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a)
b)
c)
d)

Answer:

The right answer is "105.17 ppb".

Explanation:

According to the question,

The amount of [tex]Pb^{2+}[/tex] in ppb will be:

= [tex]0.5076\times 10^{-6}\times 207.2\times 106[/tex]

= [tex]105.17 \ ppb[/tex]

Thus, the amount of [tex]Pb^{2+}[/tex] is above action limit.

Answer:

N2 but i really didn't know

The compound that would be expected to be an electrolyte is : ( A ) N₂

An electrolyte is any subsatnce which conducts electircity when dissolved in a solvent such as water. From the question the compound that can conduct electricty when dissolved in water is N₂

Hence we can conclude that The compound that would be expected to be an electrolyte is : ( A ) N₂

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Answer:

The hydrogen must be compressed to 1333.13302 kPa.

Explanation:

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case:

Replacing:

P* 10.5 L= 5.75 moles* 0.082 [tex]\frac{atm*L}{mol*K}[/tex] * 293 K

Solving:

[tex]P=\frac{5.75 moles* 0.082 \frac{atm*L}{mol*K} * 293 K}{10.5 L}[/tex]

P= 13.157 atm

If 1 atm is equal to 101.325 kPa, then 13.157 atm is equal to 1333.13302 kPa.

The hydrogen must be compressed to 1333.13302 kPa.

Explanation:

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

Molarity of stock solution (M₁) = 1.2 M

Molarity of diluted solution (M₂) = 0.5 M

Volume of diluted solution (V₂) = 100 mL

The volume of stock solution needed can be obtained by using the dilution formula as illustrated below:

1.2 × V₁ = 0.5 × 100

1.2 × V₁ = 50

Divide both side by 1.2

V₁ = 50 / 1.2

Thus, 42 mL of the stock solution is needed.

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Answer:

They need 41.7 mL of the original stock solution.

Explanation:

We can use the following equation for dilutions:

Cc x Vc = Cd x Vd

Where Cc and Vc are the concentration and volume values in the concentrated condition, whereas Cd and Vd are the concentration and volume values in the diluted condition.

The concentrated solution is the original stock solution, and it has:

Cc = 1.2 M

The diluted solution must be:

Cd = 0.500 M

Vd = 100 mL

So, we have to calculate Vc. For this, we replace the data in the equation:

[tex]V_{c} = \frac{C_{d} V_{d} }{C_{c} } = \frac{(0.500 M)(100 mL)}{1.2 M} = 41.7 mL[/tex]

Therefore, 41.7 mL of 1.2 M original stock solution are required to make 100  mL of a diluted solution with a concentration of 0.500 M.

Answer:

-76.3 kJ

Explanation:

Here is the complete question

Given the standard enthalpy changes for the following two reactions:

(1) 2Fe(s) + O₂(g) → 2FeO(s)......ΔH° = -544.0 kJ

(2) 2Zn(s) + O₂(g) → 2ZnO(s)......ΔH° = -696.6 kJ. What is the standard enthalpy change for the reaction:

(3) FeO(s) + Zn(s) → Fe(s) + ZnO(s)......ΔH° = ?

Solution

Since (1) 2Fe(s) + O₂(g) → 2FeO(s)......ΔH° = -544.0 kJ

reversing the reaction, we have

2FeO(s) → 2Fe(s) + O₂(g) ......ΔH° = +544.0 kJ (4)

Adding reactions (2) and (3), we have

2FeO(s) → 2Fe(s) + O₂(g) ......ΔH° = +544.0 kJ (4)

2Zn(s) + O₂(g) → 2ZnO(s)......ΔH° = -696.6 kJ  (2)

This gives

2FeO(s) + 2Zn(s) → 2Fe(s) + 2ZnO(s)......ΔH° =

The enthalpy change for this reaction is the sum of enthalpy changes for reaction (2) and (3) = ΔH° = +544.0 kJ + (-696.6 kJ)

= +544.0 kJ - 696.6 kJ)

= -152.6 kJ

Since the required reaction is (3) which is FeO(s) + Zn(s) → Fe(s) + ZnO(s)

we divide the enthalpy change for reaction (4) by 2 to obtain the enthalpy change for reaction (3).

So, ΔH° = -152.6 kJ/2 = -76.3 kJ

So, the standard enthalpy change for the reaction

FeO(s) + Zn(s) → Fe(s) + ZnO(s) is -76.3 kJ

Answer:

1090 Torr

Explanation:

Step 1: Given data

Step 2: Convert pO₂ to Torr

we will use the conversion factor 1 atm = 760 Torr.

1.0 atm × 760 Torr/1 atm = 760 Torr

Step 3: Calculate the total pressure of the mixture (P)

The total pressure of the mixture is the sum of the partial pressures of the gases.

P = 330 Torr + 760 Torr = 1090 Torr

Ortho and para hydrogen are nuclei forms

Answer:

1,000.000

Explanation:

Answer:

a. is released in the process

Explanation:

In fusion reaction the nucleus is unstable so it releases its binding energy resulting in decreasing its mass so it becomes more stable.

