What is the minimum pressure in kPa that must be applied at 25 °C to obtain pure water by reverse osmosis from water that is 0.690 M in sodium chloride and 0.08 M in zinc sulfate? Assume complete dissociation for electrolytes.

Answer:

Half life is the time taken by a radio active isotope to reduce by half of its original amount. Radium-226 has a half life of 1602 years meaning that it would take 1602 years for a mass of radium to reduce by half.

Number of half lives in 9612 years = 9612/1602 = 6 half lives

New mass = Original mass x (1/2)n where n is the number of half lives.

Therefore, New mass= 500 x (1/2)∧6

                                 = 500 x 0.015625

                                 = 7.8125 g

Hence the mass of radium after 9612 years will be 7.8125 grams.

Explanation:

Answer:

[tex]\boxed {\boxed {\sf 6.25 \ grams}}[/tex]

Explanation:

We are asked to find the mass of a sample of radium-226 after half-life decay. We will use the following formula:

[tex]A= A_o *\frac{1}{2}^{\frac{t}{h}}[/tex]

In this formula, [tex]A_o[/tex] is the initial amount, t is the time, and h is the half-life.

For this problem, the initial amount is 200 grams of radium-226, the time is 8,000 years, and the half-life is 1,600 years.

[tex]\bullet \ A_o= 200 \ g \\\\bullet \ t= 8,000 \ \\\bullet \ h= 1,600[/tex]

Substitute the values into the formula.

[tex]A= 200 \ g * \frac{1}{2} ^{\frac{8.000}{1,600}[/tex]

Solve the fraction in the exponent.

[tex]A= 200 \ g * \frac{1}{2}^{5}[/tex]

Solve the exponent.

[tex]A= 200 \ g *0.03125[/tex]

[tex]A= 6.25 \ g[/tex]

In addition, we can solve this another way. First, we find the number of half-lives by dividing the total time by the half-life.

Every half-life, 1/2 of the mass decays. Divide the initial mass in half, then that result in half, and so on 5 times.

After 8,000 years, 6.25 grams of radium-226 remains.

Answer:

0.1 mol

Explanation:

6/(15*3+15)

0.1 mol moles of (CH3)3NH+ are in 6.0 g of (CH3)3NH+

The mole, symbol mol, exists as the SI base unit of the amount of substance. The quantity amount of substance exists as a measure of how many elementary entities of a provided substance exist in an object or sample.A mole corresponds to the mass of a substance that includes 6.023 x 1023 particles of the substance. The mole exists the SI unit for the amount of a substance. Its symbol stands mol.

The compound trimethylamine, (CH3 )3N, exists as a  weak base when dissolved in water.

A mole exist expressed as 6.02214076 × 1023 of some chemical unit, be it atoms, molecules, ions, or others. The mole exists as a convenient unit to utilize because of the great number of atoms, molecules, or others in any substance.

To find the amount of the substance (CH3)3NH+ to calculate its molar mass:

M((CH3)3NH+) = (12+3)*3 + 14+1 = 60 g/mol

n((CH3)3NH+) = m/M

m((CH3)3NH+) = 6g

Thus,

n((CH3)3NH+) = 6g/60 g/mol = 0.1 mol

Hence,

n((CH3)3NH+) = 0.1 mol

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Answer:

Explanation:

A balanced chemical equation refers to the reaction taking place whereby the number of atoms associated in the reactants side is equivalent to the number of atoms on the products side.

From the given information, the balanced equations are as follows:

[tex]\mathbf{(a) \ \ \ Ti(s) + O_{2(g)} \to TiO_{2(s)}}[/tex]

[tex]\mathbf{(b) \ \ \ 2Ag_{2}O \to 4Ag_{(s)} + O_{2(g)}}[/tex]

[tex]\mathbf{(c) \ \ \ 2C_3H_7OH + 9O_2 \to 6CO_2+8H_2O}[/tex]

[tex]\mathbf{(d) \ \ \ 2C_5 H_{12}O \to 10 CO_2 + 12 H_2O}[/tex]

Answer:

Explanation:

1. A solution is made by dissolving 5.84g of NaCl is enough distilled water to a give a final volume of 1.00L. What is the molarity of the solution? a. 0.100 M b. 1.00 M c. 0.0250 M d. 0.400 M 2. A 0.9% NaCl (w/w) solution in water is a. is made by mixing 0.9 moles of NaCl in a 100 moles of water b. made and has the same final volume as 0.9% solution in ethyl alcohol c. a solution that boils at or above 100°C d. All the above (don't choose this one) 3. In an exergonic process, the system a. gains energy b. loses energy c. either gains or loses energy d. no energy change at all

Answer:

[tex]\boxed {\boxed {\sf 0.100 \ M }}[/tex]

Explanation:

Molarity is a measure of concentration in moles per liter.

