which bond does water atom share​

Answer:

a. 25.7mL are necessaries.

b. 1.964g of salicylic acid can be isolated.

Explanation:

a. As the solubility of salicylicic acid in boiling water is 77.8g/L, to dissolve 2.0g of salicylic acid are necessaries:

2.0g * (1L / 77.8g) = 0.0257L =

b. Then, at 10°C the solution just can dissolve 1.4g/L, that is:

0.0257L * (1.4g/L) = 0.036g of salicylic acid are dissolved.

And will precipitate:

2.0g - 0.036g =

Answer:

C. Kidneys filter wastes from the bloodstream and produce urine

Explanation:

Answer:

As the climate changes, temperatures are increasing, sea levels are rising, carbon dioxide levels in the atmosphere are increasing, the pH of the ocean is decreasing, and the amount of sea ice is decreasing.

Explanation:

Answer: The equilibrium constant for [tex]PX_5(g)\rightarrow PX_3(g)+X_2(g)[/tex] is 1.35

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as [tex]K_c[/tex]  

For the given chemical reaction:

[tex]PX_3(g)+X_2(g)\rightarrow PX_5(g)[/tex]

The expression for [tex]K_c[/tex] is written as:

[tex]K_c=\frac{[PX_5]^1}{[PX_3]^1[X_2]^1}[/tex]

[tex]0.74=\frac{[PX_5]^1}{[PX_3]^1[X_2]^1}[/tex]

For the reverse chemical reaction:

[tex]PX_5(g)\rightarrow PX_3(g)+X_2(g)[/tex]

The expression for [tex]K_c'[/tex] is written as:

[tex]K_c'=\frac{[PX_3]^1[X_2]^1}{[PX_5]^1}[/tex]

[tex]K_c'=\frac{1}{K_c}=\frac{1}{0.74}=1.35[/tex]

The equilibrium constant for [tex]PX_5(g)\rightarrow PX_3(g)+X_2(g)[/tex] is 1.35

Answer:

0.150M

Explanation:

hope it helps!!!!!!!!!!!!

The molarity of the resulting solution when 300. mL of a 0.400 M solution is diluted to 800 mL is 0.15M.

Molarity is defined as the ratio of the number of moles of solute present in a solution, to the volume of that particular solution in litres.

It is calculated by using the formula

M₁V₁ = M₂V₂

M₁= initial molarity of the solution

V₁ = initial volume of the solution

M₂ = final molarity of the solution

V₂ = final volume of the solution

Given data,

M₁ = 0.400 M (initial molarity)

V₁ = 300 mL (initial volume)

V₂ = 800 mL (final volume)

Substituting the values we get

M₁V₁ = M₂V₂

.4× 300/800 = M₂

Therefore, molarity of the resulting solution = .15

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Answer: The percent yield of the reaction is 83.5 %

Explanation:

The given balanced equation is

[tex]SiO_2+3C\rightarrow SiC+2CO[/tex]

[tex]SiO_2[/tex] is the limiting reagent as it limits the formation of product and [tex]C[/tex] is the excess reagent.

According to stoichiometry :

60.08 g of [tex]SiO_2[/tex] produce = 40.11 of [tex]SiC[/tex]

Thus 50.0 of [tex]SiO_2[/tex] will produce=[tex]\frac{40.11}{60.08}\times 50.0=33.4[/tex]  of [tex]SiC[/tex]

 Experimental yield of SiC = 27.9 g

Percent yield = [tex]\frac{\text {Experimental yield}}{\text {theoretical yield}}\times 100=\frac{27.9g}{33.4g}\times 100=83.5\%[/tex]

Thus  percent yield of the reaction is 83.5 %

The percent yield of 83.5 % of 50.0 g of silicon dioxide is heated with an excess of carbon, and 27.9 g of silicon carbide is produced in the reaction.

The chemical equations are balanced when the reactants react to form products. The reactants and products react in proper ratios and if they are not in ratio then we balance them by adding the required quantity in the reactants and the products.

