Write a balanced equation for the double-replacement precipitation reaction described, using the smallest possible integer coefficients. A precipitate forms when aqueous solutions of ammonium bromide and silver(I) nitrate are combined. (Use the lowest possible coefficients. Omit states of matter.)

Answer:

26.87g

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2Fe + O2 —> 2FeO

Next, we shall determine the masses of Fe and O2 that reacted and the mass of FeO produced from the balanced equation.

This is illustrated below:

Molar mass of Fe =56 g/mol

Mass of Fe from the balanced equation = 2 x 56 = 112 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 1 x 32 = 32 g

Molar mass of FeO = 56 + 16 =72 g/mol

Mass of FeO from the balanced equation = 2 x 72 = 144 g

From the balanced equation above,

112 g of Fe reacted with 32 g of O2 to produce 144 g of FeO.

Next, we shall determine the limiting reactant.

This is illustrated below:

From the balanced equation above,

112 g of Fe reacted with 32 g of O2.

Therefore, 20.9 g of Fe will react with = (20.9 x 32)/112 = 5.97 g of O2.

From the calculations made above, we can see that only 5.97 g out of 9.19 g of O2 given were required to react completely with 20.9 g of Fe.

Therefore, Fe is the limiting reactant and O2 is the excess reactant.

Finally, we shall determine the mass of FeO produced from the reaction.

In this case, the limiting reactant will be used, as it will give the maximum yield of the reaction since all of it is used up in the reaction.

The limiting reactant is Fe and the maximum mass of FeO produced can be obtained as follow:

From the balanced equation above,

112 g of Fe reacted to produce 144 g of FeO.

Therefore, 20.9 g of Fe will react to produce = (20.9 x 144)/112 = 26.87g of FeO.

Therefore, the maximum mass of iron(II) oxide, FeO produced is 26.87g.

Answer:

to. Turn on a heater (electrical energy that is transformed into heat energy)

yes. Throwing an apple core in the trash (chemical energy that is transformed into nutritional energy for decomposers)

C. Climbing a rope ladder (chemical energy from our food into mechanical energy that allows us to climb the ladder)

re. Starting a car (electrochemical energy in mechanical energy)

me. Turning on a flashlight (chemical energy from the battery into light energy)

Five daily actions that exemplify the transformation of energy:

Do physical activity (change of metabolic energy in mechanics)

Turn on a heater (electrical energy that is transformed into heat energy)

Lighting a stove (Chemical energy product of combustion that results in the transformation of that energy into heat)

Turn on a cell phone (chemical energy characteristic of the battery that is transformed into sound and light energy)

Riding a bicycle (nutritional energy or energy of the metabolism that is transformed into mechanical energy)

Explanation:

The energy is never lost, it is always transformed.

It is one of the great principles of physics and is the reason why the universe is infinite.

Answer:

19.07 g mol^-1

Explanation:

The computation of the molecular mass of the unknown gas is shown below:

As we know that

[tex]\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }[/tex]

where,

Diffusion rate of unknown gas = 155 mL/s

CO_2 diffusion rate = 102 mL/s

CO_2 molar mass = 44 g mol^-1

Unknown gas molercualr mass = M_unknown

Now placing these values to the above formula

[tex]\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}[/tex]

After solving this, the molecular mass of the unknown gas is

= 19.07 g mol^-1

Answer:

Increasing the concentration of A will cause the greatest change over the rate.

Explanation:

Hello,

In this case, considering the given rate law, which is third-order with respect to A, changing its concentration, the rate will be significantly modified. For instance, suppose a concentration of A and B of 1M and a symbolic rate constant (k),  this causes the rate to be:

[tex]r=k[1M]^3[1M]=1k\frac{M^4}{s}[/tex]

Then, if we change the concentration of A to 2 M holding the concentration of B in 1 M, the new rate constant will be:

[tex]r=k[2M]^3[1M]=8k\frac{M^4}{s}[/tex]

Nevertheless, if we hold the concentration of A in 1 M and the concentration of B is now 2 M (same change), the new rate constant is:

[tex]r=k[1M]^3[2M]=2k\frac{M^4}{s}[/tex]

It means that increasing the concentration of A will cause the greatest change over the rate.

