Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperatures. If 18.1 g of Nh3 is reacted with 90.4 g of CuO, which is the limiting reactant? How many grams of N2 will be formed? Explain how you solved for your answers.
Balanced Equation: 2NH3 + 3CuO → 3Cu + N2 + 3H2O

Answer:

A

Explanation:

The lithosphere represents the layer of hardened/solid rock that makes up the hard part of the earth, including the brittle upper portion of the mantle and the crust. The lithosphere is broken into pieces that are referred to as plates. The pieces move to and away from each other in a process known as plate tectonics. The movement of plates accounts for the global locations of volcanoes, earthquakes, and mountain ranges.

The lithosphere is made of largely of inorganic materials known as silicates. The weathering of the solid rocks together with the interaction of living organisms gives rise to soil with an appreciable amount of organic materials.

The correct option is, therefore A.

Answer:

Explanation:

Given:

Mass of solute = 30 gram

Volume of water = 500 ml

Density = 1.15g/ml

Find:

(i)ppm

(ii) %w/v

(iii) %w/w

Computation:

Water in gram = 500 ml × 1.15 g/ml

Water in gram = 575 gram

In ppm

Concentration of MgSO4 = [30 / 575] × 10⁶

Concentration of MgSO4 = 0.0521 × 10⁶ ppm

in % w/v

Concentration of MgSO4 = [30 / 500] × 100

Concentration of MgSO4 = 6% w/v

in % w/w

Concentration of MgSO4 = [30 / 575] × 100

Concentration of MgSO4 = 5.21% w/w

Answer:

"Both signs are possible" is the correct choice.

Explanation:

As such that when both signals could have been the enthalpy change.

If the dissolution of borax in water is spontaneous, so the change in enthalpy should be negative.

Enthalpy of the reaction tells about the total amount of energy released or absorbed during any chemical reaction.

And for the spontaneous reaction, value of standard Gibb's free energy change of the reaction should be negative and formula will be represented as:
ΔG = ΔH - TΔS, where

ΔH = change in enthalpy

For the spontaneous reaction, value of enthalpy should be negative so that we get the negative value of ΔG.

Hence change in enthalpy should be negative.

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Answer:

The pH of the solution will be 3

Explanation:

The strength of acids is determined by their ability to dissociate into ions in aqueous solution. A strong acid is any compound capable of completely and irreversibly releasing protons or hydrogen ions, H⁺. That is, an acid is said to be strong if it is fully dissociated into hydrogen ions and anions in solution.

Being pH=- log [H⁺] or pH= - log [H₃O⁺] and being a strong acid, all the HClO₃ dissociates:

HClO₄      +    H₂O        →      H₃O⁺      +      ClO₄-  

So: [HCLO₄]= [H₃O⁺]

The molar concentration is:

[tex]molar concentration=\frac{number of moles of solute}{volume solution}[/tex]

The molar mass of HClO₄ being 100 g / mole, then if 100 grams of the compound are present in 1 mole, 0.025 grams in how many moles are present?

[tex]moles of HClO_{4} =\frac{0.025 grams*1 mole}{100 grams}[/tex]

moles of HClO₄= 0.00025

Then:

[tex][HClO_{4}]=\frac{0.00025 moles}{0.25 L}[/tex]

[tex][HClO_{4}]=0.001 \frac{ moles}{ L}[/tex]

Being [HCLO₄]= [H₃O⁺]:

pH= - log 0.001

pH= 3

The pH of the solution will be 3

Answer:

The metal block had a mass of 1.50kg

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Answer:according to this picture

Explanation:

Answer:

Siδ⁺ -- Pδ⁻⁻

Clδ⁻⁻ -- Pδ⁺

Siδ⁺ -- Clδ⁻⁻

Of the mentioned bonds the most polar bond is Si -- Cl

The polarity of the bond primarily relies upon the electronegativity difference between the two atoms that forms the bond. Therefore, if the electronegativity difference between the two atoms that forms the bond is more the bond will be more polar, and if it is less then the bond will be less polar. The electronegativity of the atoms mentioned is Si = 1.8 , P = 2.1 and Cl = 3.00.  

Therefore, the Si - Cl atoms exhibit more electronegativity difference, thus, the Si - Cl bond will be the most polar bond.  

