How does the size of the atoms relate to how reactive it is?

Answer:

hydroxide ion / OH-

Explanation:

Basic solutions have a greater concentration of hydroxide ions than hydrogen (H+) ions

Answer:

22.6881 Grams of CO

Explanation:

Fe2O3 + 3CO = 3CO2 + 2Fe

Ratio

1:3 = 3:2

How many grams of CO are needed to produce 30.2 grams of Fe

Fe2O3 Molar mass: 159.69 g/mol

CO Molar mass: 28.01 g/mol

Fe Atomic mass: 55.845

Ok so our end result is 30.2 Grams of Fe, we have 2 Fe so we have 15.1 Grams per Fe

15.1/55.845 = 0.27

So then we multiply this by 3 as we we know the ratio and get get 0.81 required as this is 3CO.

So we need 0.81 Moles of CO so we need 22.6881 Grams of CO.

Answer: C. Based in previous data, ten hurricanes may occur in the year 2013.

Explanation:

Inference can be explained as conclusion made from an existing evidence or facts. Option C, best fits an inference statement.

Answer:

C

Explanation:

because everything looks like opinions

Moles: 1

Weight, g : 192.3900

Answer:

Moles: 1

Weight, g : 192.3900

Explanation:

Answer : The value of [tex]\Delta H_{rxn}[/tex] for the reaction is, -390.3 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The main chemical reaction is:

[tex]CH_4(g)+4Cl_2(g)\rightarrow CCl_4(g)+4HCl(g)[/tex]    [tex]\Delta H_{rxn}=?[/tex]

The intermediate balanced chemical reaction are:

(1) [tex]C(s)+2H_2(g)\rightarrow CH_4(g)[/tex]     [tex]\Delta H_1=-74.6kJ[/tex]

(2) [tex]C(s)+2Cl_2(g)\rightarrow CCl_4(g)[/tex]     [tex]\Delta H_2=-95.7kJ[/tex]

(3) [tex]H_2(g)+Cl_2(g)\rightarrow 2HCl(g)[/tex]     [tex]\Delta H_2=-184.6kJ[/tex]

Now we are reversing reaction 1, multiplying reaction 3 by 2 and then adding all the equations, we get :

(1) [tex]CH_4(g)\rightarrow C(s)+2H_2(g)[/tex]     [tex]\Delta H_1=74.6kJ[/tex]

(2) [tex]C(s)+2Cl_2(g)\rightarrow CCl_4(g)[/tex]     [tex]\Delta H_2=-95.7kJ[/tex]

(3) [tex]2H_2(g)+2Cl_2(g)\rightarrow 4HCl(g)[/tex]     [tex]\Delta H_2=2\times (-184.6kJ)=-369.2kJ[/tex]

The expression for enthalpy change for the reaction will be,

[tex]\Delta H_{rxn}=\Delta H_1+\Delta H_2+\Delta H_3[/tex]

[tex]\Delta H_{rxn}=(74.6kJ)+(-95.7kJ)+(-369.2kJ)[/tex]

[tex]\Delta H_{rxn}=-390.3kJ[/tex]

Therefore, the value of [tex]\Delta H_{rxn}[/tex] for the reaction is, -390.3 kJ

Answer:

D. Na2CO3

Explanation:

I'm just guessing tbh

Answer:

From left to right

Neutral

Acid

Base

Explanation:

So using HSAB theory hard and soft (Lewis) acids and bases.

OH - Is a Hard Base

H+ Strong Acid

Answer:

-1

Explanation:

Because it is the smallest sub-atomic particle

Answer:

[tex]\large \boxed{\text{528.7 g} }[/tex]

Explanation:

It often helps to write the heat as if it were a reactant or a product in the thermochemical equation.

Then you can consider it to be 11018 "moles" of "kJ"  

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

M_r:                      32.00

              2C₈H₁₈ + 25O₂ ⟶ 16CO₂ + 8H₂O + 11 018 kJ

n/mol:                                                                  7280

1. Moles of O₂

The molar ratio is 25 mol O₂:11 018 kJ

[tex]\text{Moles of O}_{2} = \text{7280 kJ} \times \dfrac{\text{25 mol O}_{2}}{\text{11 018 kJ}} = \text{16.52 mol O}_{2}[/tex]

2. Mass of O₂

[tex]\text{Mass of C$_{8}$H}_{18} = \text{16.52 mol O}_{2} \times \dfrac{\text{32.00 g O}_{2}}{\text{1 mol O}_{2}} = \textbf{528.6 g O}_{2}\\\text{The reaction requires $\large \boxed{\textbf{528.67 g O}_{2}}$}[/tex]

