How many electrons are there in a chlorine ion?

Answer:

from the Sun to the surface of the Earth, where it is absorbed or reflected

Explanation:

The solar radiation that emerges out of the sun and that reaches the surface of the Earth is sunlight. There are numerous constituents and particles present in the atmosphere that absorbs the solar radiation. These radiations are weakened by these constituents. Later, these radiations gets scattered in the atmosphere. Some amount of he solar radiation that reaches the Earth's surface is absorbed and some is reflected back.

Answer:

B

Explanation:

Answer:  

3.2 moles of carbon dioxide

Explanation:

When C₂H₂ reacts with O₂ the chemical reaction is:

C₂H₂ + 5/2O₂ → 2CO₂ + H₂O

Where 1 mole of acetylene produce 2 moles of carbon dioxide.

Based on the chemical equation of this reaction, there are produced twice the moles of acetylene in moles of carbon dioxide assuming a compete reaction.

If 1.6 moles of acetylene reacts there are produced:

1.6 moles Acetylene * (2 moles Carbon Dioxide / 1 mole Acetylene) =

3.2 moles of carbon dioxide

Answer:

C               (I think)

Explanation:

the levels are higher than they have ever been

Answer:

52.4%

Explanation:

Since a simple cubic unit cell contains only 1 atom. The packing efficiency of the simple cubic cell is 52.4 %

Answer: A property closely related to an atom's mass number is its atomic mass. The atomic mass of a single atom is simply its total mass and is typically expressed in atomic mass units or amu. By definition, an atom of carbon with six neutrons, carbon-12, has an atomic mass of 12 amu.

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Answer:

it is 7

Explanation:

dogs are 7 HOPE THIS HELPED kidding i don get the question

Explanation:

DON"T EAT MY DOG OR I"LL EAT YOU

Explanation:

Moles of Co = 24.63 / 58.9 = 0.418

Moles of O = (31.32 - 24.63) / 16 = 0.418

Since the mole ratio is 1 : 1, the empirical formula is CoO.

The empirical formula of the metal oxide is CaO₂.

The empirical formula is the simplest ratio of the number of atoms participating in the formation of a compound.

The chemical formula of any compound is based on two types of formula one is molecular formulas and another one is empirical formulas. Both the molecular formula and empirical formula represent the identity and number of the atom.

Given, the mass of the sample of the cobalt = 24.63g

The mass of the metal oxide = 31.32g

From the law of the conservation of mass:

The mass of the metal oxide - Mass of cobalt = mass of the oxygen

So the mass of the oxygen = 31.32 - 24.63 = 6.69 g

If mass 59 grams of the cobalt has moles = 1

Then 24.63 g of Cobalt will have moles = 24.63/59 = 0.417 mol

Similarly, 8 grams of the oxygen gas moles = 1

Then 6.69g of the oxygen has moles = 6.69/8 = 0.8362 mol

The simplest ratio of moles for Co = 0.417/0.417 = 1

For oxygen = 0.8362/0.417 = 2

Therefore the empirical formula of the metal oxide is CaO₂.

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The element will lose electrons to become stable if the valence shell contains 3 or fewer electrons. The production of cations is a sign of electron loss. The species having positive charge are called cations. As a result, the ion will get a positive charge.

An atom or molecule with a net electrical charge is known as an ion. Conventionally, the charge of an electron is thought to be negative; this charge is equal to and opposite to the charge of a proton, which is thought to be positive.

An atom will shed its valence electrons in order to become a charged ion and create an ionic bond with another atom if it lacks one to three of them.

A positive ion is created by taking electrons from a neutral atom, whereas a negative ion is created by adding electrons to a neutral atom. The quantity of protons remains constant during the formation of an ion.

Thus, the ion will get a positive charge.

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In every 100g of that compund there is 50.84 g of C, 8.53 g H and (100-59.37) g = 40.63 g of O.

Step 1: Convert each element's mass in moles. To do that we need to divide each element's mass by their respective molar mass.

For Carbon.

[tex]C = \frac{50.84}{12} = 4.24 mol [/tex]

For Hydrogen.

