What is the limiting reactant if 8 g of Ba reacts with 2.8 g of Al2(SO4)3?

Answer:

Gold(III) chloride

Answer:

NH4NO3 = N2O + 2(H2O)

Explanation:

there are 2 N, 4 H, 3 O

Answer:

NH4NO3=N2O+2H2O

Explanation:

N-2,O-3,H-4

Answer: Ethanol is the odd one out.

Explanation:

A polar compound is defined as the compound which is formed when there is a difference of electronegativities between the atoms. It is also defined as the bond which is formed due to the unequal sharing of electrons between the atoms.

Non-polar compound is defined as the compound which is formed when there is no difference of electronegativities between the atoms or the polarities cancel out.

Hexane [tex](C_6H_{14}), Oil (mixture of hydrocarbons) and carbon tetrachloride [tex](CCl_4)[/tex] all are non polar whereas ethanol is polar due to electronegative difference between hydrogen and oxygen.

Answer:

Concepts and reason

Lewis structure is a structure that explains the bonding between atoms of a molecule and lone pair of electrons that is present in the molecule is called a Lewis structure.

With the help  of Lewis structure the electronic geometry of a molecule can be determine.

Fundamentals

According to Lewis structure, every atom and their position in the structure of a molecule by using its chemical symbol.

Lines connecting the atoms that are bonded to them are drawn. Lone pairs are expressed by pairs of dots and are located beside the atoms.

Lewis structure  of [tex]H_{2}O[/tex] is, the total number of valence electrons is eight in [tex]H_{2}O[/tex].

Answer:

Explanation:

The objective here is mainly drawing the diagrams of every stereoisomer for 1-bromo-2-chloro-1,2-difluorocyclopentane.

Stereoisomerism is the difference of the spatial arrangement of atoms in a molecule or a compound with the same molecular formula.

For 1-bromo-2-chloro-1,2-difluorocyclopentane.

We have the stereoisomers as follows:

(1R,2S)-1-bromo-2-chloro-1,2-difluorocyclopentane.

(1S,2R)-1-bromo-2-chloro-1,2-difluorocyclopentane.

(1S,1S)-1-bromo-2-chloro-1,2-difluorocyclopentane.

(1R,1R)-1-bromo-2-chloro-1,2-difluorocyclopentane.

Their diagrams are drawn and shown in the attached file below in the order with which they are listed above.

Answer: 7.45 g of [tex]Pb(NO_3)_2[/tex] excess reagent are left over after the reaction is complete.

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

a) [tex]{\text{Number of moles of} KI}=\frac{7.55g}{166g/mol}=0.045moles[/tex]

b) [tex]{\text{Number of moles of} Pb(NO_3)_2}=\frac{9.06g}{331.2g/mol}=0.027moles[/tex]

The balanced chemical reaction is :

[tex]Pb(NO_3)_2(s)+2KI(s)\rightarrow 2KNO_3(s)+PbI(s)[/tex]

According to stoichiometry :

2 moles of [tex]KI[/tex] require = 1 mole of [tex]Pb(NO_3)_2[/tex]

Thus 0.045 moles of [tex]KI[/tex] will require=[tex]\frac{1}{2}\times 0.045=0.0225moles[/tex]  of [tex]Pb(NO_3)_2[/tex]

Thus [tex]KI[/tex] is the limiting reagent as it limits the formation of product and [tex]Pb(NO_3)_2[/tex] is the excess reagent as (0.045-0.0225) = 0.0225 moles are left

Mass of [tex]Pb(NO_3)_2=moles\times {\text {Molar mass}}=0.0225moles\times 331.2g/mol=7.45g[/tex]

Thus 7.45 g of [tex]Pb(NO_3)_2[/tex] of excess reagent are left over after the reaction is complete.

