A scuba diver goes deeper underwater the diver must be aware that the increased pressure affects the human body be increasing the

Answer:

The correct answer is : 0.025.

Explanation:

The precipitation reaction is as follows in this procedure:

NiSO4.6H20 (aq) + Na2CO3.10H2O (aq)⇒ NiCO3 (s) + Na2SO4 (aq) + 16H2O

So, the number of moles of reactants can be found by

number of moles : mass used/ molar mass

n (NiSO4×6H2O) = (6.57 g)/(262.84 g/mol) = 0.025 mol (for Ni)

n (Na2CO3×10H2O) = (7.15 g)/(286.14 g/mol) = 0.0250 mol (for CO3).

Thus, The maximum number of moles of NiCO3 that can form is 0.025 mol.

Volume ..........is the answer

Answer:

the answer will be B.1 and 3 only

Explanation:1. O2- means it has in total 10 electrons as atomic no of oxygen is 8

again F- has also 10 electrons as flourine has atomic no 9

and neon  also  has 10 electrons

incase of 3rd no S2- has 18 electrons as atomic no of sulfur is 16 as its an ion so it will take 2 more electrons

argon is an inert gas so it has 18 electrons and potassium has 19 electrons eliminating 1 we will 18

# incase of the second option lithium + contains 2 electrons and Na+ contains 10 and K+ contains 18 electrons so all of them are not same

ANS is 1 and 3 only(B.i and iii) only

thank u

Answer:

70

Explanation:

100/20 =5

5 x each day over 2 weeks (14 days) = 70

The vials are necessary for the entire treatment is 70 vials

We have to first calculate the total number of antibiotics taken by the patient in two weeks, then find the total vials neccessary for the entire treatment.

The total antibiotics for a period of two weeks = (100 mg/day)(2 weeks)

Total = 100 mg/day×14 days

Total = 1400 mg for two weeks.

If the antibiotics contain vials in 20 mg/vials of the drugs,

Then the vials necessary for the entire treatment is

vial for the entire treatment  = 1400/20

total vials = 70 vials

Hence the vials are necessary for the entire treatment is 70 vials

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Answer:

Groups

Explanation:

1 A, 3 B, and 7 A are examples of  group number on the periodic table

Group 1A are  the alkali metals includes lithium sodium and potassium.

Group 3B in most periodic tables, lanthanum and actinium are considered to be a part of Group3B.

Group 7A are the halogens includes chlorine ,bromine and iodine.

They are all soft, silver metals. Due to their low ionization energy, these metals have low melting points and are highly reactive. The group 1A elements with their ns1 valence electron configurations are very active metals. They lose their valence electrons very readily.

All the Group 3B elements are rather soft, silvery-white metals, although their hardness increases with atomic number. They have higher ionization energies than the Group 1A and 2A elements, and are ionized to form a 3+ charges.

They have very high electronegativity. They have seven valence electrons . They are highly reactive, especially with alkali metals and alkaline earths metals.

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Answer:

B, adding an indicator to a solution to test its acidity

Explanation:

Answer:

B - Adding an indicator to a solution to test its acidity

Explanation:

Took the test

Answer:

b. H-3

Explanation:

The half life of the following isotopes is given as;

a. 0-19

26.88 seconds

b. H-3

12 years

c. Zn-71

2.4 minutes

d. Rb-85

This is a stable isotope. It does not decomposes.

Comparing the half lives, the isotope with the longest half life is Hydrogen - 3. With half life of 12 years.

Answer:

Pressure of hydrogen gas = 695.2 mmHg

Explanation:

Given:

Water temperature = 22°C

Pressure inside the tube = 715 mmHg

Find:

Pressure of hydrogen gas

Computation:

Using vapor pressure of water table

Water pressure at 22°C = 19.8 mmHg

Pressure inside the tube = Pressure of hydrogen gas + Water pressure at 22°C

715 = Pressure of hydrogen gas + 19.8

Pressure of hydrogen gas = 715 - 19.8

Pressure of hydrogen gas = 695.2 mmHg

Answer:

the wavelength of radiation emitted  is [tex]\mathbf{\lambda= 2169.62 \ nm}[/tex]

Explanation:

The energy of the Bohr's hydrogen atom can be expressed with the formula:

[tex]\mathtt{E_n =- \dfrac{13.6\ ev}{n^2}}[/tex]

For n = 7:

[tex]\mathtt{E_7 =- \dfrac{13.6\ ev}{7^2}}[/tex]

[tex]\mathtt{E_7 =-0.27755 \ eV}[/tex]

For n = 4

[tex]\mathtt{E_4=- \dfrac{13.6\ ev}{4^2}}[/tex]

[tex]\mathtt{E_4 =- 0.85\ eV}[/tex]

The  electron goes from the n = 7 to the n = 4, then :

[tex]\mathtt{E_7-E_4 = (-0.27755 - (-0.85) ) \ eV}[/tex]

[tex]\mathtt{= 0.57245\ eV}[/tex]

Wavelength of the radiation emitted:

[tex]\mathtt{\lambda= \dfrac{hc}{0.57245 \ eV}}[/tex]

where;

hc  = 1242 eV.nm

[tex]\mathtt{\lambda= \dfrac{1242 \ eV.nm }{0.57245 \ eV}}[/tex]

[tex]\mathbf{\lambda= 2169.62 \ nm}[/tex]

Answer:

1.34x10⁻⁵ mol / L is the solubility of AgCl

Explanation:

Ksp of AgCl is defined as:

AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)

If 0.010M of AgCl is added, some Ag⁺ and Cl⁻ will be produced until:

Ksp = [Ag⁺] [Cl⁻]

Ksp for AgCl = 1.8x10⁻¹⁰ (Taken from ALEKS Data tab):

Some Ag⁺ and Cl⁻ are produced, you can take this "some" as X:

[Ag⁺] = X

[Cl⁻] = X

Where X is the amount of AgCl that dissolvesin water. X = solubility:

Ksp = 1.8x10⁻¹⁰ = [Ag⁺] [Cl⁻]

1.8x10⁻¹⁰ = [X] [X]

1.8x10⁻¹⁰ = X²

X =

Answer:

0.64 M

Explanation:

Step 1: Given data

Step 2: Calculate the final concentration of the dextrose solution

We want to prepare a diluted solution from a concentrated one. We can find the final concentration using the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁ / V₂

C₂ = 1.0 M × 16 mL / 25 mL

C₂ = 0.64 M

Answer:

diffraction

Explanation:

Louis de Broglie in 1924 proposed the idea of wave particle duality. In his proposition, matter could exhibit wavelike properties.

Electrons are generally regarded as particles. Electrons may also display wavelike properties such as diffraction patterns. Generally, diffraction is regarded as a property of waves.

Hence, electron diffraction effect owes to the wavelike nature of of electrons when they are passed near matter.

The volume of KOH required for the neutralization of acid has been 38.9 ml. Thus, option B is correct.

Neutralization reaction has resulted in the formation of salt and water with the reaction of acid and base.

In the neutralization reaction, the strength of the acid and base can be given by:

Molarity of acid [tex]\times[/tex] Volume of acid = Molarity of base [tex]\times[/tex] Volume of base

Given, the molarity of KOH base = 0.124 M

The volume of acid (HCl) = 23.4 ml

Molarity of acid (HCl) = 0.206 M.

Substituting the values:

0.206 [tex]\times[/tex] 23.4 = 0.124 × Volume of base (KOH)

Volume of base (KOH) = 38.874 ml.

Volume of KOH = 38.9 ml

The volume of KOH required for the neutralization of acid has been 38.9 ml. Thus, option B is correct.

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Answer:

71.3 g

Explanation:

molar mass of C₇H₆O₃ = 138.13 g

molar mass of [tex]C_{9}H_{8}O_{4}[/tex] = 180.17 g

find the moles of reactant: C₇H₆O₃

C₇H₆O₃ that reacted = mass/molar mass = 57.6 g C₇H₆O₃ / 138.13 g C₇H₆O₃

= 0.417 mol [tex]C_{7}H_{6}O_{3}[/tex]

From the reaction equation, 1 mole of C₇H₆O₃ yields one mole of aspirin

find theoretical yield of aspirin:

0.417 mol C7H6O3 x 180.17 g C9H8O4 / 1 mol C9H8O4[tex]\frac{0.417 mol C_{7}H_{6}O_{3}}{} x \frac{180.17 g C_{9}H_{8}O_{4}}{1 mol C_{9}H_{8}O_{4}}[/tex]

= 75.1 g C9H8O4

Actual yield= % yield × theoretical yield/100

Actual yield = 95.0 × 75.1/100

Actual yield = 71.3 g

Taking into account the reaction stoichiometry and actual yield, the mass of C₉H₈O₄ produced from 57.6 g of C₇H₆O₃ assuming 95.0% yield is 57.0285 grams.