The given balanced equation is

[tex]\rm SiO_2+3C---- > SiC+2CO[/tex]

[tex]SiO_2[/tex] is the limiting reagent as it limits the formation of product and is the excess reagent.

According to stoichiometry

60.08 g [tex]SiO_2[/tex] of  produce = 40.11 of [tex]SiC[/tex]

Thus 50.0 of  [tex]SiO_2[/tex] will produce=  [tex]\dfrac{40.11}{60.08} \times 50=33.4\ SiC[/tex]

The experimental yield of SiC = 27.9 g

The percentage yield will be calculated as

[tex]\rm Percentage \ Yield = \frac{Experimental\ yield}{Theoretical \ yield }\times 100[/tex]

[tex]\rm Percentage \ yield =\dfrac{27.9}{33.49} \times 100=83.5[/tex]

Thus the percent yield of 83.5 % of 50.0 g of silicon dioxide is heated with an excess of carbon, and 27.9 g of silicon carbide is produced in the reaction.

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Answer:

The correct answer is

- after deprotonation carboxylic acid stabilizes by binding -OH group with hydrogen bonding in salicylic acid and while it's not possible with its constitutional isomer (para-hydroxy benzoic acid).

The more electronegative atoms are replaced by the H atom of the carboxylic group and form more stabilize carboxylic acid which not takes place in para-hydroxy benzoic acid.

Answer:

For safety.

Explanation:

Because uncovered wires are conductors, an electric current could damage someone if touched. Plastic and rubber are insulators, meaning that current cannot pass through them.

The calculated formula masses to 133.5 amu, we find that the substance with a formula mass closest to 133.5 amu is (d) AlCl3. Therefore, the answer is option D.

To identify the substance with a formula mass of 133.5 amu, we need to calculate the formula mass of each given substance and compare it to 133.5 amu.

(a) MgCl2:

The formula mass of MgCl2 can be calculated by adding the atomic masses of magnesium (Mg) and chlorine (Cl).

Mg: atomic mass = 24.31 amu

Cl: atomic mass = 35.45 amu

Formula mass of MgCl2 = (24.31 amu) + 2(35.45 amu) = 95.21 amu

(b) SCl:

The formula mass of SCl can be calculated by adding the atomic masses of sulfur (S) and chlorine (Cl).

S: atomic mass = 32.07 amu

Cl: atomic mass = 35.45 amu

Formula mass of SCl = 32.07 amu + 35.45 amu = 67.52 amu

(c) BCl:

The formula mass of BCl can be calculated by adding the atomic mass of boron (B) and chlorine (Cl).

B: atomic mass = 10.81 amu

Cl: atomic mass = 35.45 amu

Formula mass of BCl = 10.81 amu + 35.45 amu = 46.26 amu

(d) AlCl3:

The formula mass of AlCl3 can be calculated by adding the atomic mass of aluminum (Al) and 3 times the atomic mass of chlorine (Cl).

Al: atomic mass = 26.98 amu

Cl: atomic mass = 35.45 amu

Formula mass of AlCl3 = 26.98 amu + 3(35.45 amu) = 133.78 amu. Option D

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The average atomic mass of the element R that has three isotopes is 127.805.

Average atomic mass is the weighted average of the atomic masses of the naturally occurring isotopes of an element.

The average atomic mass of an element can be calculated by summing up the product of the percent abundance and masses of each isotope as follows;

Average atomic mass of R = (0.0825 × 81) + (0.2671 × 115) + (0.6504 × 139)

Average atomic mass = 6.6825 + 30.7165 + 90.4056 = 127.805

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Answer:

A

Explanation:

The weak base ionization constant (Kb) for [tex]C_{4} H_{4} N_{2}[/tex] is equal to Option A.

A weak foundation is a base that makes ionize only slightly in an aqueous solution. Remember that a foundation can be defined as an object, which receives a hydrogen ion from another object.