Answer:

b) the Ka of NH₄⁺ to find the hydronium concentration

Explanation:

The equilbrium of NH₄⁺ (The conjugate acid of NH₃, a weak base), is:

NH₄⁺ ⇄ NH₃ + H⁺

Where Ka of the conjugate acid is:

Ka = [NH₃] [H⁺] / [NH₄⁺]

Thus, if you know Ka of NH₄⁺ and its molar concentration you can calculate  [H⁺], the hydronium concentration, to find pH (Because pH =  -log [H⁺])

Thus, right option is:

The words given are not clear, so the clear question is as follows:

Match the words below to the appropriate blanks in the sentences. Make certain each sentence is complete before submitting your answer:

A. β-1,4- and α-1,6-glycosidic

B. α-1,4-glycosidic

C. α-1,4-galactose

D. an unbranched glucose

E. a branched fructose

F. α-1,6-glycosidic

Amylose is ......... polymer of ....... units joined by ........ bonds.

Amylopectin is ....... polymer of .......units joined by ........ bonds.

Answer:

D. an unbranched glucose

C. α-1,4-galactose

B. α-1,4-glycosidic

E. a branched fructose

A. β-1,4- and α-1,6-glycosidic

F. α-1,6-glycosidic

Explanation:

Amylose and amylopectin are two types of polysaccharides that can be found in starch granules.

Amylose is linear or unbranched glucose polymer of α-1,4-galactose units that are joined by α-1,4-glycosidic.

Amylopectin is a branched fructose polymer of β-1,4- and α-1,6-glycosidic units joined by α-1,6-glycosidic bonds.

Hence, the correct answers in the sequential order are:

Amylose:

D. an unbranched glucose

C. α-1,4-galactose

B. α-1,4-glycosidic

Amylopectin:

E. a branched fructose

A. β-1,4- and α-1,6-glycosidic

F. α-1,6-glycosidic

Answer:

The gas is Methane at 340K

Answer:

OPTION C is correct

3.86 x 10-19 J

Explanation:

Energy of the line can be calculated using below formula

E= h ν.................(1)

Where E= energy

h= plank constant= 6.626 10-34 J s

c=speed of light=3 x 108 m/s

But we know that Velocity V= = c / λ

Then substitute into equation (1) we have

E = h c / λ.............(2)

We can calculate our( hc ) in nm for unit consistency

h c =( 6.626 ×10^-34)x(3×108)

h c = (1.986 x 10-16 )

hc = 1.986 x 10-16 J nm then since our (hc) and λ are in the same unit , were good to go then substitute into equation(2)

E = h c / λ = (1.986 x 10-16) / 515

E = 3.86 x 10-19 J

Therefore, the Energy is 3.86 x 10-19 J

Answer:

AgBr

Explanation:

Silver bromide has a very low solubility product constant of about 7.7 ×10^-13 in pure water hence it is not quite soluble in pure water.

However, with NaCN, the AgBr forms the complex [Ag(CN)2]^2- which has a formation constant of about 5.6 ×10^8. This very high formation constant implies that the complex is easily formed leading to the dissolution of AgBr in NaCN.

The equation for the dissolution of AgBr in cyanide is shown below;

AgBr(s) + 2CN^-(aq) ----> [Ag(CN)2]^2-(aq) + Br^-(aq)

Answer:

SCN- is a weak field ligand while CN- is a strong field ligand

Explanation:

The spectrochemical series is an arrangement of ligands according to their magnitude of crystal field splitting. Ligands that cause only a small degree of crystal field splitting are called weak field ligands while ligands that cause large crystal field splitting are called strong field ligands.

Strong field ligands often lead to the formation of low spin complexes with the least number of unpaired electrons while high spin complexes are formed by weak field ligands.

CN- is a strong field ligand as it lies towards the right hand side of the spectrochemical series.

SCN- is a weak field ligand hence it forms a high spin complex having the maximum number of unpaired electrons for Fe^3+, hence the answer.

SCN⁻ lies in the lower (weak field) region of the spectrochemical series while CN⁻ lies in the higher (stronger field) region.

CN⁻ is a strong field ligand with a large splitting constant, and it is high up in the spectrochemical series.

Conversely, SCN⁻ is a weak field ligand with a low splitting constant, and it is lower in the spectrochemical series.

Hence, SCN⁻ lies in the lower (weak field) region of the spectrochemical series while CN⁻ lies in the higher (stronger field) region.

