how many moles of helium gas occupy 22.4 L at 0 degreeC at 1 atm pressure

Explanation:

Alkali metal refers to group1 elements.

Alkali earth metal refers to group 2 elements.

Non metals refers to elements in grouos 4 to group 7.

Halogen refers to group 17 elements

Transition Metal refers to group 3 to group 12 elements

Noble gases refer to elements in group 18.

To obtain the group number from the electronic configuration, we calculate the total number of electrons in the principal quantum number (coefficient of the letters).

a. [Ne]3s1

Principal quantum number = 3

Number of electrons present = 1

This element belongs to group 1. It is an Alkali Metal.

b. [Ne]3s23p3

Principal quantum number = 3

Number of electrons present = 2 + 3 = 5

This element belongs to group 15 (5A). It is a Non metal

c. [Ar]4s23d104p5

Principal quantum number = 4

Number of electrons present = 2 + 5 = 7

This element belongs to group 17 (7A). It is an Halogen.

d. [Kr]5s24d1

This configuration belongs to the element yttrium and has an incomplete d shell. Hence it is a transition metal.

e. [Kr]5s24d105p6

Principal quantum number = 5

Number of electrons present = 2 + 6 = 8

This element belongs to group 18 (8A). It is a Noble gas.

Answer:

Canonical structures of a chemical specie explain its observed properties from a valence bond theory perspective.

Explanation:

Resonance is a valence bond concept introduced by Linus Pauling to explain the observed properties of certain chemical species such as bond lengths, bond angles, bond order , etc.

There are certain chemical species for which a single chemical structure does not suffice in explaining its observed properties. For instance, the bond order in CO3^2- is about 1.33. Its bond length, shows that the C-O bond present in CO3^2- is neither a pure C-O single bond nor a pure C-O double bond. Hence the structure of CO3^2- is 'somewhere in between' three contributing canonical structures as shown in the image attached to this answer. The resonance structures of NO3^- are also shown.

Answer:

C6H6<C7H15OH<C6H13OH<C2H5OH​

Explanation:

Organic substances are ordinarily nonpolar. This means that they do not dissolve in water. However, certain homologous series of organic compounds actually dissolve in water because they possess certain functional groups that effectively interact with water via hydrogen bonding.

A typical example of this is alcohol family. All members of this homologous series contain the -OH functional group. This group can effectively interact with water via hydrogen bonding, leading to the dissolution of low molecular weight alcohols in water.

Low molecular weight alcohols are miscible with water in all proportions. This implies that they are highly soluble in water. However, as the size of the alkyl moiety in the alcohol increases, the solubility of the alcohol in water decreases due to less effective interaction of the -OH group with water via hydrogen bonding. This explains the fact that C2H5OH​ is the most soluble alcohol in the list.

C6H6 is insoluble in water since it is purely a hydrocarbon with no -OH group capable of interaction with water via hydrogen bonding.

Answer:

1. Non-target

2. Bycatch

Explanation:

In the fishing industry, the main aim of the industry is to capture fishes that can be used or eaten and sell. A variety of fishes are captured for this purpose and since they are used therefore are known as Target catch.

But there are some species which has to be discarded because they are toxic and not useful. These non-useful species like eel which gets captured in the net while capturing other fishes are known as Non-target fish.

The eel fish which gets captured is known as bycatch fishes in the fishing operation.

Thus, Non-target and Bycatch are the correct answer.

Answer:

Its just bycatch

Explanation:

The eel species is called a(n) bycatch of the fishing operation.

Answer:

Explanation:

H₂CO₃(aq) + H₂O(l) ↔ H₃O⁺(aq) + HCO₃⁻(aq)

Let d be the degree of dissociation

.02( 1-d )                              .02d          .02d

Dissociation constant Ka₁ is given

4.3 x 10⁻⁷  = .02d x .02d / .02( 1-d )

= .004 d² / .02  ( neglecting d in denominator )

= .02 d²

d² = 215 x 10⁻⁷

d = 4.636 x 10⁻³

= .004636

concentration of H₃O⁺

= d x .02

= .004636 x .02

= 9.272 x 10⁻⁵

pH = - log [ H₃O⁺ ]

= - log ( 9.272 x 10⁻⁵ )

5 - log 9.272

= 5 - .967

= 4.033 .

Complete Question

A chemistry student weighs out 0.950 kg  of an unknown solid compound and adds it to 2.00 L of distilled water at . After minutes of stirring, only some of the has dissolved. The student drains off the solution, then washes, dries and weighs the that did not dissolve. It weighs 0.570 kg.

Required:

a. Using the information above, can you calculate the solubility of X?

b. If so, calculate it. Remember to use the correct significant digits and units. .

Answer:

a

Yes the solubility of X can be calculated this is because the solubility of a substance dissolved in a solution is the amount of that substance that is needed to saturate  1 unit volume of the solvent solution at that given temperature.

And from our question we see that substance  X saturated the solvent and there is  still remained undissolved substance X

b

The solubility of X is  [tex]S = 190 g /L[/tex]

Explanation:

From the question we are told that

    The initial mass of the unknown solid is [tex]m_i =0. 950 \ kg[/tex]

    The mass of the undissolved substance is  [tex]m_u = 0.570 \ kg[/tex]

    The volume of the solution is  [tex]V =2.00\ L[/tex]

Yes the solubility of X can be calculated this is because the solubility of a substance dissolved in a solution is the amount of that substance that is needed to saturate  1 unit volume of the solvent solution at that given temperature.