Best regards.

The change in concentration of A will cause the great­est in­crease in the reaction rate.

Given reaction has rate law ,

          [tex]rate=k[A]^{3} [B][/tex]

which is third-order with respect to concentration of A

From rate law equation, It is observed that the degree of concentration of A is three.

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Answer:

100 g of water: specific heat of water 4.18 J/g°C

Explanation:

To know the correct answer to the question, we shall determine the temperature change in each case.

For 100 g of water:

Mass (M) = 100 g

Specific heat capacity (C) = 4.18 J/g°C

Heat absorbed (Q) = 55 KJ = 55000 J

Change in temperature (ΔT) =..?

Q = MCΔT

55000 = 100 x 4.18 x ΔT

Divide both side by 100 x 4.18

ΔT = 55000/ (100 x 4.18)

ΔT = 131.6 °C

Therefore the temperature change is 131.6 °C

For 50 g of water:

Mass (M) = 50 g

Specific heat capacity (C) = 4.18 J/g°C

Heat absorbed (Q) = 55 KJ = 55000 J

Change in temperature (ΔT) =..?

Q = MCΔT

55000 = 50 x 4.18 x ΔT

Divide both side by 50 x 4.18

ΔT = 55000/ (50 x 4.18)

ΔT = 263.2 °C

Therefore the temperature change is 263.2 °C

For 50 g of lead:

Mass (M) = 50 g

Specific heat capacity (C) = 0.128 J/g°C

Heat absorbed (Q) = 55 KJ = 55000 J

Change in temperature (ΔT) =..?

Q = MCΔT

55000 = 50 x 0.128 x ΔT

Divide both side by 50 x 0.128

ΔT = 55000/ (50 x 0.128)

ΔT = 8593.8 °C

Therefore the temperature change is 8593.8 °C.

For 100 g of iron:

Mass (M) = 100 g

Specific heat capacity (C) = 0.449 J/g°C

Heat absorbed (Q) = 55 KJ = 55000 J

Change in temperature (ΔT) =..?

Q = MCΔT

55000 = 100 x 0.449 x ΔT

Divide both side by 100 x 0.449

ΔT = 55000/ (100 x 0.449)

ΔT = 1224.9 °C

Therefore the temperature change is 1224.9 °C.

The table below gives the summary of the temperature change of each substance:

Mass >>> Substance >> Temp. Change

100 g >>> Water >>>>>> 131.6 °C

50 g >>>> Water >>>>>> 263.2 °C

50 g >>>> Lead >>>>>>> 8593.8 °C

100 g >>> Iron >>>>>>>> 1224.9 °C

From the table given above we can see that 100 g of water has the smallest temperature change.

Answer:

b. Al3+ + 3Br– → Al + (3/2)Br2

Explanation:

If we look at this reaction closely, aluminum was reduced while bromine was oxidized. The reduction potential of aluminum is -1.66 V while that of bromine is + 1.087. Recall that the more negative the redox potential of a chemical specie, the greater its tendency to function as a reducing agent donating electrons in an electrochemical reaction.

However, in this reaction, aluminium is found to accept rather than donate electrons. Therefore, the process is non spontaneous and can only occur in an electrolytic cell, hence the answer.

b. [tex]Al^{3+} + 3 Br^{-}[/tex] → [tex]Al + \frac{3}{2} Br_{2}[/tex]

Electrolytic  Cell v/s Electrochemical Cell:

Out of the given reactions, [tex]Al^{3+} + 3 Br^{-}[/tex] → [tex]Al + \frac{3}{2} Br_{2}[/tex]  is carried out in an electrolytic cell rather than in an electrochemical cell.

As we know, In electrolytic cells, like galvanic cells, are composed of two half-cells

In the given reaction, aluminum is being reduced while bromine gets oxidized and the value of reduction potential of aluminum is -1.66 V while that of bromine is + 1.087. Therefore, the process is non spontaneous and can only occur in an electrolytic cell.

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Answer:

3.00 L

Explanation:

Convert the pressure to Pascals.