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Answer:

[tex]Speed = 3.30 \frac{km}{hr}[/tex]

Explanation:

Given

Distance = 1 mile

Time = 8.92 minutes

Required

Calculate Speed in km/hr

Speed is calculated as thus;

[tex]Speed = \frac{Distance}{Time}[/tex]

Substitute 1 mile for distance and 8.92 minutes for time

[tex]Speed = \frac{1\ mile}{8.92\ minutes}[/tex]

Convert Miles to Kilometres

If

[tex]1\ mile = 1609\ m[/tex];

Then

[tex]1\ mile = \frac{1609\ km}{1000}[/tex]

[tex]1\ mile = 1.609\ km[/tex] --- (1)

Convert minutes to hour

[tex]1\ minutes = 0.0167\ hour[/tex]

Multiply both sides by 8.92

[tex]8.92 * 1\ minutes = 0.0167\ hour * 8.92[/tex]

[tex]8.92 \ minutes = 0.148964\ hour[/tex] ---- (2)

By substituting (1) and (2) in [tex]Speed = \frac{1\ mile}{8.92\ minutes}[/tex], we have

[tex]Speed = \frac{1.609\ km}{0.48694\ hour}[/tex]

[tex]Speed = 3.304309 \frac{km}{hr}[/tex]

[tex]Speed = 3.30 \frac{km}{hr}[/tex] -- Approximated

Answer:

Methyl radical

Explanation:

A radical is any specie that contains an odd number of electrons. We must note that the greater the number of alkyl groups which are attached to a carbon atom that bears the odd electrons, the more the degree of delocalization of the odd electrons and consequently the more stable we expect the free radical to be.

Hence the order of free radical stability is; Methyl < Primary < Secondary < Tertiary. Hence, we can easily see that the methyl radical is the least stable free radical.

Answer: Methyl radical

Explanation:

Answer:

d) Solid water

Explanation:

because it's particles are more fixed together ( least apart ), so their mobility and conductivity is very low hence lowest kinetic energy.

Answer:

d. Solid water

Explanation:

example ice

Answer:

rate = [NO]²[H₂]

Explanation:

2NO + H2 ⟶N2 + H2O2 (slow)

H2O2 + H2 ⟶2H2O (fast)

From the question, we are given two equations.

In chemical kinetics; that is the study of rate reactions and changes in concentration. The rate law is obtained from the slowest reaction.

This means that our focus would be on the slow reaction. Generally the rate law is obtained from the concentrations of reactants in a reaction.

This means our rate law is;

rate = [NO]²[H₂]

Answer:

0.0630

Explanation:

The molar mass of urea = 60 g/mol

we all know that:

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

Then; the number of moles of urea

= [tex]\mathtt{\dfrac{4.0 \ g}{60 \ g/mol}}[/tex]

= 0.0667 mol

Similarly; the number of moles of methanol

= [tex]\mathtt{\dfrac{32 \ g}{32.04 \ g/mol}}[/tex]

= 0.9988 mol

The total number of moles = (0.0667 + 0.9988) mol

= 1.0655 mol

Finally,the mole fraction of urea  [tex]\mathtt{(X_{urea})}[/tex] = [tex]\mathtt{\dfrac{ n_{urea}}{(n_{urea}+n_{methanol})}}[/tex]

[tex]\mathtt{(X_{urea})}[/tex] = [tex]\mathtt{\dfrac{0.0667 \ mole}{1.0655 \ mole}}[/tex]

= 0.0630

The given question is incomplete. The complete question is:

Compound X has a molar mass of 153.05 g/mol and the following composition:

element mass %

carbon 47.09%

hydrogen 6.59%

chlorine 46.33%

Write the molecular formula of X.

Answer: The molecular formula of X is [tex]C_6H_{10}Cl_2[/tex]

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 47.09 g

Mass of H = 6.59 g

Mass of Cl = 46.33 g

Step 1 : convert given masses into moles.

Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{47.09g}{12g/mole}=3.92moles[/tex]

Moles of H =[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.59g}{1g/mole}=6.59moles[/tex]

Moles of Cl =[tex]\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{46.33g}{35.5g/mole}=1.30moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = [tex]\frac{3.92}{1.30}=3[/tex]

For H = [tex]\frac{6.59}{1.30}=5[/tex]

For Cl =[tex]\frac{1.30}{1.30}=1[/tex]

The ratio of C : H: Cl= 3: 5 :1

Hence the empirical formula is [tex]C_3H_5Cl[/tex]

The empirical weight of [tex]C_3H_5Cl[/tex] = 3(12)+5(1)+1(35.5)= 76.5g.

The molecular weight = 153.05 g/mole

Now we have to calculate the molecular formula.

[tex]n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{153.05}{76.5}=2[/tex]

The molecular formula will be=[tex]2\times C_3H_5Cl=C_6H_{10}Cl_2[/tex]

Answer:

It is a combustion reaction.

Answer:

222.3 ml of a 0.130 M aqueous solution of chromium (II) nitrate must be taken to obtain 5.08 grams of the salt.