To obtain an aqueous solution of a weak acid, the dissociation constant is called the acid ionization constant (Ka). Similarly, the constant equilibrium of a weak base reaction with water is the base ionization constant (Kb). In any conjugate acid-base pair, KaKb = Kw.

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Answer:

Dont completly trust me here but i think its a but id try and wait for someone else just be careful

Explanation:

Solution :

"Aldol" stands for the abbreviation, aldehyde and alcohol. When a ketone or an aldehyde's enolate reacts with the carbonyl of a  molecule at the alpha carbon, under the acidic or basic conditions so as to obtained the ketone or β-hydroxy aldehyde, is known as an aldol reaction.

For the conversion of the aldol addition product of a 3-hydroxy-3-(4-nitrophenyl)-1-(2-pyridyl)-1-propanone to an aldol condensation product of (E)-3-(4-nitrophenyl)-1-(2-pyridyl)-1-propenone, the mechanism is given in the diagram a below :

Answer:

2.0 × 10⁻³ g

Explanation:

Step 1: Given data

Step 2: Calculate the mass of Br₂ required to prepare the solution

The concentration of Br₂ is 2.0 ppm, that is, there are 2.0 mg of Br₂ per kilogram of solution. The mass of Br₂ required to prepare 1.000 kg of solution is:

1.000 kg Solution × 2.0 mg Br₂/1 kg Solution = 2.0 mg

Step 3: Convert the mass to grams

We will use the conversion factor 1 g = 1000 mg.

2.0 mg × 1 g/1000 mg = 2.0 × 10⁻³ g

HCH bond angle is ~110 degrees.

The data set is missing in the question. The data set is given in the attachment.

Solution :

a). In the table, there are four positive examples and give number of negative examples.

Therefore,

[tex]$P(+) = \frac{4}{9}$[/tex]   and

[tex]$P(-) = \frac{5}{9}$[/tex]

The entropy of the training examples is given by :

[tex]$ -\frac{4}{9}\log_2\left(\frac{4}{9}\right)-\frac{5}{9}\log_2\left(\frac{5}{9}\right)$[/tex]

= 0.9911

b). For the attribute all the associating increments and the probability are :

  [tex]$a_1$[/tex]   +   -

  T   3    1

  F    1    4

Th entropy for   [tex]$a_1$[/tex]  is given by :

[tex]$\frac{4}{9}[ -\frac{3}{4}\log\left(\frac{3}{4}\right)-\frac{1}{4}\log\left(\frac{1}{4}\right)]+\frac{5}{9}[ -\frac{1}{5}\log\left(\frac{1}{5}\right)-\frac{4}{5}\log\left(\frac{4}{5}\right)]$[/tex]

= 0.7616

Therefore, the information gain for [tex]$a_1$[/tex]  is

  0.9911 - 0.7616 = 0.2294

Similarly for the attribute [tex]$a_2$[/tex]  the associating counts and the probabilities are :

  [tex]$a_2$[/tex]  +   -

  T   2    3

  F   2    2

Th entropy for   [tex]$a_2$[/tex] is given by :

[tex]$\frac{5}{9}[ -\frac{2}{5}\log\left(\frac{2}{5}\right)-\frac{3}{5}\log\left(\frac{3}{5}\right)]+\frac{4}{9}[ -\frac{2}{4}\log\left(\frac{2}{4}\right)-\frac{2}{4}\log\left(\frac{2}{4}\right)]$[/tex]

= 0.9839

Therefore, the information gain for [tex]$a_2$[/tex] is

  0.9911 - 0.9839 = 0.0072

[tex]$a_3$[/tex]     Class label      split point       entropy        Info gain

1.0         +                        2.0            0.8484        0.1427

3.0        -                         3.5            0.9885        0.0026

4.0        +                        4.5            0.9183         0.0728

5.0        -

5.0        -                        5.5            0.9839        0.0072

6.0        +                       6.5             0.9728       0.0183

7.0        +

7.0        -                        7.5             0.8889       0.1022

The best split for [tex]$a_3$[/tex]  observed at split point which is equal to 2.

c). From the table mention in part (b) of the information gain, we can say that [tex]$a_1$[/tex] produces the best split.