P = 82 atm × (101325 Pa/atm)

P = 8,308,650 Pa

Convert temperature to Kelvins.

T = 27°C + 273

T = 300 K

Use ideal gas law:

PV = nRT

(8,308,650 Pa) V = (10 mol) (8.314 J/mol/K) (300 K)

V = 0.00300 m³

If desired, convert to liters.

V = (0.00300 m³) (1000 L/m³)

V = 3.00 L

Answer:

[tex]\large \boxed{\text{3.0 L}}[/tex]

Explanation:

[tex]\begin{array}{rcl}pV &=& nRT\\\text{82 atm} \times V & = & \text{10 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{300.15 K}\\82V & = & \text{246 L}\\V & = & \textbf{3.0 L} \\\end{array}\\\text{The volume of the balloon is $\large \boxed{\textbf{3.0 L}}$}[/tex]

Answer:

-second

Explanation:

6-ethyl-2-octyne is an unsaturated compound with a triple bond.

6-ethyl-2-octyne will have a triple bound attached to the second carbon. The suffix -yne suggests that compound carry a triple bond and the number  "2" before suffix refers to the position of triple bond that is second carbon.

Hence, the correct option is  "-second ".

Answer:

–36 KJ.

Explanation:

The equation for the reaction is given below:

2B + C —› D + E. ΔH = – 24 KJ

From the equation above,

1 mole of D required – 24 KJ of energy.

Now, we shall determine the energy change associated with 1.5 moles of D.

This can be obtained as illustrated below:

From the equation above,

1 mole of D required – 24 KJ of energy

Therefore,

1.5 moles of D will require = 1.5 × – 24 = –36 KJ.

Therefore, –36 KJ of energy is associated with 1.5 moles of D.

Answer:

6.9 Kg/mol•dL

Explanation:

To convert 6.9×10⁴ g/mol•L to kg/mol•dL,

First, we shall convert to kg/mol•L.

This can be achieved by doing the following:

Recall: 1 g = 1×10¯³ Kg

1 g/mol•L = 1×10¯³ Kg/mol•L.

Therefore,

6.9×10⁴ g/mol•L = 6.9×10⁴× 1×10¯³

6.9×10⁴ g/mol•L = 69 Kg/mol•L

Finally, we shall convert 69 Kg/mol•L to Kg/mol•dL.

This is illustrated below:

Recall: 1 L = 10 dL

1 Kg/mol•L = 1×10¯¹ Kg/mol•dL

Therefore,

69 Kg/mol•L = 69 × 1×10¯¹

69 Kg/mol•L = 6.9 Kg/mol•dL

Therefore, 6.9×10⁴ g/mol•L is equivalent to 6.9 Kg/mol•dL.

Answer:

Explanation:

[tex]n = 5.01\\V = 70L\\P = ?\\T = 303K\\R = 0.08206\\\\PV =nRT\\Make \:P \:Subject\:of\:the\:formula\\P = \frac{nRT}{V} \\\\P = \frac{5.01\times0.08206\times303}{70} \\\\P =\frac{124.5695}{70}\\ P = 1.7795[/tex]

First, we find the molar mass of KCl which is AK+ACl=39+35.5=74.5g/mole

now we find out the number of moles by dividing the given mass to the molar mass n=m/M=19.9/74.5=0.26 moles. The molarity of a solution is equal to the number of moles divided by the volume of the solution. =0.26moles/0.75liters=0.346M

Answer:

Explanation:

[tex]2x^2+10x-48\\=2\left(x^2+5x-24\right)\\x^2+5x-24\\=\left(x^2-3x\right)+\left(8x-24\right)\\=x\left(x-3\right)+8\left(x-3\right)\\=\left(x-3\right)\left(x+8\right)\\=2\left(x-3\right)\left(x+8\right)\\2\left(x-3\right)\left(x+8\right)=0\\x-3=0\\x = 0+3\\x = 3\\x+8=0\\x+8-8=0-8\\x=-8\\x=3,\:x=-8[/tex]

Answer:

Percent yield: 78.2%

Explanation:

Based on the reaction:

4Al + 3O₂ → 2Al₂O₃

4 moles of Al produce 2 moles of Al₂O₃

To find percent yield we need to find theoretical yield (Assuming a yield of 100%) and using:

(Actual yield (6.8g) / Theoretical yield) × 100

Moles of 4.6g of Al (Molar mass: 26.98g/mol) are:

4.6g Al × (1mol / 26.98g) = 0.1705 moles of Al.