[tex]molarity = \frac{moles \ of \ solute}{liters \ of \ solution}}[/tex]

The solution has 5.84 grams of sodium chloride or NaCl and a volume of 1.00 liters.

We are given the mass of solute in grams, so we must convert to moles. This requires the molar mass, or the mass of 1 mole of a substance. These values are found on the Periodic Table as the atomic masses, but the units are grams per mole, not atomic mass units.

We have the compound sodium chloride, so look up the molar masses of the individual elements: sodium and chlorine.

The chemical formula (NaCl) contains no subscripts, so there is 1 mole of each element in 1 mole of the compound. Add the 2 molar masses to find the compound's molar mass.

There are 58.4397693 grams of sodium chloride in 1 mole. We will use dimensional analysis and create a ratio using this information.

[tex]\frac {58.4397693 \ g\ \ NaCl} {1 \ mol \ NaCl}[/tex]

We are converting 5.84 grams to moles, so we multiply by that value.

[tex]5.84 \ g \ NaCl *\frac {58.4397693 \ g\ NaCl} {1 \ mol \ NaCl}[/tex]

Flip the ratio. It remains equivalent and the units of grams of sodium chloride cancel.

[tex]5.84 \ g \ NaCl *\frac {1 \ mol \ NaCl}{58.4397693 \ g\ NaCl}[/tex]

[tex]5.84 *\frac {1 \ mol \ NaCl}{58.4397693 }[/tex]

[tex]0.09993194823 \ mol \ NaCl[/tex]

We can use the number of moles we just calculated to find the molarity. Remember there is 1 liter of solution.

[tex]molarity= \frac{moles \ of \ solute}{liters \ of \ solution}[/tex]

[tex]molarity= \frac{ 0.09993194823 \ mol \ NaCl}{1 \ L}[/tex]

[tex]molarity= 0.09993194823 \ mol \ NaCl/L[/tex]

The original measurements of mass and volume have 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandths place. The 9 in the ten-thousandths place tells us to round the 9 to a 0, but then we must also the next 9 to a 0, and the 0 to a 1.

[tex]molarity \approx 0.100 \ mol \ NaCl/L[/tex]

1 mole per liter is 1 molar or M. We can convert the units.

[tex]molarity \approx 0.100 \ M \ NaCl[/tex]

The molarity of the solution is 0.100 M.

Answer:

Hence the solids that should test to investigate the effects of cations on pH is

[tex]AlBr_{3}[/tex] (Cation is Al 3+)  

[tex]MgCl_{2}[/tex]  ( Cation is Mg 2+)  

[tex]CH_{3} NH_{3} Br[/tex] ( Cation is NH2+).

Explanation:

The solids in the following should you test to investigate the effects of cations on pH.  

[tex]AlBr_{3}[/tex] contains (Cation is Al 3+)  

[tex]MgCl_{2}[/tex] contains ( Cation is Mg 2+)  

[tex]CH_{3} NH_{3} Br[/tex] contains( Cation is NH2+ )

The atoms or the molecules containing the positive charge that gets attracted to the cathode are called cations. The compounds a. [tex]\rm AlBr_{3}[/tex], c. [tex]\rm MgCl_{2}[/tex] and e. [tex]\rm CH_{3}NH_{3}Br[/tex] should be investigated.

Cations are the positive charge containing molecules and atoms that have more protons in their nucleus than the number of electrons in their shells. They are formed when they lose one or more electrons to another atom.

The addition or release of the electrons of the cations and anions affects the pH system as absorption of the cation decreases the pH and absorption of the anions increases the pH.

Hence, [tex]\rm Al^{3+}[/tex], [tex]\rm Mg^{2+}[/tex] and [tex]\rm NH^{2+}[/tex] are the cation that should be investigated. The addition of the cations will reduce the pH of the reaction.

Therefore, absorption of the cation reduces the pH.

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An isotope is an element with the same atomic number but different mass number due to differences in number of neutrons.

electron configuration is 2,8,6.