Answer:

$0.0238

Explanation:

The energy you need to increase the temperature of water from 20°C to 100°C is obtained from:

Q = C×m×ΔT

Where Q is the energy, C is specific heat of water (4.184J/g°C), m is mass of water (2.00x10²g - Density of water 1g/mL), ΔT is change in temperature (100.0°C - 20.0°C)

Replacing:

Q = 4.184J/g°C × 2.00x10²g × 80.0°C

Q = 66944J = 66.944kJ

As you are assuming the energy of combustion will be just 50.0% to heat the water the energy you need is 66.944kJ × 2 = 133.888kJ

The combustion of methane is:

CH4(g) + 2O2(g) ⟶ CO2(g) + 2H2O(l) ΔH = −890.8kJ

That means 1 mole of methane produce 890.8kJ. As you need 133.888kJ, moles of methane are:

133.888kJ × (1 mol CH₄ / 890.8kJ) = 0.150 moles of CH₄.

Using PV = nRT, moles of 15.0ft³ (424.8L) at 20.0°C (293.15K) and 1.00atm:

1.00atmₓ424.8L = moles CH₄ₓ0.082atmL/molKₓ293.15K

17.67 = moles CH₄

As 17.67 moles of CH₄ cost $2.80, the cost of 0.150 moles of CH₄ is:

0.150 moles CH₄ ₓ ($2.80 / 17.67 moles) =

Answer:

1. PEP is a feedback inhibitor of phosphofructokinase.

4. PEP inhibition of phosphofructokinase yields a sigmoidal velocity versus substrate curve.

6. The binding of PEP to one phosphofructokinase subunit causes a conformation change that affects the ability of the substrate to bind to the other subunits.

Explanation:

Phosphofructokinase-1, PFK-1, is an allosteric enzymes composed of four protein subunits.

Allosteric enzymes are enzymes that function through non-covalent binding of allosteric modulators which may be activators or inhibitors. They produce a characteristic velocity versus substrate sigmoidal curve. PFK-1 has a separate binding site for its substrate, fructose-6-phosphate and it's allosteric modulators: ATP, ADP or phosphoenolpyruvate, PEP.

The enzyme can exist in two conformations, the T-state (tense) or the R-state (resting). Binding of substrate causes a conformational change from T-state to R-state, whereas binding of allosteric inhibitors returns it to the T-state.

PEP, the product of step 9 in glycolysis, is an allosteric inhibitor of PFK-1. When it binds to the the allosteric site, it leads to conformational changes in PFK-1 from the R-state to the T-state which reduces the enzymes ability to bind the substrate. These changes are responsible for the sigmoidal velocity/substrate curve in allosteric enzymes.

Therefore, the true statements from the options above are 1, 4, 6.

Options 2,3 and 5 are wrong because PEP is a negative effector of PFK-1, thus its binding reduces the affinity of PFK-1 for its substrate. Also, PFK-1 being an allosteric enzyme has separate binding sites for its substrate and its modulators. Thus, there is no competition for active site binding by substrate and modulators.

Answer:

The molarity is 0.359[tex]\frac{moles}{L}[/tex]

The molality is 0.354 [tex]\frac{moles}{kg}[/tex]

The mass percent of the solution es 3.45%

Explanation:

Molarity is a unit of concentration that indicates the amount of moles of solute that appear dissolved in each liter of the mixture. It is determined by:

[tex]Molarity (M)=\frac{number of moles of solute}{volume}[/tex]

Being:

The molar mass of KNO₃ is:

KNO₃=  39 g/mole + 14 g/mole + 3*16 g/mole= 101 g/mole

You can apply the following rule of three: if 101 grams of KNO₃ are present in 1 mole, 72.5 grams in how many moles are present?

[tex]moles of KNO_{3}=\frac{72.5 grams*1 mole}{101 grams}[/tex]

moles of KNO₃= 0.718

So you have:

Applying this quantity in the definition of molarity:

[tex]molarity=\frac{0.718 moles}{2 L}[/tex]

Molarity= 0.359[tex]\frac{moles}{L}[/tex]

The molarity is 0.359[tex]\frac{moles}{L}[/tex]

Molality is a way of measuring the concentration of solute in solvent and indicates the amount of moles of solute in each kilogram of solvent.