The balanced reaction is:

22 C₇H₆O₃ (s) + 32 C₄H₆O₃ (s) → 24 C₉H₈O₄ (s) + 33 HC₂H₃O₂ (aq)

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

The mass molar of each compound is:

Then by reaction stoichiometry, the following amounts of mass of each compound participate in the reaction:

Then you can apply the following rule of three: if by stoichiometry 3036 grams of C₇H₆O₃ produce 3264 grams of C₉H₈O₄, 57.6 grams of C₇H₆O₃ will produce how many mass of C₉H₈O₄?

[tex]mass of C_{9} H_{8} O_{4} =\frac{57.6 grams of C_{7} H_{6} O_{3} x3264 gramsof C_{9} H_{8} O_{4} }{3036 grams of C_{7} H_{6} O_{3}}[/tex]

mass of C₉H₈O₄= 60.03 grams

On the other side, actual yield is the amount of product actually obtained from a reaction. Assuming 95.0% yield  and considering the previously calculated mass the maximum amount of product that can be produced in the reaction, you can apply the following rule of three:  If 100% equals 60.03 grams of the compound produced, 95% equals how much mass of the compound?

[tex]mass of C_{9} H_{8} O_{4} =\frac{95 percentx60.03gramsof C_{9} H_{8} O_{4} }{100 percent}[/tex]

mass of C₉H₈O₄= 57.0285 grams

Finally, the mass of C₉H₈O₄ produced from 57.6 g of C₇H₆O₃ assuming 95.0% yield is 57.0285 grams.

Learn more:

Answer:

7.89 g

Explanation:

Step 1: Write the balanced equation

S₈ + 16 F₂(g) → 8 SF₄

Step 2: Calculate the moles corresponding to 2.34 g of S₈

The molar mass of S₈ is 256.52 g/mol.

[tex]2.34g \times \frac{1mol}{256.52g} = 9.12 \times 10^{-3} mol[/tex]

Step 3: Calculate the moles of SF₄ produced from 9.12 × 10⁻³ mol of S₈

The molar ratio of S₈ to SF₄ is 1:8. The moles of SF₄ produced are 8/1 × 9.12 × 10⁻³ mol = 0.0730 mol

Step 4: Calculate the mass corresponding to 0.0730 moles of SF₄

The molar mass of SF₄ is 108.07 g/mol.

[tex]0.0730 mol \times \frac{108.07 g}{mol} = 7.89 g[/tex]

Answer:

0.088 mole of Al.

Explanation:

First, we shall determine the number of mole in 23.6 g of AlBr₃.

This is illustrated below:

Mass of AlBr₃ = 23.6 g

Molar Mass of AlBr₃ = 27 + 3(80) = 267 g/mol

Mole of AlBr₃ =.?

Mole = mass/Molar mass

Mole of AlBr₃ = 23.6 / 267

Mole of AlBr₃ = 0.088 mol

Next, we shall writing the balanced equation for the reaction.

This is given below:

2Al(s) + 3Br₂(l) → 2AlBr₃(s)

From the balanced equation above,

2 moles of Al reacted with 3 mole of Br₂ to 2 moles AlBr₃.

Finally, we shall determine the number of mole of Al needed for the reaction as follow:

From the balanced equation above,

2 moles of Al reacted to 2 moles AlBr₃.

Therefore, 0.088 mole of Al will also react to produce 0.088 mole of AlBr₃.

0.085 moles of Al are required to form 23.6 g of AlBr₃.

Let's consider the following balanced equation for the synthesis reaction of AlBr₃.

2 Al(s) + 3 Br₂(l) → 2 AlBr₃(s)

First, we will convert 23.6 g to moles using the molar mass of AlBr₃ (266.69 g/mol).