Belongs to group 6 and period group 3.

It forms an ion by accepting 2 electrons

Answer:

See explanation and image attached

Explanation:

Aromatic compounds undergo electrophilic aromatic substitution reactions in which the aromatic ring is maintained.

Substituted benzenes may be more or less reactive towards electrophilic aromatic substitution than benzene depending on the nature of the substituent present in the ring.

Substituents that activate the ring towards electrophilic substitution such as -OCH3 are ortho-para directing.

The major products of the bromination of anisole are p-bromoanisole and o-bromoanisole. The resonance structures leading to these products are shown in the image attached.

Answer:

Dung dịch nào sau đây chỉ chứa các ion (bỏ qua sự điện li của nước, các chất điện li mạnh phân li hoàn toàn)?

A. HBr, Na2S, Mg(OH)2, Na2CO3.

B. H2SO4, NaOH, NaCl, HF.

C. HNO3, H2SO4, KOH, K2SiO3.

D. Ca(OH)2, KOH, CH3COOH, NaCl.

Answer:

1.12 mol/L.

Explanation:

From the question given above, the following data were obtained:

Mole of LiCl = 3.75 moles

Volume = 3.36 L

Molarity =?

Molarity is simply defined as the mole of solute per unit litre of the solution. Mathematically, it is expressed as:

Molarity = mole / Volume

With the above formula, we can obtain the molarity of the solution as follow:

Mole of LiCl = 3.75 moles

Volume = 3.36 L

Molarity =?

Molarity = mole /Volume

Molarity = 3.75 / 3.36

Molarity = 1.12 mol/L

Thus, the molarity of the solution is 1.12 mol/L

Answer:

Biphenyl

Explanation:

The reaction of bromo benzene with magnesium-ether solution yields a Grignard reagent.

The byproduct of this reaction is biphenyl. It is formed when two unreacted bromobenzene molecules are coupled together.

Hence, It is advised that the bromobenzene solution be added slowly to the magnesium-ether solution so that it isn't present in a high concentration, thus reducing the amount of biphenyl by-product formed.

Answer:

Explanation:

418.4kj is the correct answer

Answer:

Hence the correct option is an option (b) Sr4, Cl,Br−,Na+.

Explanation:    

Bromine and chlorine belong to an equivalent group. As we go down the group the dimensions increases which too there's a charge on the bromine atom. therefore the size of the Br- is going to be larger in comparison to the chlorine atom.

Sr atom is within the second group, and also it's below the above-mentioned atoms.so Sr is going to be the larger one among all the atoms.

Sodium and chlorine belong to an equivalent period .size decrease from left to right. but due to the charge on sodium its size decreases and there's an opportunity that Na+ size could be adequate for Cl.      

Here we finally assume that two atoms are of an equivalent size (Na+ and Cl) which are less in size compared to the opposite two(Sr and Br-) during which one is greater (Sr)and the opposite is smaller(Br-).

Answer:

5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺

Explanation:

We will balance the redox reaction through the ion-electron method.

Step 1: Identify both half-reactions

Reduction: Br₂ ⇒ Br⁻

Oxidation: S₂O₃²⁻ ⇒ SO₄²⁻

Step 2: Perform the mass balance, adding H⁺ and H₂O where appropriate

Br₂ ⇒ 2 Br⁻

5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺

Step 3: Perform the charge balance, adding electrons where appropriate

2 e⁻ + Br₂ ⇒ 2 Br⁻

5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻

Step 4: Make the number of electrons gained and lost equal

5 × (2 e⁻ + Br₂ ⇒ 2 Br⁻)

1 × (5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻)

Step 5: Add both half-reactions

5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺

Answer:

yes it is corrwect iyt is absolitle correct

Explanation:

The equilibrium constant for the reaction is K = 0.137

We obtain the equilibrium constant considering the following equilibria and their constants:

COO(s) + H₂(g) → Co(s) + H₂O(g)    K₁ = 67

COO(s) + CO(g) → Co(s) + CO₂(g)   K₂ = 490

We write the first reaction in the forward direction because we need H₂(g) in the reactants side:

(1)     COO(s) + H₂(g) → Co(s) + H₂O(g)    K₁ = 67

Then, we write the second reaction in the reverse direction because we need CO₂(g) in the reactants side. Thus, the equilibrium constant for the reaction in the reverse direction is the reciprocal of the constant for the reaction in the forward direction (K₂):

(2)   Co(s) + CO₂(g) → COO(s) + CO(g)   K₂ = 1/490

From the addition of (1) and (2), we obtain:

COO(s) + H₂(g) → Co(s) + H₂O(g)    K₁ = 67

+

Co(s) + CO₂(g) → COO(s) + CO(g)   K₂ = 1/490

-------------------------------------------------

H₂(g) +  CO₂(g) → CO(g) + H₂O(g)

Notice that Co(s) and COO(s) are removed that appear in the same amount at both sides of the chemical equation.