Then the molality is calculated by:

[tex]Molality=\frac{moles of solute}{mass of solvent in kilograms}[/tex]

Density is defined as the property that matter, whether solid, liquid or gas, has to compress in a given space. That is, it is the amount of mass per unit volume. So, if the density of 1.05 g / mL for the solution indicates that in 1 mL of solution there are 1.05 grams of solution, in 2000 mL (where 2L = 2000 mL, because 1 L = 1000mL) how much mass is there?

[tex]mass=\frac{2000 mL*1.05 grams}{1 mL}[/tex]

mass= 2100 grams

Since mass solution = mass water + mass KNO₃

then mass water = mass solution - mass KNO₃

Being mass solution 2100 grams and mass KNO₃ 72.5 grams, and replacing you get: mass water= 2100 grams - 72.5 grams

mass water= 2,027.5 grams

Then, being:

Replacing in the definition of molality:

[tex]molality=\frac{0.718 moles}{2.0275 kg}[/tex]

molality= 0.354 [tex]\frac{moles}{kg}[/tex]

The molality is 0.354 [tex]\frac{moles}{kg}[/tex]

The mass percent of a solution is the number of grams of solute per 100 grams of solution. Then the mass percent is the mass of the element or solute divided by the mass of the compound or solute and the result of which is multiplied by 100 to give a percentage.

[tex]mass percent=\frac{mass of solute}{mass of solution} *100[/tex]

So, in this case:

[tex]mass percent=\frac{72.5 grams}{2100 grams} *100[/tex]

mass percent= 3.45 % KNO₃ by mass

The mass percent of the solution es 3.45%

The molarity of the solution is 0.36 mol/L. The molality of the solution is 0.34 m. The mass percent of the solution is   3.33%.

Number of moles of  KNO3 = mass/molar mass =  72.5 g/101 g/mol = 0.72 moles

Molarity = Number of moles / volume =  0.72 moles/ 2.00 L = 0.36 mol/L

The molality = Number of moles of solute/Mass of solution in kilograms

mass of solution =  1.05 g/mL × 2000 mL = 21000 g or 2.1Kg

Molality of solution =  0.72 moles/2.1 Kg = 0.34 m

Mass percent of solution = mass of solute/mass of solution × 100/1

Mass percent of solution =  72.5 g/ (72.5 g + 2100 g) × 100/1

= 3.33%

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Answer:

C3H6.

Explanation:

Data obtained from the question:

Mass of the compound = 8g

Mass of CO2 = 24.01g

Mass of H2O = 13.10g

Next, we shall determine the mass of C, H and O present in the compound. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of C in compound = Mass of C/Molar Mass of CO2 x 24.01

=> 12/44 x 24.01 = 6.5g

Mass of H in the compound = Mass of H/Molar Mass of H2O x 13.1

=> 2x1/18 x 13.1 = 1.5g

Mass of O in the compound = Mass of compound – (mass of C + Mass of H)

=> 8 – (6.5 + 1.5) = 0

Next, we shall determine the empirical formula of the compound. This is illustrated below:

C = 6.5g

H = 1.

Divide by their molar mass

C = 6.5/12 = 0.54

H = 1.4/1 = 1.

Divide by the smallest

C = 0.54/0.54 = 1

H = 1/0.54 = 2

Therefore, the empirical formula is CH2

Finally, we shall determine the molecular formula as follow:

The molecular formula of a compound is a multiple of the empirical formula.