Solution :

Given :

Partial pressure of HCl, [tex]$P_{HCl}$[/tex] = 84.4 atm

Partial pressure of [tex]H_2[/tex], [tex]$P_{H_2}$[/tex] = 77.9 atm

Partial pressure of [tex]Cl_2[/tex], [tex]$P_{Cl_2}$[/tex] = 54.4 atm

Reaction :

[tex]$2HCl (g) \leftrightharpoons H_2(g) + Cl_2(g)$[/tex]

Using equilibrium concept,

[tex]$k_p=\frac{(P_{H_2})(P_{Cl_{2}})}{(P_{HCl})^2}$[/tex]

[tex]$k_p=\frac{77.9 \times 54.4}{(84.4)^2}$[/tex]

[tex]$k_p=0.594$[/tex]

[tex]k_p=0.59[/tex]  (in 2 significant figures)

or [tex]k_p=5.9 \times 10^{-1}[/tex]

Answer:

Two methods which help us to conserve water are:

Sprinkler irrigation system: this irrigation has an arrangement of vertical pipes with rotating nozzles on the top. It is more useful in the uneven and sandy land where sufficient water is not available.

Drip irrigation system: this irrigation system has an arrangement of pipes or tubes with very small holes in them to water plants drop by drop just at the base of the root. It is very efficient as water is not wasted at all.

Answer:

Explanation:

hope this helps you

The mass of the marble is greater than that of the water. The marble weighs more than an equivalent volume of the water. The force from dropping the marble breaks the surface tension of the water. The marble has greater mass and volume than the water.

Answer:

That involve the complete transfer of an electron from one atom of an element to another

Answer:

8.65 atm

Explanation:

Using ideal law equation;

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (Latm/molK)

T = temperature (K)

According to the information given in this question;

V = 18.0 L

n = 6.20 moles

R = 0.0821 Latm/molK

T = 33°C = 33 + 273 = 306K

P = ?

Using PV = nRT

P × 18 = 6.20 × 0.0821 × 306

18P = 155.76

P = 155.76/18

P = 8.65 atm

Answer:

d. Prevents a drastic change in pH when an acid or base enters a solution.

Explanation:

The purpose of a buffer is to resist pH change and keep the solution relatively stable.

Answer:

2.21% wt/v

Explanation:

The mass/volume percentage, %wt/v, is an unit of concentration used in chemistry defined as 100 times the ratio of the mass of solute in g (In this case, CaCl2 = 1.77g) and the volume of solution in mL = 80mL

The %wt/v of this solution is:

%wt /v = 1.77g / 80mL * 100

%wt/v = 2.21% wt/v

Answer:

false, it is not an example of facilitated diffusion

Answer:

Explanation:

When a drop of ink added into the water gradually moves in the whole quantity of water due to this entire water turns into blue color. This is nothing but the diffusion of ink particles into the water molecules. This is because water, as well as ink molecules, are in random motion due to the motion of ink substance.

Solution :

[tex]H_2CO_3[/tex] is considered a diprotic acid.

Sp it can dissociate in solution by giving two protons.

Chemical equations for the first step of carbonic acid is :

First ionization

[tex]$H_2CO_3(aq) + H_0(1) \rightleftharpoons H_.O^+(aq) + HCO_3^-(aq)$[/tex]

Equilibrium constant expression is

[tex]$K_{a}_{1}=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}$[/tex]

Second ionization -

[tex]$HCO_3^-(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + CO_3^{2-}(aq)$[/tex]

Equilibrium constant expression is

[tex]$K_{a2}=\frac{[H_3O^+][CO_3^{2-}]}{[HCO_3^-]}$[/tex]

Answer:

19.6 x 10⁵

1.96 x 10⁶

I hope it's helps you

Product B and E are toxic because they contain "BROMIDE" or ingredients that end in 'ol'

Based on the directions given in the information of this question, any product containing "bromide" or containing an ingredient that ends in "ol" is assumed to be TOXIC.

After carefully evaluating the ingredient contents of each product in the image attached to this question, it was realized that product B contains "pyridostigmine bromide" as an ingredient while product E contains "butorphanol" as an ingredient. Hence, in accordance to the guide given in this question, products B and E are toxic.

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Explanation:

Zinc is the anode (solid zinc is oxidised). Silver is the cathode (silver ions are reduced).

By convention in standard cell notation, the anode is written on the left and the cathode is written on the right. So, in this cell: Zinc is the anode (solid zinc is oxidised). Silver is the cathode (silver ions are reduced).

Answer:

Moles of water are 0.868

Explanation:

Answer:

See explanation

Explanation:

The molecular mass is the sum of the relative atomic masses of all the atoms in the molecule.