Answer:

Theoretical Yield = 0.4882 grams

Explanation:

Potassium Chromate = K2CrO4 Molar mass: 194.1896 g/mol

Silver Nitrate = AgNO3 Molar mass: 169.87 g/mol

Silver(I) Chromate = Ag2CrO4 Molar mass: 331.73 g/mol

Potassium Nitrate = KNO3  Molar mass: 101.1032 g/mol

K2CrO4 + 2AgNO3 = Ag2CrO4 + 2KNO3

Potassium Chromate + Silver Nitrate = Silver(I) Chromate + Potassium Nitrate

Ratio 1:2 = 1:2

0.5g AgNO3

AgNO3 Molar mass: 169.87 g/mol

2AgNO3 = 339.74grams

0.5/339.74 = 0.0014717

Ag2CrO4 Molar mass: 331.73 g/mol

0.4882 grams

Theoretical Yield = 0.4882 grams

Answer:

Check Explanation

Explanation:

In this case the Mushroom would get energy from decomposing dead plants and animals in the soil/ground

The plant would get energy from the sun (simplified)

the rabbit would get energy from eating the plant

in this case the wolf would energy from eating the rabbit but in a normal setting the wolf could get energy from eating any animal

Answer:

Explanation: A wolf gets its energy from food and water

A rabbit gets its energy from food and water too

A plant get it energy from sunlight, rain

A mushroom also get its energy from dead plant/animal’s

If this wasn’t the answer you wanted I’m sorry

Answer:

Foliation is developed by stress and fire while layering is developed by the embedding of fine and coarse deposits. Foliation is caused due to an alteration of minerals from pressure and heat while layering is developed by seasonal changes

Explanation:

Banding occurs when two different mineral compositions alternate in thin layers (typically 1 mm to 1 cm). The two types of layers typically contain the same minerals, but in different proportions, giving the rock a striped appearance. A foliation is defined by its banding.

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Answer:

CrO

Explanation:

0.765g Cr

0.235g O

O = 16g/mol

Cr = 52g/mol

O     2-

Cr  2+, 3+, 6+

0.765 + 0.235 = 1

so 76.5% Cr

23.5% O

76.5/52 = 1.47/1.47 = 1

23.5/16 = 1.47/1.47 = 1

CrO

Answer:

[tex]\large \boxed{\text{-827.4 kJ}}[/tex]

Explanation:

We have three equations:

1. 2H₂S(g)            + O₂(g)   ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ

2. S(s, rhombic)  + O₂(g)   ⟶ SO₂(g);                                 ∆H = -296.8 kJ

3. PbO(s)             + H₂S(g) ⟶ PbS(s)               + SO₂(g);    ∆H =  -104.3 kJ

From these, we must devise the target equation:

4. 2PbS(s)            + 3O₂(g) ⟶2PbO(s)             + 2SO₂(g); ΔH = ?

The target equation has PbS(s) on the left, so you reverse Equation 3 and double it.

When you reverse an equation, you reverse the sign of its ΔH.

When you double an equation, you double its ΔH.

5. 2PbS(s) + 2H₂O(g) ⟶ 2PbO(s) + 2H₂S(g); ∆H = 208.6 kJ

Equation 5 has 2H₂O on the left. That is not in the target equation.

You need an equation with 2H₂O on the right, so you copy Equation 1.  

6. 2H₂S(g) + O₂(g) ⟶ 2S(s, rhombic) + 2H₂O(g) ; ∆H = -442.4 kJ  

Equation 6 has 2S(s, rhombic) on the right. That is not in the target equation.

You need an equation with 2S(s, rhombic) on the left, so you double Equation 2.  

7. 2S(s, rhombic)  + 2O₂(g) ⟶ 2SO₂(g); ∆H = -593.6 kJ

Now, you add equations 5, 6, and 7, cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

You get the target equation 4:

5. 2PbS(s)  + 2H₂O(g)  ⟶ 2PbO(s) + 2H₂S(g);  ∆H =  208.6 kJ

6. 2H₂S(g) + O₂(g)        ⟶ 2S(s)     + 2H₂O(g) ; ∆H = -442.4 kJ

7. 2S(s)      + 2O₂(g)      ⟶ 2SO₂(g);                   ∆H = -593.6 kJ

4 . 2PbS(s) + 3O₂(g)      ⟶ 2PbO(s) + 2SO₂(g); ΔH = -827.4 kJ

[tex]\Delta H \text{ for the reaction is $ \large \boxed{\textbf{-827.4 kJ}}$}[/tex]

Answer:it b

Explanation:

I've done that before

Answer:

The Answer is C or 3

Explanation:

I took the test

Answer:

None of them, answer seems to be 24 unless I messed up

Explanation:

How many electrons appear in the balanced equation

What the heck do they mean, do they mean transferred?