Molecular formula = [CH2]n

[CH2]n = 44

[12 + (2x1)]n = 44

14n = 44

Divide both side by 14

n = 44/14

n = 3

Molecular formula = [CH2]n = [CH2]3 = C3H6

Therefore, the molecular formula of the compound is C3H6

Answer:

0.053moles

Explanation:

Hello,

To calculate the number of moles of gas remaining in his after he exhale, we'll have to use Avogadro's law which states that the volume of a given mass of gas is directly proportional to its number of moles provided that temperature and pressure are kept constant. Mathematically,

V = kN, k = V / N

V1 / N1 = V2 / N2= V3 / N3 = Vx / Nx

V1 = 1.7L

N1 = 0.070mol

V2 = 1.3L

N2 = ?

From the above equation,

V1 / N1 = V2 / N2

Make N2 the subject of formula

N2 = (N1 × V2) / V1

N2 = (0.07 × 1.3) / 1.7

N2 = 0.053mol

The number of moles of gas in his lungs when he exhale is 0.053 moles

Answer:

1188.976 mmHg

Explanation:

Initial pressure P1= 755 mmHg

Initial volume V1 = 6.00 litres

Final volume V2 = 3.81 litres

Final pressure P2= the unknown

Now applying Boyle's Law,we have;

P1V1 = P2V2

Since P2 is the unknown then it has to be made the subject of the formula.

P2=P1V1/ V2

P2= 755 × 6.00/ 3.81

P2= 1188.976 mmHg

Therefore, the new pressure is; 1188.976 mmHg

Answer: The equation for the reaction that goes with this equilibrium constant is [tex]5.56\times 10^{-10}=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]

Explanation:

[tex]CH_3COOH\rightarrow CH_3COO^-+H^+[/tex]

Here [tex]CH_3COOH[/tex] donates a proton and thus behaves as an acid and forms [tex]CH_3COO^-[/tex] which is called as the conjugate base of [tex]CH_3COOH[/tex]

The dissociation constant of acids is given by the term [tex]K_a[/tex] and the dissociation constant of bases is given by the term [tex]K_b[/tex] and is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.

[tex]K_a[/tex] for  [tex]CH_3COOH[/tex] :

[tex]K_a=\frac{[CH_3COO^-]\times [H^+]}{[CH_3COOH]}[/tex]

[tex]CH_3COO^-+H^+\rightarrow CH_3COOH[/tex]

[tex]K_b=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]

[tex]5.56\times 10^{-10}=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]

The equation for the reaction that goes with this equilibrium constant is [tex]K_b=\frac{[CH_3COOH]}{[CH_3COO^-]\times [H^+]}[/tex]

Compounds formed from non-metals consist of molecules. The atoms in a molecule are joined together by covalent bonds. These bonds form when atoms share pairs of electrons.

Answer:

(1). Mole fraction = 0.152 for sulfur tetrafluoride gas.

Mole fraction = 0.848 For dinitrogen monoxide gas.

(2). Partial Pressure for dinitrogen monoxide gas = 187 kPa

Partial Pressure for sulfur tetrafluoride gas = 33.4 kpa.

(3). Total Partial Pressure = 220.4 kpa.

Explanation:

So, we are given the following data or parameters or information in the question above;

• Volume of the tank = 5.00L per tank;

• Temperature of the tank = 7.03°C;

• The mass of the content in the tank =

17.7g of dinitrogen monoxide gas and

7.77g of sulfur tetrafluoride gas.

So, we will be making use of the formulae below to calculate the MOLE FRACTION:

Moles, n= mass/molar mass and mole fraction = n(1)/ n(1) + n(2) per each constituents.

Moles, n1 = 17.7g of dinitrogen monoxide gas/ 44 grams per mole. =0.4023 moles.

Moles, n2 = 7.77g of sulfur tetrafluoride gas/ 108.1 grams per mole. = 0.07188 moles.

Total numbers of moles = n1 + n2 = 0.47415 moles

Mole fraction =0.4023 / 0.47415 = 0.848 of dinitrogen monoxide gas.

Mole fraction = 0.07188/0.47415 = 0.152 of sulfur tetrafluoride gas.