The relative atomic mass of reactants and products are calculated as follows;

Benzoic acid is C7H6O2 hence the molar mass of benzoic acid is ;

7(12) + 6(1) + 2(16) = 84 + 6 + 32 = 122.0 g/mol

Methyl benzoate is C8H8O2

8(12) + 8(1) + 2(16) = 96 + 8 + 32 = 136.0 g/mol

Explanation:

1. Nucleus is a memberane bound organelle that contains cell,s chromosomes.

Answer:

21.4g of HBr is the minimum mass that could be left over.

Explanation:

Based on the reaction:

HBr + NaOH → NaBr + H2O

1 mole of HBr reacts per mole of NaOH

To solve this question we need to find the moles of both reactants. If moles NaOH > moles HBr, the difference in moles represents the minimum moles of HBr that could be left over because this reaction is 1:1. Using the molar mass we can find the minimum mass of HBr that could be left over, as follows:

Moles NaOH -40.0g/mol-

17g * (1mol/40.0g) = 0.425 moles NaOH

Moles HBr -Molar mass: 80.91g/mol-

55.8g * (1mol/80.91g) = 0.690 moles HBr

The difference in moles is:

0.690 moles - 0.425 moles =

0.265 moles of HBr could be left over

The mass is:

0.265 moles * (80.91g/mol) =

Follow the rules of Bronsted Lowry theory

So

#a

NH_2-

Add a H

#b

HnO_3

Take a H

#1

Conjugate acid means one H+ is added

sw

#2

Conjugate base means donate a H+

Let's consider the neutralization reaction between HCl and NaOH.

NaOH + HCl ⇒ NaCl + H₂O

To determine the pH of the resulting mixture, we need to determine the reactant in excess. First, we will calculate the reacting moles of each reactant.

NaOH: 0.0500 L × 0.500 mol/L = 0.0200 mol

HCl: 0.0750 L × 0.250 mol/L = 0.0188 mol

Now, let's determine the reactant in excess and the remaining moles of that reactant.

                    NaOH    +     HCl ⇒ NaCl + H₂O

Initial           0.0200       0.0188

Reaction    -0.0188       -0.0188

Final         1.20 × 10⁻³          0

The volume of the mixture is 50.0 mL + 75.0 mL = 125.0 mL. Then, 1.20 × 10⁻³  moles of NaOH are in 125.0 mL of solution. The concentration of NaOH is:

[NaOH] = 1.20 × 10⁻³ mol/0.1250 L = 9.60 × 10⁻³ M

NaOH is a strong base according to the following equation.

NaOH ⇒ Na⁺ + OH⁻

The concentration of OH⁻ is 1/1 × 9.60 × 10⁻³ M = 9.60 × 10⁻³ M.

The pOH is:

pOH = -log [OH⁻] = -log 9.60 × 10⁻³ = 2.02

We will calculate the pH using the following expression.

pH = 14.00 - pOH = 14.00 - 2.02 = 11.98

The pH is 11.98. Since pH > 7, the solution is basic.

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Answer:

a. 1 mole of acid is equal to one equivalent.

b. 1.00 moles of HCl are found.

c. 1L of 2.00M NaOH is needed to reach the equivalence point

Explanation:

HCl reacts with NaOH as follows:

HCl + NaOH → NaCl + H2O

Where 1 mole of HCl reacts with 1 mole of NaOH. The reaction is 1:1

a. As the reaction is 1:1, 1 mole of acid is equal to one equivalent

b. The initial moles of HCl are:

1.00L * (2.00moles HCl / 1L) = 2.00 moles of HCl

At the halfway point, the moles of HCl are the half, that is:

1.00 moles of HCl are found

c. At equivalence point, we need to add the moles of NaOH needed for a complete reaction of the moles of HCl. As the moles of HCl are 2.00 and the reaction is 1:1, we need to add 2.00 moles of NaOH, that is:

2.00moles NaOH * (1L / 2.00mol) =

1L of 2.00M NaOH is needed to reach the equivalence point

Answer:

Explanation:

The moles of H2 and N2 are as follows respectively, 0.3915mol of H2 and 0.1305 mol of N2.

0.2175,0.0725 moles of [tex]H_2[/tex] and [tex]N_2[/tex] can be formed by the decomposition of 0.145 mol of ammonia, [tex]NH_3[/tex]

The reaction for the decomposition of ammonia is as follows:-

[tex]N_2+3H_2 \rightarrow 2NH_3[/tex]

Calculate the mole  of [tex]H_2[/tex] and [tex]N_2[/tex] as follows:-

[tex]Mole\ of \ H_2=0.145\ mol\ NH_3\times\frac{3\ mol\ H_2}{2\ mol\ NH_3} \\\\=0.2175\ mol\ H_2[/tex]

[tex]Mole\ of \ N_2=0.145\ mol\ NH_3\times\frac{1\ mol\ N_2}{2\ mol\ NH_3} \\\\=0.0725\ mol\ H_2[/tex]

Hence, the number of moles of [tex]H_2[/tex] and [tex]N_2[/tex]  are 0.2175 mol, and 0.0725 mol.

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Answer:

Compounds are substances made from atoms of different elements joined by chemical bonds. Common examples are water (H2O), salt (sodium chloride, NaCl), methane (CH4).