Explanation:

Being:

the molar mass of chromium (II) nitrate, Cr(NO₃)₂ is:

Cr(NO₃)₂ = 52 g/mole + 2* (14 g/mole + 3* 16 g/mole)= 176 g/mole

So: if 176 grams are present in 1 mole of the compound, 5.08 grams in how many moles of the compound will be present?

[tex]amount of moles=\frac{5.08 grams* 1 mole}{176 grams}[/tex]

amount of moles=0.0289 moles

Molarity (M) is the number of moles of solute that are dissolved in a given volume. It is then calculated by dividing the moles of the solute by the volume of the solution:

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

Molarity is expressed in [tex]\frac{moles}{liter}[/tex]

So in this case:

Replacing:

[tex]0.130 M=0.130 \frac{moles}{liter} =\frac{0.0289 moles}{volume}[/tex]

Solving:

[tex]volume=\frac{0.0289 moles}{0.130 \frac{moles}{liter} }[/tex]

volume=0.2223 liters

Being 1 L= 1,000 mL:

volume=0.222 liters= 222.3 mL

222.3 ml of a 0.130 M aqueous solution of chromium (II) nitrate must be taken to obtain 5.08 grams of the salt.

Answer:

[tex]91°C[/tex]

Explanation:

CHECK THE COMPLETE QUESTION BELOW;

Suppose that 0.95 g of water condenses on a 75.0 g block of iron that is initially at 22 °c. if the heat released during condensation is used only to warm the iron block, what is the final temperature (in °c) of the iron block? (assume a constant enthalpy of vaporization for water of 44.0 kj/mol.)

Heat capacity which is the amount of heat required to raise the temperature of an object or a substance by one degree

From the question, it was said that that 0.95 g of water condenses on the block thenwe know that Heat evolved during condensation is equal to the heat absorbed by iron block.

Then number of moles =given mass/ molecular mass

Molecular mass of water= 18 g/mol

Given mass= 0.95 g

( 0.95 g/18 g/mol)

= 0.053 moles

Then Heat evolved during condensation = moles of water x Latent heat of vaporization

Q= heat absorbed or released

H=enthalpy of vaporization for water

n= number of moles

Q=nΔH

Q = 0.053 moles x 44.0 kJ/mol

= 2.322 Kj

=2322J

We can now calculate Heat gained by Iron block

Q = mCΔT

m = mass of substance

c = specific heat capacity

=change in temperature

m = 75 g

c = 0.450 J/g/°C

If we substitute into the above formula we have

Q= 75 x 0.450 x ΔT

2322 = 75 x 0.450 x ΔT

ΔT = 68.8°C

Since we know the difference in temperature, we can calculate the final temperature

ΔT = T2 - T1

T1= Initial temperature = 22°C

T2= final temperature

ΔT= change in temperature

T2 = T1+ ΔT

= 68.8 + 22

= 90.8 °C

=91°C

Therefore, final temperature is [tex]91°C[/tex]

The final temperature of the iron block is 91∘C.

Given that;

Heat lost during condensation of the water = Heat gained by iron block

Mass of water(mw) = 0.95 g

Latent heat of vaporization =  Latent heat of condensation(L) = 44.0 kJ/mol

Mass of iron(mi) = 75.0 g

Initial temperature of iron(T1) =  22∘C

Final temperature of iron(T2) = ?

Heat capacity of iron(ci) =  0.449 J⋅g−1⋅∘C−1

So;

mwL = mici(T2 - T1)

Substituting values;

(0.95g/18g/mol) ×  44.0 × 10^3(J/mol) = 75.0(g) × 0.449 J⋅g−1⋅∘C−1 (T2 - 22∘C)

2322.2 = 33.7T2 - 741.4

2322.2 +  741.4 = 37.4T2

T2 = (2322.2 +  741.4)/ 33.7

T2 =91∘C

Missing parts;

Suppose that 0.95 g of water condenses on a 75.0 g block of iron that is initially at 22 °c. if the heat released during condensation is used only to warm the iron block, what is the final temperature (in °c) of the iron block? (assume a constant enthalpy of vaporization for water of 44.0 kj/mol.)

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Answer:

123.96 g Na₂O

Explanation:

4 Na  +  O₂  ⇒  2 Na₂O

You first need to find the limiting reagent.  Convert the reactants to moles and see which produces the least amount of product using the mole ratios in the chemical equation.

(91.96 g Na)/(22.99 g/mol Na) = 4 mol Na

(4 mol Na) × (2 mol Na₂O/4 mol Na) = 2 mol Na₂O

(32.0 g O₂)/(32.0 g/mol) = 1 mol O₂

(1 mol O₂) × (2 mol Na₂O/1 mol O₂) = 2 mol Na₂O

Since they both produce the same amount of product, you don't need to pick a limiting reagent.  Now, convert moles of Na₂O to grams.