[tex]H = \frac{8.53}{1} = 8.53 mol [/tex]

For Oxygen.

[tex]O = \frac{40.63}{16} = 2.54 mol [/tex]

Step 2: Divide each of the numbers by the smallest number.

For Carbon.

[tex]C = \frac{4.24}{2.54} = 1.7 [/tex]

For Hydrogen.

[tex]H = \frac{8.53}{2.54} = 3.36 [/tex]

For Oxygen.

[tex]O = \frac{2.54}{2.54} = 1 [/tex]

Step 3: So the empirical formula will be.

[tex] C_{1.7} H_{3.36} O_1 [/tex]

But using decimal will be messy. So we multiply the numbers by 3. The right empirical formula will be.

[tex] C_{5} H_{10} O_3 [/tex]

The empirical formula of an organic compound that contains 50.84% C, 8.53% H, and the rest oxygen is C₅H₁₀O₃.

First, we have to determine the mass percentage of oxygen, knowing that C, H and O amount to the 100% of the compound.

[tex]C + H + O = 100\%\\O = 100\% - C - H\\O = 100\% - 50.84\% - 8.53\% = 40.63\%[/tex]

Then, we divide each mass percentage by the atomic mass of the element.

[tex]C: 50.84/12.01 = 4.23\\H: 8.53/1.01 = 8.45\\O: 40.63/16.00 = 2.54[/tex]

Afterwards, we divide all the numbers by the smallest one, i.e. 2.54.

[tex]C: 4.23/2.54 = 1.66\\H: 8.45/2.54 = 3.33\\O: 2.54/2.54 = 1[/tex]

Finally, we multiply all the numbers by 3, so that they are all integers.

[tex]C: 1.66 \times 3 \approx 5\\H: 3.33 \times 3 \approx 10\\O: 1 \times 3 = 3[/tex]

The empirical formula of the organic compound is C₅H₁₀O₃.

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Explanation:

a. Write the molecular formula of the compound.

Acetaminophen contains 8 carbon atoms, 9 Hydrogen atoms, 1 Nitrogen atom and 2 Oxygen atoms.

The molecular formular is given as;  C8H9NO2

b. What is the hybridization state of each C, N, and O atom?

From the image attached;

All the carbon atoms in the benzene ring has hybridization of sp2.

The carbon that is doubly bonded (carbonyl) to the oxygen has hybridization of sp2

The last carbon to the right has hybridization of sp3

The hybridization of nitrogen in the compound is sp3 hbridized.

Oxygen atom in alcohol  is sp3 hybridised

Oxgen in the carbonyl group is sp2 hybridized.

Answer:

ca you attach these two pdf ?? or provide the link to this document

Answer:

0.98 M

Explanation:

Given data:

Mass of sugar = 20 g

Mass of water = 59 g

Concentration of solution = ?

Solution:

1 g = 0.01 L

59 g× 0.01 L/1 g

0.059 L

Concentration formula:

C = m/v

m = moles of solute

v = volume of solution

Number of moles of sugar:

Number of moles = mass/molar mass

Number of moles = 20 g/ 342.3 g/mol

Number of moles = 0.058 mol

Now we will calculate the concentration:

C = 0.058 mol/0.059 L

C = 0.98 mol/L

C = 0.98 M

A.

H₃C-CH₃

this is called ethane

B.

H₃C-CH₂-CH₂-CH₃

this is called butane

C.H₃C-CH₂-CH₂-CH₂-CH₂-CH₃

this is called hexane

D.

H₃C-CH₂-CH₂-CH₂-CH₂-CH₂-CH₃

this is called heptane

Answer:

1. a) Ethane (C2H6)

2. b) Butane (C4H10 )

3. c) Hexane (C6H14)

4. d) Heptane (C7H16)

Answer:  The oxidation number of chlorine in [tex]Cl_2[/tex] is zero

Explanation:

Oxidation number or oxidation state : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Rules for Oxidation Numbers are :

1.  The oxidation number of a free element is always zero.

2. The oxidation number of a monatomic ion equals the charge of the ion.

3. The oxidation number of  oxygen (O)  in compounds is usually -2.

4. The oxidation number of a Group 17 element in a binary compound is -1.

5. The oxidation number of  Hydrogen (H)  is +1, but it is -1 in when combined with less electronegative elements.

6. The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.

7. The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

As [tex]Cl_2[/tex] exists as a free molecule, its oxidation state is zero.

