Identify the labeled structures.

Answer:

The height of A is 900

Explanation: Beside A, there is a number which explains what the height is so the answer is A.

Answer:

e. Dilute a solution of commercial vinegar to prepare it for titration.

a. Perform a scout colorimetric titration of diluted vinegar with NaOH.

b. One pair performs a careful colorimetric titration of diluted vinegar with NaOH and the other pair performs a potentiometric titration of diluted vinegar with NaOH.

c. Perform a scout colorimetric titration of HCL with NaOH.

d. One pair performs a careful colorimetric titration of HCL with NaOH and the other pair performs a potentiometric titration of HCL with NaOH.

Explanation:

Titration is a process in which acid and base are introduced together to determine unknown concentration of acid or base. This is also known as neutralization reaction. The chemical process form salt and water as by products because acid molecules react with base molecules.

The experimental steps for the Acid-Base Equilibria in the correct order is e, a, b, c and d.

Titration is a process in which we determine the concentration of any unknown acid and base by titrating them with the standard known concentration of base or acid.

In potentiometric titration, in the place of indicator potential difference is used across the analyte of electrolytic solution. And in colorimetric titration, absorbance of light is measured by the solution. Process of method which is involved in the titration is:
e. Dilute a solution of commercial vinegar to prepare it for titration.

a. Perform a scout colorimetric titration of diluted vinegar with NaOH.

b. One pair performs a careful colorimetric titration of diluted vinegar with NaOH and the other pair performs a potentiometric titration of diluted vinegar with NaOH.

c. Perform a scout colorimetric titration of HCL with NaOH.

d. One pair performs a careful colorimetric titration of HCL with NaOH and the other pair performs a potentiometric titration of HCL with NaOH.

Hence correct sequence is e, a, b, c and d.

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Answer:

1.3-bromo,2-chloro,4-floro hexane

3.bromocyclo pentane

Answer:

Lead  sulfide.

Explanation:

Answer: 1. [tex]CO[/tex] is the limiting reagent and [tex]O_2[/tex] is the excess reagent.

2. 2.91 g of excess reagent

3. 4.66 g of carbon dioxide produced.

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} CO=\frac{2.98g}{28g/mol}=0.106moles[/tex]

[tex]\text{Moles of} O_2=\frac{4.62g}{32g/mol}=0.144moles[/tex]

[tex]O_2+2CO\rightarrow 2CO_2[/tex]  

According to stoichiometry :

2 moles of [tex]CO[/tex] require = 1 mole of [tex]O_2[/tex]

Thus 0.106 moles of [tex]CO[/tex] will require=[tex]\frac{1}{2}\times 0.106=0.053moles[/tex]  of [tex]O_2[/tex]

Thus [tex]CO[/tex] is the limiting reagent as it limits the formation of product and [tex]O_2[/tex] is the excess reagent.

Amount of excess reagent = ( 0.144-0.053) moles = 0.091 moles

Mass of excess reagent [tex]O_2=moles\times {\text {Molar mass}}=0.091moles\times 32g/mol=2.91g[/tex]

As 2 moles of [tex]CO[/tex] give = 2 moles of [tex]CO_2[/tex]

Thus 0.106 moles of [tex]CO[/tex] give =[tex]\frac{2}{2}\times 0.106=0.106moles[/tex]  of [tex]CO_2[/tex]

Mass of [tex]CO_2=moles\times {\text {Molar mass}}=0.106moles\times 44g/mol=4.66g[/tex]

Explanation:

[tex]6.42 \times \frac{1}{30} \times \frac{1}{1} \times \frac{46}{1} = 9.844[/tex]

[tex]8.53 \div 9.844 \times 100 = 86.65\%[/tex]

Answer:

the wind and the temperature of water both drive ocean currents

Answer:

B. Write 'T' for true or 'F' for false statements :

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Answer:

-4.88 × 10⁴ kJ/mol

Explanation:

Step 1: Calculate the heat gained by the calorimeter (Qcal)

We will use the following expression.

Qcal = C × ΔT

where,

Qcal = 9.001 kJ/°C × 25.0°C = 225 kJ

Step 2: Calculate the heat released by the fuel

According to the law of conservation of energy, the sum of the heat released by the fuel (Qf) and the heat absorbed by the calorimeter is zero.