N2O5 --> NH4+

Left Side

N 5+  Electrons 2 in central shell, 5 in outer

O 2- Electrons 2 in central shell, 6 in outer

Right Side

H  1+ Electrons 1 in central shell

N 3- Electrons 2 in central shell, 5 in outer

Hmm, ok so need to balance the half equation first, this is a redox reaction.

Hmm, so it's going to be something like

H2 + N2O5 → NH + O3  

Step 1. Write down the unbalanced equation ('skeleton equation') of the chemical reaction. All reactants and products must be known. For a better result write the reaction in ionic form.

H2 + N2O5 → NH + O3  

Step 2. Separate the process into half reactions. A redox reaction is nothing but both oxidation and reduction reactions taking place simultaneously.

a) Assign oxidation numbers for each atom in the equation. Oxidation number (also called oxidation state) is a measure of the degree of oxidation of an atom in a substance (see: Rules for assigning oxidation numbers).

H02 + N+52O-25 → N-1H+1 + O03  

b) Identify and write out all redox couples in reaction. Identify which reactants are being oxidized (the oxidation number increases when it reacts) and which are being reduced (the oxidation number goes down). Write down the transfer of electrons. Carefully, insert coefficients, if necessary, to make the numbers of oxidized and reduced atoms equal on the two sides of each redox couples.

When one member of the redox couple is oxygen with an oxidation state of -2 or hydrogen with an oxidation state of +1, it is best to replace it with a water molecule.

O:3H+12O-2 → O03 + 6e-(O)

H02 → H+12O-2 + 2e-(H)

R:N+52O-25 + 12e- → 2N-1H+1(N)

c) Combine these redox couples into two half-reactions: one for the oxidation, and one for the reduction (see: Divide the redox reaction into two half-reactions).

O:3H+12O-2 + H02 → O03 + H+12O-2 + 8e-  

R:N+52O-25 + 12e- → 2N-1H+1  

Step 3. Balance the atoms in each half reaction. A chemical equation must have the same number of atoms of each element on both sides of the equation. Add appropriate coefficients (stoichiometric coefficients) in front of the chemical formulas to balance the number of atoms. Never change any formulas.

a) Balance all other atoms except hydrogen and oxygen. We can use any of the species that appear in the skeleton equations for this purpose. Keep in mind that reactants should be added only to the left side of the equation and products to the right.

O:3H+12O-2 + H02 → O03 + H+12O-2 + 8e-  

R:N+52O-25 + 12e- → 2N-1H+1  

b) Balance the charge. For reactions in an acidic solution, balance the charge so that both sides have the same total charge by adding a H+ ion to the side deficient in positive charge.

O:3H+12O-2 + H02 → O03 + H+12O-2 + 8e- + 8H+  

R:N+52O-25 + 12e- + 12H+ → 2N-1H+1  

c) Balance the oxygen atoms. Check if there are the same numbers of oxygen atoms on the left and right side, if they aren't equilibrate these atoms by adding water molecules.

O:3H+12O-2 + H02 + H2O → O03 + H+12O-2 + 8e- + 8H+  

R:N+52O-25 + 12e- + 12H+ → 2N-1H+1 + 5H2O  

Balanced half-reactions are well tabulated in handbooks and on the web in a 'Tables of standard electrode potentials'. These tables, by convention, contain the half-cell potentials for reduction. To make the oxidation reaction, simply reverse the reduction reaction and change the sign on the E1/2 value.

Step 4. Make electron gain equivalent to electron lost. The electrons lost in the oxidation half-reaction must be equal the electrons gained in the reduction half-reaction. To make the two equal, multiply the coefficients of all species by integers producing the lowest common multiple between the half-reactions.