The atomic mass of element R is 151 (no decimals).

To calculate the atomic mass of element R, we need to consider the relative abundance of each isotope and its corresponding mass. The atomic mass is the weighted average of the masses of all the isotopes, taking into account their relative abundance.

Given:

Isotope 1: Relative abundance = 0.0398, Mass = 191

Isotope 2: Relative abundance = 0.1614, Mass = 180

Isotope 3: Relative abundance = 0.7988, Mass = 143

To calculate the atomic mass, we multiply the relative abundance of each isotope by its mass, and then sum up the results.

Atomic mass = (Relative abundance of Isotope 1 * Mass of Isotope 1) + (Relative abundance of Isotope 2 * Mass of Isotope 2) + (Relative abundance of Isotope 3 * Mass of Isotope 3)

Atomic mass = (0.0398 * 191) + (0.1614 * 180) + (0.7988 * 143)

Calculating the values:

Atomic mass = 7.6098 + 29.0256 + 114.6872

Atomic mass = 151.3226

Rounding to the nearest whole number, the atomic mass of element R is 151.

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Answer:

they can track the weather which can show if it would be posssible to live or have life on mars

Explanation:

Answer:  Formula: Mass = (Volume)(Density)

Iron Density = 7.87 g/cm^3

Volume of Iron = 55.2 cm^3

Mass=(V)(D)

Mass= (55.2 cm^3) x (7.87 g/cm^3)

Mass=  434,42 g

Explanation:

Iron has a density of 7.87 g/cm³. 434,42 g is the mass of 55.2 cm³ of iron.

The term density is defined as the measurement of how closely a material is packed together.

It is also defined as the mass per unit volume. Density Symbol is D or ρ Density Formula is ρ = m/V, where ρ is the density, m is the mass of the object and V is the volume of the object.

Density is an important because it allows us to find out what substances will float and what substances will sink when placed in a liquid.

Formula:

Mass = (Volume)(Density)

Given:

Iron Density = 7.87 g/cm³

Volume of Iron = 55.2 cm³

Mass=(V)(D)

Mass= (55.2 cm³) x (7.87 g/cm³)

Mass= 434,42 g

Thus, Iron has a density of 7.87 g/cm³. 434,42 g is the mass of 55.2 cm³ of iron.

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Answer:

1.5 mol

Explanation:

Step 1: Given data

Step 2: Calculate the moles corresponding to 33 L of argon at standard temperature and pressure (STP)

At STP, 1 mole of argon gas occupies 22.4 L.

33 L × 1 mol/22.4 L = 1.5 mol

Answer:

Explanation:

There are 7 protons in nitrogen

7 TYPE OF NITRONGEN

. . .

Answer:

3.03 x 10²³ formula units

Explanation:

First we convert 48.0 grams of magnesium chloride into moles, using its molar mass:

Then we convert 0.504 moles into formula units, using Avogadro's number:

Answer:

78.75L

Explanation:

According to Charle's Law:

V1/T1=V2/T2

35/200=V2/450

V2=(35*450)200

V2=78.75L

Answer:

125000mg

Explanation:

1ml = 1000mg

125(1000) = 125000mg

Answer: The discovery of three buried lakes. Scientists think that a long time ago there were lakes and rivers, etc on Mars. Now of course, you can't see any visible water sources on the surface.

Answer:

Almost all water on Mars today exists as ice, though it also exists in small quantities as vapor in the atmosphere.[5] What was thought to be low-volume liquid brines in shallow Martian soil, also called recurrent slope lineae may be grains of flowing sand and dust slipping downhill to make dark streaks.The only place where water ice is visible at the surface is at the north polar ice cap. Abundant water ice is also present beneath the permanent carbon dioxide ice cap at the Martian south pole and in the shallow subsurface at more temperate conditions. More than 5 million km3 of ice have been detected at or near the surface of Mars, enough to cover the whole planet to a depth of 35 meters. Even more ice is likely to be locked away in the deep subsurface.