PART TWO: CALCULATE THE PARTIAL PRESSURE AND TOTAL PRESSURE BY USING THE FORMULA BELOW;

pressure × volume = number of moles × gas constant, R × temperature.

Pressure = n × R × T/ V.

For dinitrogen monoxide gas. ;

Partial Pressure = 0.4023 × 8.314 × 280.03 / 5 × 10^-3 = 187 kPa.

For sulfur tetrafluoride gas

Partial Pressure = 0.07188 × 8.314 ( × 280.03 / 5 × 10^-3. = 33.4 kpa.

(3). Total pressure = (187 + 33.4)kpa = 220.4 kpa

Answer:

A. 1.54 mole.

B. 55.39g of carbon

Explanation:

A. Determination of the number of mole in 67.7g of C3H8.

Mass of C3H8 = 67.7g

Molar mass of C3H8 = (3x12) + (8x1) = 44g/mol

Number of mole of C3H8 =..?

Number of mole = Mass/Molar Mass

Number of mole of C3H8 = 67.7/44

Number of mole of C3H8 = 1.54 mole

B. Determination of the mass of carbon in the compound.

This is illustrated below:

The mass of C in compound can be obtained as follow:

=> 3C/C3H8 x 67. 7

=> 3x12 / 44 x 67.7

=> 36/44 x 67.7

=> 55.39g

Therefore, 55.39g of carbon is present in the compound.

Answer:

Number of proton emmitted by laser=[tex]2.48*10^17proton[/tex]

Explanation:

Energy is the ability to cause change; power is directly proportional to energy and its the rate energy is utilized.

Power=energy/time.

First we need to calculate the total energy used which is equal to the total power utilized.

E(total)= P( total) = 1.4W × 0.070 s =[tex]0.098J[/tex]

CHECK THE ATTACHMENT FOR THE REMAINING DETAILED CALCULATION

Answer:

The correct answer is 12.43 Liters.

Explanation:

Based on the given question, the volume V₁ occupied by the sample of carbon dioxide gas is 14.2 liters at temperature (T₁) 65 degree C or 65+273 K = 338 K.  

The gas is cooled at a temperature (T₂) 23 degree C or 273+23 K = 296 K

The volume of the gas (V₂) after cooling can be determined by using the formula,  

V₁/T₁ = V₂/T₂

14.2/338 = V₂/296

0.0420 = V₂/296

V₂ = 0.0420 * 296  

V₂ = 12.43 Liters.  

Explanation:

P(aq)+Q(aq)⇌3R(aq)

This problem involves applying LeChatelier's principle.

LeChatelier's principle states that whenever a system in equilibrium is disturbed, the equilibrium position would change in order to annul that change.

1) Increase [P]

This would cause the equilibrium position to shift to the right. This is because more reactions have been added, to annul that change more products have to be formed.

2) Increase [Q]

This would cause the equilibrium position to shift to the right. This is because more reactions have been added, to annul that change more products have to be formed.

3) Increase [R]

This would cause the equlibrium position to shift to the left. This is because more products have been formed, to annul that change more reactants have to be formed.

4) Decrease [P]

This would cause the equlibrium position to shift to the left. This is because there are now less reactants, to annul that change more reactants have to be formed.

5) Decrease [Q]

This would cause the equilibrium position to shift to the left. This is because there are now less reactants, to annul that change more reactants have to be formed.

6) Decrease [R]

This would cause the equilibrium position to shift to the right. This is because there are now less products, to annul that change more products have to be formed.

7) Triple [P] and reduce [Q] to one third

No shift in the direction of the net reaction because both changes cancels each other.

8) Triple both [Q] and [R]

No shift in the direction of the net reaction because both changes cancels each other.