Answer:

0.00752L of 0.305M K₃PO₄

Explanation:

Based on the balanced chemical equation, 2 moles of K₃PO₄ react with 3 moles of NiCl₂.

To find the volume of K₃PO₄ we need to determine the moles of this compound required for a complete reaction finding moles of NiCl₂ and using the chemical equation:

Moles NiCl₂:

187mL = 0.187L * (0.0184mol / L) = 0.00344 moles NiCl₂

Moles K₃PO₄:

As 2 moles of K₃PO₄ react with 3 moles of NiCl₂:

0.00344 moles NiCl₂ * (2 mol K₃PO₄ / 3 moles NiCl₂) =

0.002294 moles of K₃PO₄

Volume of 0.305M K₃PO₄:

0.002294 moles of K₃PO₄ * (1L / 0.305moles K₃PO₄) =

Answer:

Glucose is combined with oxygen and releases usable energy, carbon dioxide, and water. Cells can use that extra energy to power their functions. The energy isn't just floating around. It's stored in an excitable compound called ATP (adenosine triphosphate).

Explanation:

Answer: ATP(adenosine triphosohate) is released with water and carbon dioxide as byproducts.

Explanation: Remember that water, CO2,and ATP are the products of cellular respiration.

Answer:

receptors which detect stimuli changes in the environment

Explanation:

Answer:

major enantiomer = 90.5 %

minor enantiomer = 9.5 %

Explanation:

Assuming that the  major isomer is in excess;

From, ee/2 + 50

ee = enantiomeric excess = 81%

%  major =81/2 + 50 = 90.5 %

%  minor = 100% - 90.5 %

%  minor = 9.5 %

Answer:

35.1 g

Explanation:

Given data:

Mass of Al(OH)₃ = 100 g

Mass of water formed = ?

Solution:

Chemical equation:

2Al(OH)₃      →        Al₂O₃ + 3H₂O

Number of moles of Al(OH)₃:

Number of moles = mass/molar mass

Number of moles = 100 g/ 78 g/mol

Number of moles = 1.3 mol

now we will compare the moles of Al(OH)₃ with water.

                          Al(OH)₃         :          H₂O

                             2                :            3

                             1.3              :           3/2×1.3 = 1.95

Mass of water produced:

Mass = number of moles ×molar mass

Mass = 1.95 mol × 18 g/mol

Mass = 35.1 g

The oxidation number for S in the compound SO₃ : +6

The formula for determining Oxidation Numbers :

1. Single element atomic oxidation number = 0.

Group IA (Li, Na, K, Rb, Cs, and Fr): +1

Group IIA (Be, Mg, Ca, Sr and Ba): +2

H in compound = +1, except metal hydride compounds (Hydrogen which binds to groups IA or IIA) oxidation number H = -1, for example, LiH, MgH₂, etc.

2. Oxidation number O in compound = -2, except OF₂ = + 2 and in peroxide (Na₂O₂, BaO₂) = -1 and superoxide, for example KO₂ = -1/2.

3 The oxidation number in an uncharged compound = 0,

Total oxidation number in ion = ion charge, Example NO₃⁻ = -1

The oxidation number for S in SO₃ :

[tex]\tt oxidation~number~of~S+3\times oxidation~number~of~O=0\\\\oxidaton~number~of~S+3\times(-2)=0\\\\oxidation~number~of~S=+6[/tex]

Answer:

The molar concentration of a solution made with 3.744 g of Mg(NO₃)₂ dissolved in enough water to make 50.0 mL of solution is [tex]0.5 \frac{moles}{L}[/tex]

Explanation:

Molarity or Molar Concentration is the number of moles of solute that are dissolved in a certain volume.