Now, the equilibrium constant K for the reaction that is the sum of other two reactions is calculated as the product of the equilibrium constants, as follows:

K = K₁ x K₂ = 67 x 1/490 = 67/490 = 0.137

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Answer:

cm3 = 2500.0 g / 10.5 g/cm3 = 238 cm3

Answer: The layers are ordered by density, with the least dense layer on top, and the densest layer on the bottom.

Explanation:

Plato

Answer:

B. The atom

Explanation:

Cells are the most basic unit of structure and the smallest unit of matter is the atom.

Answer:

B. The atom

Explanation:

yeee it was right

The measurement of body fluid compartments can be achieved by the dilution of chemical compounds that only circulate and disperse in the region of selected areas in the body. The dilution process is dependent on how the concentration is defined.

Given that:

∴

Concentration C  = 0.5 mg/100

Concentration C = 0.005 mg/ml

Since the mass amount of the chemical compound(i.e. insulin) and the concentration is known.

The volume of the body fluid compartment can be calculated as:

[tex]\mathbf{volume = \dfrac{\text{mass of the marker }}{concentration }}[/tex]

[tex]Volume = \dfrac{80 \ mg}{0.005 \ mg/ml}[/tex]

Volume = 16000 ml

Thus, it is known that insulin is generally utilized for the measurement of the extracellular fluid volume and serves as a cell impermeant marker.

As a result;

The volume of the extracellular fluid compartment is 16000 ml.

However, the tritium-labeled water is a good marker for the entire body fluid compartment due to the fact that:

Given that:

plasma activity of tritium = 20 cpm/ml

i.e.

In 1 ml of plasma, 20 cpm of tritium is present.

As such, in 960,000 counts per min (cpm) of tritium-labeled water, the volume of the whole body compartment is:

[tex]\mathbf{= \dfrac{960000}{20} ml \plasma}[/tex]

= 48000 ml of plasma

Therefore, we can conclude that the volumes of the body compartment that can be determined are:

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1.42liters, which is equivalent to 3pints, of blood is required for the surgery

1 US pint = 0.473 liters

3 × 0.473

= 1.419 liters of blood for the surgery

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Answer:

Solution temperature.

Explanation:

Hello there!

In this case, considering this question about chemical kinetics, it will be possible for us to analyze two perspectives:

1. Molecular: here, we infer that the solution temperature will provide more energy to the molecules in order to undergo more effective crashes which will make more products and therefore, increase the rate constant.

2. Mathematical: by means of the Arrhenius equation, it will be possible to tell that the increase in the temperature of the system, the negative of the exponent present in such equation will increase and therefore turn the rate constant bigger.

In such way, we infer the answer is solution temperature.

Regards!

Answer:

Calcular la molaridad de una solución que se preparó disolviendo 14 g de KOH en suficiente  

agua para obtener 250 mL de solución. (masa molar del KOH = 56 g/mol).

Resolución: de acuerdo a la definición de “molaridad” debemos calcular primero, el número de mol de soluto (KOH) que  

se han disuelto en el volumen dado, es decir, “se transforma g de soluto a mol de soluto” por medio de la masa molar,  

así:

56 g de KOH 14 g de KOH

----------------- = ------------------- X = 0,25 mol de KOH

1 mol X

Ahora, de acuerdo con la definición de molaridad, el número de mol debe estar contenido en 1000 mL (o 1 L) de  

solución, que es el volumen estándar para esta unidad de concentración, lo que se determina con el siguiente planteamiento:

0,25 mol X

----------------------- = ------------------------- X = 1 mol de KOH

250 mL de solución 1000 mL de solución

Explanation:

Answer:

Solution given:

1 mole of KCl[tex]\rightarrow [/tex]22.4l

1 mole of KCl[tex]\rightarrow [/tex]74.55g

we have

0.14 mole of KCl[tex]\rightarrow [/tex]74.55*0.14=10.347g

74.55g of KCl[tex]\rightarrow [/tex]22.4l

10.347 g of KCl[tex]\rightarrow [/tex]22.4/74.55*10.347=3.11litre

volume of each solution contains 0.14 mol of KCl contain 3.11litre.