As 4 moles of Al produce 2 moles of Al₂O₃, theoretical moles of Al₂O₃ obtained from 0.1705 moles of Al are:

0.17505 moles Al × (2 moles Al₂O₃ / 4 moles Al) = 0.0852 moles of Al₂O₃,

In grams (Molar mass Al₂O₃ = 101.96g/mol):

0.0852 moles of Al₂O₃ × (101.96g / mol) =

Thus, Percent yield is:

(6.8g / 8.7g) × 100 =

Answer:

7.37 mL of KOH

Explanation:

So here we have the following chemical formula ( already balanced ), as HNO3 reacts with KOH to form the products KNO3 and H2O. As you can tell, this is a double replacement reaction,

HNO3 + KOH → KNO3 + H2O

Step 1 : The moles of HNO3 here can be calculated through the given molar mass (  0.140 M HNO3 ) and the mL of this nitric acid. Of course the molar mass is given by mol / L, so we would have to convert mL to L.

Mol of NHO3 = 0.140 M [tex]*[/tex] 30 / 1000 L = 0.140 M [tex]*[/tex] 0.03 L = .0042 mol

Step 2 : We can now convert the moles of HNO3 to moles of KOH through dimensional analysis,

0.0042 mol HNO2 [tex]*[/tex] ( 1 mol KOH / 1 mol HNO2 ) = 0.0042 mol KOH

From the formula we can see that there is 1 mole of KOH present per 1 moles of HNO2, in a 1 : 1 ratio. As expected the number of moles of each should be the same,

Step 3 : Now we can calculate the volume of KOH knowing it's moles, and molar mass ( 0.570 M ).

Volume of KOH = 0.0042 mol [tex]*[/tex] ( 1 L / 0.570 mol ) [tex]*[/tex] ( 1000 mL / 1 L ) = 7.37 mL of KOH

Answer:

B. CH3Br

Explanation:

Dipole -Dipole interactions take place in polar molecules.

CH3Br exhibits dipole -dipole forces as its strongest attraction between molecules because it is a polar molecule due to the slightly negative dipole present on the Br molecule.

While O2 is a nonpolar molecule due to its linear structure, CCl4 has zero resultant dipole moment, Helium is non-polar and BrCH2CH2OH is a non polar compound having net dipole moment is zero.

Hence, the correct option is B. CH3Br.

The compound that exhibits dipole -dipole forces is CH3Br

It should be taken place in polar molecules. Also, CH3Br should be the strongest attraction that lies between the molecules since it is treated as the polar molecule because of the slightly negative. While on the other hand, O2 should be non-polar molecule because of the linear structure.

Therefore, The compound that exhibits dipole -dipole forces is CH3Br

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Answer:

A) Chlorine (Cl)

B) Cobalt (Co)

C) Caesium (Cs)

Hope this helps.

The abbreviated electron configurations that was given in the question belongs to

Chlorine (Cl)

Cobalt (Co)

Caesium (Cs) respectively.

Therefore, electron configurations  explain orbitals of an atom when it is in it's ground state.

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Answer:

3.62×10³ L/mol

Explanation:

Beer-Lambert law relates the absorbance of a sample and its concentration. Its formula is:

A = ε×C×l

Where A is absorbance of the sample, ε is molar absorptivity (A constant f each sample), C its concentration and l is path length

Now, the formula obtained was:

y = (3.62×10³ L/mol) x

Where Y ia absorbance = A, x its concentration = C and 1cm is path length.

You can write:

A = (3.62×10³ L/mol)×C×l

That means, molar absorptivity of your sample under the meaured conditions is:

Answer:

The peaks are registered from tetramethyl silane (TMS)

Explanation:

Tetramethyl silane (TMS) is used as internal reference in proton nmr (H NMR) spectrometry.