[tex]23.6 g \times \frac{1mol}{266.69g} = 0.0885 mol[/tex]

The molar ratio of Al to AlBr₃ is 2:2. The moles of Al required to form 0.0885  moles of AlBr₃ are:

[tex]0.0885molAlBr_3 \times \frac{2molAl}{2molAlBr_3} = 0.0885molAl[/tex]

0.085 moles of Al are required to form 23.6 g of AlBr₃.

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Answer: The mole fraction of benzene will be 0.34 and mole fraction of toluene is 0.66

Explanation:

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

[tex]p_1=x_1p_1^0[/tex] and [tex]p_2=x_2P_2^0[/tex]

where, x = mole fraction in solution

[tex]p^0[/tex] = pressure in the pure state

According to Dalton's law, the total pressure is the sum of individual pressures.

[tex]p_{total}=p_1+p_2\\[/tex][tex]p_{total}=x_{benzene}p_{benzene}^0+x_{toluene}P_{toluene}^0[/tex]

[tex]x_{benzene}=x[/tex],

[tex]x_{toluene}=1-x_{benzene}=1-x[/tex],

[tex]p_{benzene}^0=75torr[/tex]

[tex]p_{toluene}^0=22torr[/tex]

[tex]p_{total}=40torr[/tex]

[tex]40=x\times 75+(1-x)\times 22[/tex]

[tex]x=0.34[/tex]

Thus (1-x0 = (1-0.34)=0.66

Thus the mole fraction of benzene will be 0.34 and that of toluene is 0.66

Answer:

Explanation:

Amidation is a reaction with the formation of an amide. It involves a process where series of reaction between amine and carboxylic acid takes place.

The main objective of this question is to draw with the aid of a diagram.

the structure of the amine and carboxylic acid reactants required to form the following amide in an amidation reaction.

But before that,

we are to draw the starting amine and to be sure to include the nonbonding electron pairs.

Also, to draw  starting carboxylic acid and to be sure to include the nonbonding electron pairs.

The whole series of the diagrammatic expression for the amine formation can be found in the diagram attached below.

Explanation:

Moles of 0.042 L of Pure water = 0.042 / 22.4

[tex]Moles \: of \: 0.042 L \: of \: Pure \: water \: = 1.87 \times {10}^{ - 3} [/tex]

[tex]Molecules \: in \: 0.042 L \: of \: Pure \: water \: = moles \times avogadro \: number[/tex]

Molecules in 0.042 L of Pure Water

= 1.126 * 10^21

Answer:

d. 0.121 M HC2H3O2 and 0.116 M NaC2H3O2

Explanation:

Hello,

In this case, since the pH variation is analyzed via the Henderson-Hasselbach equation:

[tex]pH=pKa+log(\frac{[Base]}{[Acid]} )[/tex]

We can infer that the nearer to 1 the ratio of of the concentration of the base to the concentration of the acid the better the buffering capacity. In such a way, since the sodium acetate is acting as the base and the acetic acid as the acid, we have:

a. [tex]\frac{[Base]}{[Acid]}=\frac{0.497M}{0.365M}=1.36[/tex]

b. [tex]\frac{[Base]}{[Acid]}=\frac{0.217M}{0.521M}=0.417[/tex]

c. [tex]\frac{[Base]}{[Acid]}=\frac{0.713M}{0.821M}=0.868[/tex]

d. [tex]\frac{[Base]}{[Acid]}=\frac{0.116M}{0.121M}=0.959[/tex]

Therefore, the d. solution has the best buffering capacity.

Regards.

Answer:

1.pp

2.cp

3.pp

4.pp

5.pp

6.pp

Answer:

milligram, centigram, decigram

Hope this answer correct :)

Answer:

C

Explanation:

hope it helps.

Answer:

1.66 moles.

Explanation:

We'll begin by calculating the number of mole in 23.2 g of nitrogen gas, N2.

This is illustrated below:

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 = 23.2 g

Mole of N2 =.?

Mole = mass /Molar mass

Mole of N2 = 23.2/28

Mole of N2 = 0.83 mole

Next, we shall determine the number of mole in 23.2 g of Hydrogen gas, H2.

This is illustrated below:

Molar mass of H2 = 2x1 = 2 g/mol

Mass of H2 = 23.2 g

Mole of H2 =?