[tex]\:[/tex]

1 mole of KCl → 22.4l

1 mole of KCl → 74.55g

we have

0.14 mole of KCl → 74.55*0.14=10.347g

74.55g of KCl  → 22.4l

10.347 g of KCl → 22.4/74.55*10.347=3.11litre

volume of each solution contains 0.14 mol of KCl contain 3.11litre.

Answer:

Specific Heat Capacity

Answer:

"[tex]6.7\times 10^{-4} \ atm[/tex]" is the right answer.

Explanation:

Given:

Partial pressure of [tex]N_2[/tex],

= 0.20 atm

Partial pressure of [tex]H_2[/tex],

= 0.15 atm

[tex]K_p = 1.5\times 10^3[/tex] at [tex]400^{\circ} C[/tex]

As we know,

⇒ [tex]K_p = \frac{pN_2\times pH_2^3}{pNH_3^2}[/tex]

By putting the values, we get

    [tex]1.5\times 10^3=\frac{0.20\times (0.15)^3}{pNH_3^2}[/tex]

        [tex]pNH_3^2 = \frac{0.000675}{1.5\times 10^3}[/tex]

                    [tex]=6.7\times 10^{-4} \ atm[/tex]

Answer:

Mg+F2= Mgf2

Explanation:

F 2 is an oxidizing agent, Mg is a reducing agent. ; Pale-yellow to greenish gas with a pungent, irritating odor.

Answer: Uranium enrichment. Uranium is used to fuel nuclear reactors; however, uranium must be enriched before it can be used as fuel. Enriching uranium increases the amount of uranium-235 (U235) that can sustain the nuclear reaction needed to release energy and produce electricity at a nuclear power plant.

385 x 42.13 x 0.079 = 1281.38395

Chemistry 11 Solutions

978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 8 Solutions and Their Properties • MHR | 85

Amount in moles, n, of the NaCl(s):

NaCl

2.5 g

m

n

M

58.44 g

2

4.2778 10 m l

ol

o

/m

u

Molar concentration, c, of the NaCl(aq):

–2 4.2778 × 10 mol

0.100

0.42778 mol/L

0.43 mol

L

/L

n

c

V

The molar concentration of the saline solution is 0.43 mol/L.

Check Your Solution

The units are correct and the answer correctly shows two significant digits. The

dilution of the original concentrated solution is correct and the change to mol/L

seems reasonable.

Section 8.4 Preparing Solutions in the Laboratory

Solutions for Practice Problems

Student Edition page 386

51. Practice Problem (page 386)

Suppose that you are given a stock solution of 1.50 mol/L ammonium sulfate,

(NH4)2SO4(aq).

What volume of the stock solution do you need to use to prepare each of the

following solutions?

a. 50.0 mL of 1.00 mol/L (NH4)2SO4(aq)

b. 2 × 102 mL of 0.800 mol/L (NH4)2SO4(aq)

c. 250 mL of 0.300 mol/L NH4

+

(aq)

What Is Required?

You need to calculate the initial volume, V1, of (NH4)2SO4(aq) stock solution

needed to prepare each given dilute solution.

The dilution gives the relationship between the molarity and the volume of the solution. The volume of stock solution with a molarity of 1.50 mol/L is 50 mL.

Dilution is said to be the addition of more volume to the concentrated solution to make it less in molar concentration. This tells about the inverse and indirect relationship between the volume and the molar concentration of the solution.

Given,

Initial volume = V₁

Initial molar concentration (M₁) = 1.50 mol/L

Final volume (V₂) = 125 mL = 0.125 L

Final molar concentration (M₂)= 0.60 mol/L

The dilution is calculated as:

M₁V₁ = M₂V₂

V₁ = M₂V₂ ÷ M₁

Substituting the values in the above formula as

V₁ = M₂V₂ ÷ M₁

V₁ = (0.60 mol/L × 0.125 L) ÷ 1.50 mol/ L

V₁ = 0.05 L

= 50 mL

Therefore, 50 mL of stock solution is needed to make a 0.60 mol/L solution.

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