Answer:

The reaction is not in equilibrium and  must move in a backward manner i.e towards the reactant so that it will attain equilibrium

Explanation:

The complete question is as follows;

Consider the following reaction where Kc = 2.90×10-2 at 1150 K: 2 SO3 (g) 2 SO2 (g) + O2 (g) A reaction mixture was found to contain 4.71×10-2 moles of SO3 (g), 5.00×10-2 moles of SO2 (g), and 4.53×10-2 moles of O2 (g), in a 1.00 liter container.

Is the reaction at equilibrium? If not, what direction must it run in order to reach equilibrium? The reaction quotient, Qc, equals . The reaction A. must run in the forward direction to reach equilibrium. B. must run in the reverse direction to reach equilibrium. C. is at equilibrium.

Solution

The first thing to do here is to calculate the pressure of each of the gases. This would be useful in the equilibrium calculations. We calculate this by dividing the respective number of moles by the volume of the container.

Now, since the volume of the container is 1L, then the number of moles will be equal to the pressure of the gaseous substances, although units will be different.

So, [SO3] = 4.71 * 10^-2 mol/L

[SO2] = 5.00 * 10^-2 mol/L

[O2] = 4.53 * 10^-2 mol/L

The equation of the reaction is as follows;

[tex]2SO_{3(g)}[/tex]    ⇆    [tex]2SO_{2(g)}[/tex]   +    [tex]O_{2(g)}[/tex]

We proceed to calculate the reaction quotient Qc

Mathematically Qc for this reaction = [[tex]SO_{2}[/tex]]^2 × [[tex]O_{2}[/tex]]/ [[tex]SO_{3}[/tex]]^2

Qc = {(5 * 10^-2)^2 * (4.53 * 10^-2)}/ (4.71 * 10^-2)^2 = 5.11 × 10^-2 mol/L

Now, we are given that the value of Kc = 2.9 * 10^-2 which is less than Qc

Since Kc < Qc, the backward reaction is favored.

Now to the question;

The reaction is not in equilibrium and  must move in a backward manner i.e towards the reactant  so that it will attain equilibrium

Answer:

A seismic wave

Explanation:

It requires a medium for its propagation.

Answer:

The correct answers are 4.93 %, 4.72 % and 4.93 %.

Explanation:

Based on the given question, 14.41 g per ml is the actual density of the object. However, the density determined by three different students of the object is 15.12 g per ml, 15.09 g per ml, and 15.12 g per ml. The percent error can be calculated by using the formula,  

% error = (actual value - calculated value) / actual value * 100

By 1st student, the calculated value is 15.12 g per ml, the percent error will be,  

% error = (14.41 - 15.12) / 14.41 * 100  

= 0.71/14.41 * 100

= 4.93 %

By 2nd student, the calculated value is 15.09 g per ml, the percent error will be,  

% error = (14.41-15.09)/14.41 * 100

= 0.68/14.41 * 100

= 4.72 %

By 3rd student, the calculated value is 15.12 g per ml, the percent error will be,  

% error = (14.41-15.12)/14.41 * 100

= 0.71/14.41 * 100

= 4.93 %

Answer:

a. Put a piece of fresh sliced yam with a bore into it into the soup.

Explanation:

b. Osmosis may occur

The chef can put a slice of yam in the soup with a hole in it as it will absorb excess of salt by process of diffusion.

Diffusion is defined as the process of movement of molecules which takes place under concentration gradient. It helps in movement of substances in and out from the cell.The molecules move from lower concentration region to a higher concentration region till the concentration becomes equal.

There are 2 main types of diffusion:

1) simple diffusion-process in which substances move through a semi-permeable membrane without the aid of transport proteins.

2) facilitated diffusion- It is a passive movement of molecules across cell membrane from higher concentration region to lower concentration.

There are 2 types of facilitated diffusion one is osmosis and dialysis.

Learn more about diffusion,here:

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Answer:

simple cations (monatomic cations of elements of only one possible charge)

Explanation:

Simple cations (monatomic cations of elements of only one possible charge)  are named using the unmodified element name and adding the word "ion"

For example, the Na+ is named the sodium ion.