Mole = mass /Molar mass

Mole of H2 = 23.2/2

Mole of H2 = 11.6 moles

Next, the balanced equation for the reaction. This is given below:

N2 + 3H2 —> 2NH3

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2 to produce 2 moles of NH3.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2.

Therefore, 0.83 moles will react with = (0.83 x 3) = 2.49 moles of H2.

From the calculations made above, we can see that only 2.49 moles out of 11.6 moles of H2 is required to react completely with 0.83 mole of N2.

Therefore, N2 is the limiting reactant.

Finally, we shall determine the maximum amount of NH3 produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of NH3 since all of it is consumed in the reaction.

The limiting reactant is N2 and the maximum amount of NH3 produced can be obtained as follow:

From the balanced equation above,

1 mole of N2 reacted to produce 2 moles of NH3.

Therefore, 0.83 mole of N2 will react to produce = (0.83 x 2) = 1.66 moles of NH3.

Therefore, the maximum amount of NH3 produced from the reaction is 1.66 moles.

Answer:

D) in chemical changes and physical changes.

Explanation:

Mass and energy are conserved in chemical changes and physical changes. Therefore, option D is correct.

Physical modifications are those that affect a chemical substance's form but not its chemical content. Physical changes may normally be used to separate compounds into chemical elements or simpler compounds, but they cannot be used to separate mixtures into their component compounds.

A change in physical attributes accompanies a physical change. Melting, turning into a gas, changing strength or durability, changing crystal structure or texture, and changing shape, size, color, volume, or density are a few examples of physical qualities.

In a physical change, the substance's shape or appearance changes, but the type of matter it contains stays the same. In contrast, when matter undergoes a chemical transformation, at least one new substance with novel features is created.

Thus, option D is correct.

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Answer:

\large \boxed{\text{150 g TiCl}_{4}}  

Explanation:

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

Mᵣ:     189.68                  79.87

           TiCl₄ + 2H₂O ⟶ TiO₂ + 4HCl

m/g:                                 50.0

To solve this stoichiometry problem, you must

1. Theoretical yield of TiO₂

[tex]\text{Theoretical yield} = \text{50.0 g actual} \times \dfrac{\text{100 g theoretical}}{\text{78.9 g actual}} = \text{63.37 g theoretical}[/tex]

2.  Moles of TiO₂

[tex]\text{Mass of TiO}_{2} = \text{63.37 g TiO}_{2} \times \dfrac{\text{1 mol TiO}_{2}}{\text{79.87 g TiO}_{2} } = \text{0.7934 mol TiO}_{2}[/tex]

3,  Moles of TiCl₄

The molar ratio is 1 mol TiO₂:1 mol TiCl₄.

[tex]\text{Moles of TiCl}_{4} = \text{0.7934 mol TiO}_{2} \times \dfrac{\text{1 mol TiCl}_{4}}{\text{1 mol TiO}_{2}} = \text{0.7934 mol TiCl}_{4}[/tex]

4.  Mass of TiCl₄

[tex]\text{Mass of TiCl}_{4} = \text{0.7934 mol TiCl}_{4} \times \dfrac{\text{189.98 g TiCl}_{4}}{\text{1 mol TiCl}_{4}} =\textbf{150 g TiCl}_{\mathbf{4}} \\\\\text{You must use $\large \boxed{\textbf{150 g TiCl}_{\mathbf{4}}}$}[/tex]

Sir Joseph John Thomson OM PRS (18 December 1856 – 30 August 1940) was a British physicist and Nobel Laureate in Physics, credited with the discovery of the electron, the first subatomic particle to be discovered.

Answer:

a. minimum voltage that must be supplied for a redox reaction to occur

c. always equal to Eanode - Ecathode

Explanation:

In an electrolytic cell;  The electromotive force(the maximum standard potential difference) of the cell formed by the system is defined as the standard electrode potential of the right handed electrode minus the standard electrode potential of the left hand electrode. (i.e [tex]\mathbf{E^{\theta}_{cell}=E_{anode} - E_{cathode}}[/tex] )

As we all known that the process by which chemical energy is being converted to electrical energy is called the Electrochemical cell. It consists of two half cells , an oxidation half cell reaction and a reduction half-cell reaction.The overall redox reaction results in a flow of electrons in an electric current which is produced by a minimum voltage.

Therefore, option a and c are both correct.