O:3H+12O-2 + H02 + H2O → O03 + H+12O-2 + 8e- + 8H+| *3

R:N+52O-25 + 12e- + 12H+ → 2N-1H+1 + 5H2O| *2

O:9H+12O-2 + 3H02 + 3H2O → 3O03 + 3H+12O-2 + 24e- + 24H+  

R:2N+52O-25 + 24e- + 24H+ → 4N-1H+1 + 10H2O  

Step 5. Add the half-reactions together. The two half-reactions can be combined just like two algebraic equations, with the arrow serving as the equals sign. Recombine the two half-reactions by adding all the reactants together on one side and all of the products together on the other side.

9H+12O-2 + 2N+52O-25 + 3H02 + 24e- + 3H2O + 24H+ → 3O03 + 4N-1H+1 + 13H2O + 24e- + 24H+

Step 6. Simplify the equation. The same species on opposite sides of the arrow can be canceled. Write the equation so that the coefficients are the smallest set of integers possible.

2N+52O-25 + 3H02 → 3O03 + 4N-1H+1 + H2O

Answer:

Intake, compression, power, and exhaust

Explanation:

A four-stroke cycle engine is an internal combustion engine that utilizes four distinct piston strokes (intake, compression, power, and exhaust) to complete one operating cycle. The piston make two complete passes in the cylinder to complete one operating cycle.

The sequence of a single complete engine cycle is Intake, compression, power, exhaust. Therefore option 1 is correct.

1. Intake: In this step, the intake valve opens, allowing a mixture of air and fuel to enter the combustion chamber.

2. Compression: After the intake stroke, the intake valve closes, and the piston moves back up, compressing the air-fuel mixture. This compression increases the pressure and temperature inside the cylinder, preparing it for combustion.

3. Power: Once the air-fuel mixture is compressed, the spark plug ignites it. The burning mixture rapidly expands, generating a high-pressure force that pushes the piston down.

4. Exhaust: After the power stroke, the exhaust valve opens, and the piston moves back up, pushing the burned gases out of the cylinder and into the exhaust system.

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Answer: The molarity of [tex]Br^-[/tex] anions in the chemist's solution is 0.0084 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]

where,

n = moles of solute

[tex]V_s[/tex] = volume of solution in ml

moles of [tex]FeBr_3[/tex] = [tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{0.0411g}{295.6g/mol}=0.00014mol[/tex]

Now put all the given values in the formula of molality, we get

[tex]Molarity=\frac{0.00014\times 1000}{50}=0.0028[/tex]

As 1 mole of [tex]FeBr_3[/tex] gives = 3 moles of [tex]Br^-[/tex]

0.0028 moles of [tex]FeBr_3[/tex] gives = [tex]\frac{3}{1}\times 0.0028=0.0084 moles[/tex] of [tex]Br^-[/tex]

Thus the molarity of [tex]Br^-[/tex] anions in the chemist's solution is 0.0084 M

Answer:

The substances on the right side of a chemical equation are the products as reactions typically move from left to right.

A chemical equation is a representation of chemical reaction in the form of symbols and formulae where the reactants are on left side and the products on right side.

Eg. CaCO3(s)→CaO(s)+CO2(g

Chemical equation is written "with the reactants on the left side of an arrow and the products of the chemical reaction on the right".

Reactants are "the starting materials in a chemical reaction".

What is a product?

A product is "a chemical species resulting from a chemical reaction".

Hence, the substances on the right side of a chemical equation are products.

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Answer:

It is calcualted based ion the  yield from the limiting reactant

Explanation:

Answer:Agas loses kinetic energy to become a liquid and then becomes a solid

i believe i had this before but where is the graph

Explanation:

please mark this answer as brainlest

B. A solid gains kinetic energy to become a liquid and then becomes a gas

Answer:

It's the bottom layer.  

Explanation:

Scientists use the Law of Superposition to determine the relative age of a layer of sedimentary rock:

The oldest rock layer is at the bottom of an undisturbed bed.

Thus, Layer A in the figure below was deposited first.

Answer:

Fuels made from oil mixtures containing large hydrocarbon molecules are not efficient as they do not flow easily and are difficult to ignite. Crude oil often contains too many large hydrocarbon molecules and not enough small hydrocarbon molecules to meet demand. This is where cracking comes in

Explanation:

Answer:

d)carbon(iv)oxide/carbon(ii)oxide

e)Calcium,carbon,oxygen

f)carnonhydride

g)Carbonhydrate

h)hydrogen+oxygen

I)Iron sulphide

j)Magnesiumoxide

m)magnesium hydroxide+hydrogen gas

d)carbon dioxide

Answer: 1500 kg•m/s north

Explanation:

A PE X