Qf + Qcal = 0

Qf = -Qcal = -225 kJ

Step 3: Calculate the heat of reaction in kJ/mol

3.00 g of a fuel (MW = 650.0 g/mol) released 225 kJ of heat. The heat of reaction is:

-225 kJ/3.00 g × 650.0 g/mol = -4.88 × 10⁴ kJ/mol

Answer: The value of the equilibrium constant for this reaction is [tex]2.8\times 10^{-5}[/tex]

Explanation:

To calculate the moles, we use the equation:

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

Moles of  [tex]Ni(CO)_4[/tex] = [tex]\frac{0.597g}{170.7g/mol}=0.0035moles[/tex]

Moles of  [tex]Ni[/tex] =  [tex]\frac{12.7g}{58.7g/mol}=0.216moles[/tex]

Moles of  [tex]CO[/tex] =  [tex]\frac{1.98g}{28.01g/mol}=0.071moles[/tex]

Volume of solution = 2.7 L

Equilibrium concentration of [tex]Ni(CO)_4[/tex] = [tex]\frac{0.0035mol}{2.7L}=1.29\times 10^{-3}M[/tex]

Equilibrium concentration of [tex]Ni[/tex] = [tex]\frac{0.216mol}{2.7L}=0.08M[/tex]

Equilibrium concentration of [tex]CO[/tex] = [tex]\frac{0.071mol}{2.7L}=0.026M[/tex]

The given balanced equilibrium reaction is,

                            [tex]Ni(CO)_4\rightleftharpoons Ni+4CO[/tex]

The expression for equilibrium constant for this reaction will be,

[tex]K_c=\frac{[Ni]^1\times [CO]^4}{[Ni(CO_4]^1}[/tex]

Now put all the given values in this expression, we get :

[tex]K_c=\frac{(0.08)^1\times (0.026)^4}{(1.29\times 10^{-3})^1}[/tex]

[tex]K_c=2.8\times 10^{-5}[/tex]

Thus the value of the equilibrium constant for this reaction is [tex]2.8\times 10^{-5}[/tex]

Answer:

4.255 J/ K

Explanation:

Given data :

mass of Gallium = 16 gm

molar mass = 69.723 g/mol

hence no of moles = 16 / 69.723 = 0.23 moles --------- ( 1 )

Δh ( heat fusion ) = 5.59 KJ/mol

Temperature = 29.8°C = 302.8 k

Determine the entropy change of gallium

Δs ( entropy change ) =  heat fusion / temperature

 = ( 5.59 * 1000) J/mol / 302.8 = 18.46 J/k*mol

Hence entropy change of gallium

= Δs * no of Gallium moles

= 18.46 * 0.23 = 4.255 J/ k

I hope this is right haha

Answer:

Q = -22.9 kJ

Explanation:

Given that,

Mass of water, m = 150.3 g

Water gets cool from 25.60°C to -10.70°C.

The specific heat of water, c = 4.2 J/g°C

The formula for heat needed is given by :

[tex]Q=mc\Delta T\\\\Q=150.3\times 4.2 \times (-10.7-25.6)\\\\Q=-22914.738\\\\or\\\\Q=22.9\ kJ[/tex]

So, 22.9 kJ of heat is needed to be removed to cool.

Answer:

See explanation

Explanation:

Let us see what happens when each solution is mixed;

a) AlCl3(aq) + K3PO4(aq) ------> 3KCl(aq) + AlPO4(s)

A precipitate is formed here

b) RbCO3(aq) + NaCl(aq) -------> This is an impossible reaction hence no solid precipitate is formed here

c) MnCl2(aq) + Na2CO3(aq) → 2NaCl(aq) + MnCO3(s)

A precipitate is formed.

d) K2S(aq) + 2NH4Cl(aq) ------> 2KCl(aq) + (NH4)2S(aq)

No solid precipitate is formed

e) CaCl2(aq) + (NH4)2CO3(aq) → CaCO3(s) + 2NH4Cl(aq)

A solid precipitate is formed

Answer: You have a solution of potassium chlorate containing 4 g at 65º C. How many additional grams of solute must be added to it, to make the solution saturated ? approximately 26 more grams [the solubility of potassium chlorate is ... water. The solution is allowed to cool. At what new temperature would crystals begin to.

Explanation:

Answer:

[tex]\boxed {\boxed {\sf 8.07 *10^{24} \ particles}}[/tex]

Explanation:

To convert from moles to representative particles, Avogadro's number: 6.02*10²³ is used. This number tells us the amount of particles in 1 mole of a substance. We can use it as a ratio.

[tex]\frac {6.02*10^{23} \ particles}{1 \ mol \ CO_2}}[/tex]

Multiply by the given number of moles.

[tex]13.4 \ mol \ CO_2 *\frac {6.02*10^{23} \ particles}{1 \ mol \ CO_2}}[/tex]

The moles of carbon dioxide will cancel out.

[tex]13.4 *\frac {6.02*10^{23} \ particles}{1}}[/tex]

The denominator of 1 can be ignored and this becomes a simple multiplication problem.

[tex]13.4 * {6.02*10^{23} \ particles[/tex]

[tex]8.0668*10^{24} \ particles[/tex]

The original number of moles has 3 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place. The 6 in the thousandth place tells us round the 6 to a 7.

[tex]8.07 *10^{24} \ particles[/tex]

13.4 moles of carbon dioxide has approximately 8.07*10²⁴ particles.

Answer: The main difference between amylopectin and glycogen is that amylopectin is an insoluble form whereas glycogen is a soluble form. The 1,4-alpha glycosidic bonds form the linear chain of both amylopectin and glycogen while 1,6-alpha glycosidic bonds form the branches.