The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution:

[tex]molarity=\frac{number of moles}{volume}[/tex]

In this case:

So, the molar mass of Mg(NO₃)₂ is:

Mg(NO₃)₂= 24.3 g/mole + 2*(14 g/mole + 3*16 g/mole)= 148.3 g/mole

So, if you have 3.744 g of Mg(NO₃)₂, you can apply the following rule of three: if 148.3 grams of Mg(NO₃)₂ are present in 1 mole, 3.744 grams in how many moles are present?

[tex]moles=\frac{3.744 grams*1mole}{148.3 grams}[/tex]

moles= 0.025

Then you have:

Replacing in the definition of molarity:

[tex]molarity=\frac{0.025 moles}{0.05 L}[/tex]

you get:

[tex]molarity=0.5 \frac{moles}{L}[/tex]

The molar concentration of a solution made with 3.744 g of Mg(NO₃)₂ dissolved in enough water to make 50.0 mL of solution is [tex]0.5 \frac{moles}{L}[/tex]

The molar concentration of the solution is 0.506 M

The molar concentration of a solution is known as molarity. The molarity of a solution is defined as the moles of a solute per liter of a solution.

It can be expressed as:

[tex]\mathbf{Molarity = \dfrac{moles \ of \ solute}{volume \ of \ the \ solution}}[/tex]

The relation used in determining the number of moles is:

[tex]\mathbf {number \ of \ moles = \dfrac{mass}{molar \ mass}}[/tex]

[tex]\mathbf {number \ of \ moles = \dfrac{3.744 \ g}{148.3 \ g/mol}}[/tex]

[tex]\mathbf {number \ of \ moles = 0.0253 \ g/mol}[/tex]

∴

[tex]\mathbf{Molarity = \dfrac{moles \ of \ solute}{volume \ of \ the \ solution}}[/tex]

[tex]\mathbf{Molarity = \dfrac{0.0253 \ g/mol}{0.05 \ L}}[/tex]

[tex]\mathbf{Molarity =0.506 \ M}[/tex]

Therefore, we can conclude that the molar concentration of the solution is 0.506 M

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Answer:

A heterogeneous catalyst can be easily separated from reactants.

Explanation:

Answer:

The four elements most commonly found in living things are carbon, nitrogen, hydrogen, and oxygen.

Explanation:

The four elements most commonly found in living things are carbon, nitrogen, hydrogen, and oxygen.

B. Sodium hydroxide and Hydrochloric acid

Chemical equations can be expressed in terms of

In a chemical reaction, there are several components, including reactants, products, and compounds that react in the form of solids, gases, and liquids

Reactants are compounds that react and form products

The reactants are generally to the left of the chemical equation and the products are to the right of the reaction

The statement of the word reaction above can be expressed by the reaction formula

[tex]\tt NaOH+HCl\rightarrow NaCl+H_2O[/tex]

This reaction is called a neutralization reaction because it involves acids(HCl) and bases(NaOH) and produces salt(NaCl)and water(H₂O)

From the reaction above, the reactants on the left are NaOH and HCl compounds

Answer:

Chemistry can be applicable in many aspects which are

i) At school, through getting knowledge you can assess different aspect of chemistry, eg production of Chemicals.

ii) At hospital, also chemistry can be applicable in hospital because most of medicine and other materials are made up from chemistry application.

iii) Agriculture, through agricultural materials it simply because you can get more efficient through planted by using manures and fertilizer which are made up from chemistry application.

Answer:

9.4 mol CO₂

General Formulas and Concepts:

Chemistry - Gas Laws

Explanation:

Step 1: Define

210 L CO₂ at STP

Step 2: Identify Conversions

22.4 L = 1 mol at STP

Step 3: Convert

[tex]210 \ L \ CO_2(\frac{1 \ mol \ CO_2}{22.4 \ L \ CO_2} )[/tex] = 9.375 mol CO₂

Step 4: Check

We are given 2 sig figs. Follow sig fig rules and round.

9.375 mol CO₂ ≈ 9.4 mol CO₂