An atom or molecule with a net electric charge as a result of the loss or gain of one or more electrons is known as an ion.

Answer:

0.134 moles of H₂ can be formed if a 3.25g sample of Mg reacts with excess HCl

Explanation:

The balanced reaction is:

Mg + 2 HCl → MgCl₂ + H₂

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles react:

Being:

the molar mass of the compounds participating in the reaction is:

Then, by stoichiometry of the reaction, the following quantities of mass participate in the reaction:

Then you can apply the following rule of three: if by stoichiometry 24.31 grams of Mg form 1 mole of H₂, 3.25 grams of Mg how many moles of H₂ will they form?

[tex]moles of H_{2} =\frac{3.25 grams of Mg*1 mole of H_{2} }{24.31 grams of Mg}[/tex]

moles of H₂= 0.134

0.134 moles of H₂ can be formed if a 3.25g sample of Mg reacts with excess HCl

0.134 moles of H₂ are formed by the reaction of 3.25 g of Mg with excess HCl.

Let's consider the balanced equation between Mg and HCl.

Mg + 2 HCl ⇒ MgCl₂ + H₂

The molar mass of Mg is 24.3 g/mol. The moles corresponding to 3.25 g of Mg are:

[tex]3.25 g \times \frac{1mol}{24.3g} = 0.134 mol[/tex]

The molar ratio of Mg to H₂ is 1:1. The moles of H₂ formed by 0.134 moles of Mg are:

[tex]0.134 mol Mg \times \frac{1molH_2}{1molMg} = 0.134molH_2[/tex]

0.134 moles of H₂ are formed by the reaction of 3.25 g of Mg with excess HCl.

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Answer:

A balanced equation is an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total charge is the same for both the reactants and the products.In other words, the mass and the charge are balanced on both sides of the reaction.

Explanation:

Answer:

In 0 group of the periodic table

Explanation:

So they will not react with other atoms because they have a full outer shell of electrons and an overall charge of 0.

Hope it helps.

Answer: Both Mike and Bobby are correct.

Explanation:

Petrifcation can be defined as the process in which the organic material of the dead living being becomes fossil by the replacement of mineral deposition in the bony, hard material.

Thus although the body components gets decomposed wiped out due to this process. The body shape of the dead organism remains the same as that was in living.

Thus the statements made by Mike and Bobby both are correct. The fossils are hard and have the same appearance as when they were alive.

Answer:

\large \boxed{\text{2.0 atm}}  

Explanation:

We can use Dalton's Law of Partial Pressures:

Each gas in a mixture of gases exerts its pressure separately from the other gases.

0.25 mol of O₂ exerts 0.50 atm.

If you add 0.75 mol of CO, the total amount of gas is  

0.25 mol + 0.75 mol = 1.00 mol

[tex]p_{\text{total}} = \text{1.00 mol} \times \dfrac{\text{0.50 atm}}{\text{0.25 mol}}= \textbf{2.0 atm}\\\\\text{The total pressure in the flask is $\large \boxed{\textbf{2.0 atm}}$}[/tex]

The pressure of the closed flask after the addition of 0.75 moles of CO has been 2 atm.

Partial pressure can be defined as the pressure exerted by each gas in a given solution.

The total moles of gas in the container by the addition of CO has been:

Total moles = moles of oxygen + moles of CO

Total moles = 0.25 + 0.75

Total moles = 1 mol.

By using Dalton's law of partial pressure:

Total pressure = total moles [tex]\rm \times\;\dfrac{pressure\;of\;oxygen}{moles\;of\;oxygen}[/tex]

Total pressure = 1 [tex]\rm \times\;\dfrac{0.50}{0.25}[/tex]

Total pressure = 2 atm.

The pressure of the closed flask after the addition of 0.75 moles of CO has been 2 atm.

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