Some liquid water may occur transiently on the Martian surface today, but limited to traces of dissolved moisture from the atmosphere and thin films, which are challenging environments for known life. No large standing bodies of liquid water exist on the planet's surface, because the atmospheric pressure there averages just 600 pascals , a figure slightly below the vapor pressure of water at its melting point; under average Martian conditions, pure water on the Martian surface would freeze or, if heated to above the melting point, would sublime to vapor. Before about 3.8 billion years ago, Mars may have had a denser atmosphere and higher surface temperatures, allowing vast amounts of liquid water on the surface, possibly including a large ocean that may have covered one-third of the planet.Water has also apparently flowed across the surface for short periods at various intervals more recently in Mars' history. Aeolis Palus in Gale Crater, explored by the Curiosity rover, is the geological remains of an ancient freshwater lake that could have been a hospitable environment for microbial life.Many lines of evidence indicate that water ice is abundant on Mars and it has played a significant role in the planet's geologic history.The present-day inventory of water on Mars can be estimated from spacecraft images, remote sensing techniques (spectroscopic measurements, radar, etc.), and surface investigations from landers and rovers.Geologic evidence of past water includes enormous outflow channels carved by floods, ancient river valley networks, deltas and lakebeds,and the detection of rocks and minerals on the surface that could only have formed in liquid water. Numerous geomorphic features suggest the presence of ground ice (permafrost)and the movement of ice in glaciers, both in the recent past and present. Gullies and slope lineae along cliffs and crater walls suggest that flowing water continues to shape the surface of Mars, although to a far lesser degree than in the ancient past.Although the surface of Mars was periodically wet and could have been hospitable to microbial life billions of years ago, the current environment at the surface is dry and subfreezing, probably presenting an insurmountable obstacle for living organisms. In addition, Mars lacks a thick atmosphere, ozone layer, and magnetic field, allowing solar and cosmic radiation to strike the surface unimpeded. The damaging effects of ionizing radiation on cellular structure is another one of the prime limiting factors on the survival of life on the surface. Therefore, the best potential locations for discovering life on Mars may be in subsurface environments. Large amounts of underground ice have been found on Mars; the volume of water detected is equivalent to the volume of water in Lake Superior. In 2018, scientists reported the discovery of a subglacial lake on Mars, 1.5 km (0.93 mi) below the southern polar ice cap, with a horizontal extent of about 20 km (12 mi), the first known stable body of liquid water on the planet.Understanding the extent and situation of water on Mars is vital to assess the planet’s potential for harboring life and for providing usable resources for future human exploration. For this reason, "Follow the Water" was the science theme of NASA's Mars Exploration Program (MEP) in the first decade of the 21st century. NASA and ESA missions including 2001 Mars Odyssey, Mars Express, Mars Exploration Rovers (MERs), Mars Reconnaissance Orbiter (MRO), and Mars Phoenix lander have provided information about water's abundance and distribution on Mars.Mars Odyssey, Mars Express, MRO, and Mars Science Lander Curiosity rover are still operating, and discoveries continue to be made.

Answer:

The sky looks different in each nighttime section of the artifact because the artifact sections represent different seasons. ... Different constellations are visible on different nights throughout the year because of the earth's orbit. The Earth orbits around the sun. A full orbit is 365 days or one year.

Explanation:

Answer:

0.01804 M KMnO4 Solution.

Explanation:

3 Percent(w/w) H2O2 Are Sold In Drug Stores As A Disinfectant. ... H2O2 In A Bottle Of Peroxide Solution Was Done By Diluting 1.50 ML To 100 ML With Water, Acidifying With Dilute H2SO4 And Titrating With A 0.01804 M KMnO4 Solution.

Answer:

All forms of energy are either potential or kinetic energy. Potential refers to stored energy while kinetic is energy in motion.

Explanation:

hope help you pls thanks...

D is correct option