Calculate the percent composition of C in carbon dioxide.

Answer:

SEE EXPLANATION

Explanation:

We must first obtain the molar mass of CHCl3 as follows;

CHCl3 = ( 12.01 * 1 )+ (1.008 * 1 ) + ( 35.45 * 3 ) => 119.37 g/mol

Then we obtain the percentage by mass of each element

For Carbon =  ( atomic mass C / molar mass CHCl3 ) * 100

C  =  (12.01 / 119.37 ) * 100

C = ( 0.1006 * 100 )

C =  10.06 %

For Hydrogen :

H = ( atomic mass H / molar mass CHCl3 ) * 100

H = ( 1.008 / 119.37 ) * 100

H = 0.008444 * 100

H = 0.8444 %

For Chlorine :

Cl  ( molar mass Cl3 / molar mass CHCl3 ):

Cl =  ( 3 * 35.5 / 119.37 ) * 100

Cl =  ( 106.5 / 119.37 ) * 100

Cl = 0.8921 * 100

Cl = 89.92%

Answer:

This happened because,the comb was charged with static electricity when it was rubbed on your hair.

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Answer:

C. Kidneys filter wastes from the bloodstream and produce urine

Explanation:

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:

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

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:

A metalloid is a chemical element that exhibits some properties of metals and some of nonmetals.The closer you get to the zone, the more blurred the properties become. Metalloids tend to be semiconductors; silicon is the best known example of a semiconductor. Most microchips and microprocessors are made with silicon.

Explanation:

That is what i know on the answer

To solve this we must be knowing each and every concept related to metalloid. Therefore, in above given ways semiconductor can be defined.

Metalloids are chemical elements with physical and chemical characteristics that fall between between metal and nonmetal.

A metalloid is indeed a chemical element with characteristics of both metals and nonmetals. The properties grow increasingly blurred as you move closer to the zone. The seven most well-known metalloids are boron, germanium, silicon, antimony, arsenic, tellurium, and polonium.

Metalloids are often semiconductors, with silicon being the most well-known example. Silicon is used to make the majority of microchips and microprocessors.

Therefore, in above given ways semiconductor can be defined.

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Convection currents in the mantle, I got it right on a p e x

Explanation:

32

2H

2

+O

2

→2H

2

O

Molecular mass of H

2

=2 g/mol

Molecular mass of O

2

=32 g/mol

From the balanced chemical equation,

2×2=4 g of hydrogen requires 32 g of Oxygen to react completely

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:

D. 0.063 mol

Explanation:

We can calculate the number of moles by using the definition of molarity:

As we are given both the molarity and the volume:

We can calculate the number of moles:

Thus the answer is option D.

Explanation:

by dissolving the mixture of lead sulphate and lead chloride in water we can separate the two. after dissolving the mixture on water lead sulphate can be obtained as the solid that 's left behind lead chloride can be recovered by evaporating.

Answer:

15 is it

Explanation:

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: 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:

Explanation:

There are 7 protons in nitrogen

7 TYPE OF NITRONGEN

. . .

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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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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Answer:

helium is the answer

Explanation:

helium is the smallest element in francium is the largest hope this helps

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

Answer:

125000mg

Explanation:

1ml = 1000mg

125(1000) = 125000mg

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.

Answer:

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

Explanation:

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:

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.

HCH bond angle is ~110 degrees.

Answer:

1 km an hour

Explanation:

i mean its pretty easy like you said if at 4 hours its traveled 4 km then its going 1 km an hour and at 10 its gone 10 km

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

Explanation: where the article????