Explanation:

Answer:

2

Explanation:

Answer:

D

Explanation:

I believe it is D. your kinetic energy would be at b. A, the cart would be going at a constant rate, because there is no hill or steep slope.

Answer:

2.9 g of water are produced by 9.6 g of ammonium perchlorate

Note: The question is incomplete. The complete question is given below:

Ammonium perchlorate (NH₄ClO₄) is the solid rocket fuel used by the U.S. Space Shuttle. It reacts with itself to produce nitrogen gas (N₂) , chlorine gas (Cl₂), oxygen gas (O₂), water (H₂O) , and a great deal of energy. What mass of water is produced by the reaction of 9.6 g of ammonium perchlorate? Be sure your answer has the correct number of significant digits.

Explanation:

The reaction of ammonium perchlorate (NH₄ClO₄) to produce nitrogen gas (N₂) , chlorine gas (Cl₂), oxygen gas (O₂), water (H₂O) is shown in the balanced chemical equation given below:

2 NH₄ClO₄ →  N₂ +  Cl₂ + 2 O₂ + 4 H₂O

From the equation, 2 moles of ammonium perchlorate produces 4 moles of water, i.e. mole ratio of NH₄ClO₄ to H₂O = 2 : 4 = 1 : 2

molar mass of ammonium perchlorate, NH₄ClO₄ = (14 + 4 * 1 + 35.5 +16 * 4) = 117.5

molar mass of water,  H₂O = (2 * 1 + 16) = 18.0 g

mass of water produced = moles of ammonium perchlorate * 2 * molar mass of water

moles of perchlorate = mass / molar mass = 9.6/117.5

mass of water produced = 9.6/117.5 * 2 * 18.0 g  = 2.94 g of water

Therefore, 2.9 g of water are produced by 9.6 g of ammonium perchlorate

If greenhouse gases in the atmosphere continues to increase, then the  

average temperature of Earth will increase too.

Greenhouse gases such as carbon dioxide and methane deplete the ozone

layer which shields the earth from radiations from the Sun. An increase in

these gases will therefore allow for more radiations to get to the earth.

This therefore brings about an increase in the average temperature of the

earth.

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Answer:

A

∆H = 28 kJ/mol, ∆S = 0.109 kJ/(mol-k)

Explanation:

A P E X

The spontaneous reaction at 298 K is the one having

ΔH = 28 kJ/mol, ΔS = 0.109 kJ/(mol-K)

We know the formula of Gibb's free energy as-

ΔG = ΔH - TΔS

ΔG = 28 - (298 × 0.109)

ΔG = -4.48

If a reaction is spontaneous, change in Gibb's free energy is negative and change in entropy is positive.

Therefore, The spontaneous reaction at 298 K is the one having

ΔH = 28 kJ/mol, ΔS = 0.109 kJ/(mol-K)

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Explanation:

Water is called the "universal solvent" because it is capable of dissolving more substances than any other liquid. This is important to every living thing on earth. It means that wherever water goes, either through the air, the ground, or through our bodies, it takes along valuable chemicals, minerals, and nutrients.

Answer:

10 core electrons

Explanation:

Answer:

Magnesium has 10 core electrons

Answer:

false

Explanation:

Answer:

the answer is very correct as AgNO3 and Cu(NO3)2

Answer:

A,C

Explanation:

the person above me is correct

Answer:

1. Cu(s) 2. Cu(NO3)2 (aq) 3. Cu(OH)2 (s) 4. CuO (s) 5. CuSO4 (aq)  6. Cu(s)

Explanation:

Answer: 1. [tex]2C_8H_{18}(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(g)[/tex]

2. The volume of carbon dioxide gas that is produced is 537 L

Explanation:

1. Combustion is defined as the type of chemical reaction where a hydrocarbon is combusted in the presence of oxygen to give carbon dioxide and water.

The balanced reaction for combustion of octane is:

[tex]2C_8H_{18}(l)+25O_2(g)\rightarrow 16CO_2(g)+18H_2O(g)[/tex]

2.

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]   [tex]\text{Moles of octane}=\frac{0.330\times 1000g}{114.23g/mol}=2.89moles[/tex]

According to stoichiometry:

2 moles of octane produce = 16 moles of carbon dioxide

2.89 moles of octane produce = [tex]\frac{16}{2}\times 2.89=23.1moles[/tex] of carbon dioxide

According to ideal gas equation:

[tex]PV=nRT[/tex]

P = pressure of gas = 1 atm

V = Volume of gas = ?

n = number of moles = 23.1

R = gas constant =[tex]0.0821Latm/Kmol[/tex]

T =temperature =[tex]10.0^0C=(10+273)K=283K[/tex]

[tex]V=\frac{nRT}{P}[/tex]

[tex]V=\frac{23.1\times 0.0821Latm/K mol\times 283K}{1atm}=537L[/tex]

Thus the volume of carbon dioxide gas that is produced is 537 L

Answer:

the radius is half the diameter

The radius of a circle is equal to half the diameter.