Its peak is usually registered at about a 2.0 chemical shift means that the hydrogen atoms which caused that peak need a magnetic field two millionths less than the field needed by TMS to produce resonance. This is not affected by the chemical shift of the sample analysed.

I hope this helped.

Answer:

See explanation

Explanation:

The noble gas core electron configuration involves writing the inert gas core of an atom followed by the valence electrons. This is shown for the following atoms;

Bismuth;

[Xe]4f14 5d10 6s2 6p3

Chromium;

[Ar]4s1 3d5

Strontium;

[Kr]5s2

Phosphorus;

[Ne]3s2 3p3

2.

Bi

6p- n=6, l= 1, ml= 1, ms= 1/2

Cr

3d- n=3, l=2, ml=2,ms=1/2

Sr

5s- n=5, l=0, ml=0, ms=1/2

P

3p- n=3, l= 1, ml= 1, ms=1/2

3.

a) Tin (Sn) - [Kr] 5s2 4d10 5p2

b) Caesium (Cs)- [Xe] 6s1

c) Copper (Cu)- [Ar] 4s1 3d10

Answer: They use other organisms for energy

Explanation:

Answer:

they use other organisms for energy

Explanation: d

Answer:

2.81mL

Explanation:

Based on the reaction:

C₆H₃COCl + 2NH₃ → C₆H₅CONH₂ + NH₄Cl

Benzoyl chloride + ammonia → Benzamide

1 mole of benzoyl chloride in excess of ammonia produce 1 mole of Benzamide.

Thus, assuming a theoretical yield, to produce 24.1mmoles of benzamide you require 24.1mmoles of benzoyl chloride.

As molar mass of benzoyl chloride is 141g/mol, mg you require are:

mg Benzoyl chloride: 24.1mmol × (141mg / 1mmol) = 3398.1mg = 3.3981g of benzoyl chloride.

to convert this mass to mL, you require density of Benzoyl chloride (1.21g/mL). Thus, mL you need are:

3.3981g × (1mL / 1.21g) =

Answer:

The correct answer is 16.61 grams methanol and 57.38 grams water.

Explanation:

The mole fraction (X) of methanol can be determined by using the formula,  

X₁ = mole number of methanol (n₁) / Total mole number (n₁ + n₂)

X₁ = n₁/n₁ + n₂ = 0.14

n₁ / n₁ + n₂ = 0.14 ---------(i)

n₁ mole CH₃OH = n₁ mol × 32.042 gram/mol (The molecular mass of CH₃OH is 32.042 grams per mole)

n₁ mole CH₃OH = 32.042 n₁ g

n₂ mole H2O = n₂ mole × 18.015 g/mol  

n₂ mole H2O = 18.015 n₂ g

Thus, total mole number is,  

32.042 n₁ + 18.015 n₂ = 74 ------------(ii)

From equation (i)

n₁/n₁ + n₂ = 0.14

n₁ = 0.14 n₁ + 0.14 n₂

n₁ - 0.14 n₁ = 0.14 n₂

n₁ = 0.14 n₂ / 1-0.14

n₁ = 0.14 n₂/0.86 ----------(iii)

From eq (ii) and (iii) we get,  

32.042 × 0.14/0.86 n₂ + 18.015 n₂ = 74

n₂ (32.042 × 0.14/0.86 + 18.015) = 74

n₂ = 74 / (32.042 × 0.14/0.86 + 18.0.15)

n₂ = 3.1854 mol

From equation (iii),  

n₁ = 0.14/0.86 n₂

n₁ = 0.14/0.86 × 3.1854

n₁ = 0.5185 mol

Now, presence of water in the mixture is,  

= 3.1854 mole × 18.015 gram per mole  

= 57.38 grams

Methanol present in the mixture is,  

= 0.5185 mol × 32.042 gram per mole

= 16.61 grams

Answer:

The balanced equation is given below:

UO2(s) + 4HF(l) —› UF4(s) + 2H2O(l)

The coefficients are: 1, 4, 1, 2

Explanation:

UO2(s) + HF(l) —› UF4(s) + H2O(l)

The above equation can be balance as follow:

There are 2 atoms of O on the left side and 1 atom on the right side. It can be balanced by writing 2 in front of H2O as shown below:

UO2(s) + HF(l) —› UF4(s) + 2H2O(l)

There are 4 atoms of F on the right side and 1 atom on the left side. It can be balanced by writing 4 in front of HF as shown below:

UO2(s) + 4HF(l) —› UF4(s) + 2H2O(l)

Now, the equation is balanced.