(2 mol Na₂O) × (61.98 g/mol Na₂O) = 123.96 g Na₂O

Answer:

C. 0.191 M

Explanation:

Our goal for this question, is to calculate the concentration of the HCl solution. For this, in the experiment, a solution of NaOH was used to find the moles of HCl. Therefore, our first step is to know the reaction between HCl and NaOH:

[tex]HCl~+~NaOH~->~NaCl~+~H_2O[/tex]

The "titrant" in this case is the NaOH solution. If we know the concentration of NaOH (0.100M) and the volume of NaOH (38.2 mL=0.0382 L), we can calculate the moles using the molarity equation:

[tex]M=\frac{mol}{L}[/tex]

[tex]0.100~M=\frac{mol}{0.0382~L}[/tex]

[tex]mol=0.100~M*0.0382~L=0.0382~mol~of~NaOH[/tex]

Now, in the reaction, we have a 1:1 molar ratio between HCl and NaOH (1 mol of HCl is consumed for each mole of NaOH added). Therefore we will have the same amount of moles of HCl in the solution:

[tex]0.0382~mol~of~NaOH\frac{1~mol~HCl}{1~mol~NaOH}=0.0382~mol~HCl[/tex]

If we want to calculate the molarity of the HCl solution we have to divide by the litters of HCl used in the experiment (20 mL= 0.02 L):

[tex]\frac{0.0382~mol~HCl}{0.02~L}~=~0.191~M[/tex]

The concentration of the HCl solution is 0.191 M

I hope it helps!

Answer:

2,3 Dimethyl hexane

Explanation:

First, start the count from which side is given the CH3 smallest number

first; the longest carbon chain in this compound is 6

and you don't have any double and triple bonds or functional groups so it is Hexane

you start to count from the right side to give the branch molecules the smallest number ..

CH3 = methyl

and you have 2 methyl in this compound ..

and 2 mean you must write ( Di )

you write the name in this way

2,3 Dimethyl hexane

hope this helps you.

stay safe ...

Answer:

Explanation:

MnO₂(s) + 4 HCl(aq)  = MnCl₂(aq) + 2 H₂O(l) + Cl₂

87 g                                                                     22.4 x 10³ mL

volume of given chlorine gas at NTP or at 760 Torr and 273 K

=  175 x ( 273 + 25 ) x 715 / (273 x 760 )

= 179.71 mL

22.4 x 10³ mL of chlorine requires 87 g of MnO₂

179.4 mL of chlorine will require    87 x 179.4 / 22.4 x 10³ g

= 696.77 x 10⁻³ g

= 696.77 mg .

a)

HSO₃⁻ + OH⁻ → H₂SO₄⁽²⁻⁾

In this reaction, we know that OH will act as a base. which means that the bisulfite ion will act as an acid

also because SO₃⁽²⁻⁾ will be relative more stable than O⁽²⁻⁾ because of larger valence shell. so HSO₃ will protonate. making it the acid

b)

HSO₃⁻ + H₃O⁺ → H₂SO₃

Here, the hydronium ion will be the one to protonate because that will help it form water, which is a much stabler compound

so the hydronium ion will protonate, making it the acid and the bisulfite will take that proton, making it the base

Answer: It's the anode broski (B)

Explanation: I'm taking the Chem summer course too broski, this was the correct answer. Cheers broski

Answer: The calculated molar enthalpy change of reaction for his second trial was -108 kJ.

Explanation:-

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

[tex]\text{no of moles}={\text{Molarity}\times {\text{Volume in L}}[/tex]

Thus [tex]\text{no of moles}of HCl={1.0M}\times {0.05L}=0.05moles[/tex]

Thus [tex]\text{no of moles}of NaOH={1.0M}\times {0.05L}=0.05moles[/tex]

[tex]HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O(l)[/tex]

Given for second trial:

[tex]\text{no of moles}of HCl={1.0M}\times {0.1L}=0.1moles[/tex]

[tex]\text{no of moles}of NaOH={1.0M}\times {0.1L}=0.1moles[/tex]

0.05 moles of [tex]HCl[/tex] reacts with 0.05 moles of [tex]NaOH[/tex] to release  heat = 54 kJ

0.1 moles of [tex]HCl[/tex] reacts with 0.05 moles of [tex]NaOH[/tex] to release  heat  =[tex]\frac{54}{0.05}\times 0.1=108kJ[/tex]

Thus calculated molar enthalpy change of reaction for his second trial was -108 kJ.