×10⁻⁵ angstrom

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