The coefficients are: 1, 4, 1, 2.

Answer:

2.30 × 10⁻⁶ M

Explanation:

Step 1: Given data

Concentration of Mg²⁺ ([Mg²⁺]): 0.039 M

Solubility product constant of Mg(OH)₂ (Ksp): 2.06 × 10⁻¹³

Step 2: Write the reaction for the solution of Mg(OH)₂

Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

Step 3: Calculate the minimum [OH⁻] required to trigger the precipitation of Mg²⁺ as Mg(OH)₂

We will use the following expression.

Ksp = 2.06 × 10⁻¹³ = [Mg²⁺] × [OH⁻]²

[OH⁻] = 2.30 × 10⁻⁶ M

Answer:

Decreases

Explanation:

Fatty acid which have the double bond or triple bond are called unsaturated fatty acids. Because of the double or triple bond, unsaturated fatty acids are loosely packed and form some distance among molecules which lowers the melting point of unsaturated fatty acids.

So, if the unsaturation of fatty acid will increase, it leads to more branched and loosely packed molecules and decreases the melting temperature accordingly.

Answer:

Pictures that we receive from space are of the past because it takes time for light to reach Earth.

Explanation:

For example, Mars is so far away that, depending on its position in orbit, a picture from Mars takes between 4 min and 24 min to reach Earth.

Answer: Pictures that we receive from space are of the

past

because it takes time for

light

to reach Earth.

Explanation:

Answer:

b) 99 kPa

Explanation:

According to Daltons law of partial pressure, the total pressure of a mixture of two or more non reactive gases is the sum of their individual pressures. Let the total pressure of a mixture of n number of gases be [tex]P_{total}[/tex] and their individual pressure be [tex]P_1,P_2,P_3,\ .\ .\ .\ ,\ P_n[/tex], According to Daltons partial pressure law:

[tex]P_{total}=P_1+P_2+P_3+.\ .\ .+P_n[/tex]

Since A glass cylinder contains 2 gases at a pressure of 106 kPa, therefore n = 2. Also one gas ([tex]P_1[/tex]) is at 7 kPa. Using Daltons partial pressure law:

[tex]P_{total}=P_1+P_2+P_3+.\ .\ .+P_n\\P_{total}=P_1+P_2\\106\ kPa=7\ kPa+P_2\\P_2=106\ kPa-7\ kPa\\P_2=99\ kPa[/tex]

Answer:

pH = 5.22

Explanation:

As you can see, your initial concentration of sodium acetate (NaCH₃COO) is 0.02M (0.02mol /L). When you add HCl, the reaction is:

NaCH₃COO + HCl → CH₃COOH + NaCl.

If you add HCl, and final concentration of NaCH₃COO is 0.015M, the concentration of CH₃COOH is 0.005M.

You can know the pH of this solution using H-H equation:

pH = pKa + log {NaCH₃COO} / {CH₃COOH}

pH = 4.74 +  log {0.015M} / {0.005M}

Answer:

25.99mL is the volume internal volume of the flask

Explanation:

To complete the question:

The temperature of the water was measured to be 21ºC. Use this data to find the internal volume of the stoppered flask

The flask was filled with water, that means the internal volume of the flask is equal to the volume that the water occupies.

To find the volume of the water you need to find the mass and by the use of density of water at 21ºC (0.997992g/mL), you can find the volume of the flask, thus:

Mass water = Mass filled flask - Mass of clean flask

Mass water = 60.167g - 34.232g

Mass water = 25.935g of water.

To convert this mass to volume:

25.935g × (1mL / 0.997992g) =