And from our question we see that substance  X saturated the solvent and there is  still remained undissolved substance X

The mass of the substance that dissolved ([tex]m_d[/tex] ) is mathematically represented as

    [tex]m_d = m_i - m_u[/tex]

  [tex]m_d = 0.95 - 0.570[/tex]

    [tex]m_d = 0.38 \ kg = 0.38 *1000 = 380 g[/tex]

The solubility of this substance (X) is mathematically represented as

      [tex]S = \frac{m_d}{V}[/tex]

substituting values

     [tex]S = \frac{ 380}{2}[/tex]

     [tex]S = 190 g /L[/tex]

Answer:

The wolf gets energy from other Animals through Cellular respiration. it's a carnivore

The rabbit gets energy from Carbohydrates,Fats.... obtained through different sources. A common example is the grass. It's an herbivore

The plant gets energy from the sun during photosynthesis. It's Autotrophic.

The mushroom gets energy from the decomposition of other organic matter. It's heterotrophic.

Explanation:

In a food chain; The Wolf eats the rabbit, when the Wolf dies, decomposers such as mushrooms breaks down its body returning it to the soil, where it provides nutrients for plants

Answer:

The empirical formula of dimethylmercury is C2H6Hg

Explanation:

Dimethylmercury, as it says in the name, presents not only the mercury metal in its structure (Hg) but also two radical groups called methyl, which is why its name begins with the prefix DI, referring to the fact that there are two methyl.

Answer:

Exothermic reaction.

Explanation:

Answer:

vaporization

Explanation:

The molar enthapy of _vaporization______ is the heat required to vaporize one mole of a liquid”

Answer:

simple cations (monatomic cations of elements of only one possible charge)

Explanation:

Simple cations (monatomic cations of elements of only one possible charge)  are named using the unmodified element name and adding the word "ion"

For example, the Na+ is named the sodium ion.

An atom or molecule with a net electric charge as a result of the loss or gain of one or more electrons is known as an ion.

Answer:

0.031 m

HgO(s) ⇒ Hg(l) + 1/2 O₂(g)

Chemical change

Element

Explanation:

A 75 gram solid cube of mercury (II) oxide has a density of 2.4 × 10³ kg/m³.  What is the length of one side of the cube in cm?

Step 1: Convert the mass to kilograms

We will use the relationship 1 kg = 1,000 g.

[tex]75g \times \frac{1kg}{1,000g} = 0.075kg[/tex]

Step 2: Calculate the volume (V) of the cube

[tex]0.075kg \times \frac{1m^{3} }{2.4 \times 10^{3} kg} = 3.1 \times 10^{-5} m^{3}[/tex]

Step 3: Calculate the length (l) of one side of the cube

We will use the following expression.

[tex]V = l^{3} \\l = \sqrt[3]{V} = \sqrt[3]{3.1 \times 10^{-5} m^{3} }=0.031m[/tex]

The mercury (II) oxide completely dissociates and forms liquid mercury and oxygen gas. Write a balanced chemical equation and indicate if this process is a chemical or physical change?

The balanced chemical equation is:

HgO(s) ⇒ Hg(l) + 1/2 O₂(g)

This is a chemical change because new substances are formed.

The oxygen gas escapes and now you are left with liquid grey substance. Is this grey substance a compound, element, homogeneous mixture or heterogeneous mixture?

The liquid gray substance is Hg(l), which is an element because it is formed by just one kind of atoms.

Answer:

Explanation:

The objective here is to prepare  1 liter of LB + Kan (100 μg/ml final concentration) + Amp (50 μg /ml final concentration) liquid media.

Given that :

the Stock concentration of Amp: 100 mg/ml (since 1 mg/ml = 1000 μg/ml)

it is required that we convert stock concentration in  μg/ml since 100 mg/ml = 100000  μg/ml

However, using formula C₁V₁=C₂V₂ (Ampicilin),

where:

C₁ = 100000 μg/ml,

V₁=?,

C₂= 50  μg/ml,

V₂=1000 ml

100000 μg/ml × V₁ = 50  μg/ml × 1000 ml

V₁ =  50  μg/ml × 1000 ml/100000 μg/ml

V₁ =   0.5 ml

Given that:

the Stock concentration of Kan: 25 mg/ml (since 1 mg/ml = 1000 μg/ml)

it is required that we convert stock concentration in  μg/ml , 25 mg/ml = 25000 μg/ml

Now by using formula C₁V₁=C₂V₂ (Kanamycin),

C₁ = 25000 μg/ml,

V₁=?,

C₂= 100  μg/ml,

V₂=1000 ml

25000 μg/ml × V₁ = 100  μg/ml × 1000 ml

V₁ =  100  μg/ml × 1000 ml/25000 μg/ml

V₁ =   4 ml

Thus; in 1 lite of Lb+ Kan+Amp preparation;

0.5 ml of Amp & 4 ml of kanamycin is used for their stock preparation.

Finally;

Sterilization step should be carried out in flask (Clean dry glass wares) for media in an autoclave, the container size should be twice the volume of media which is prepared.

Answer:

21 mL of NaOH is required.

Explanation:

Balanced reaction: [tex]HBr+NaOH\rightarrow NaBr+H_{2}O[/tex]

Number of moles of HBr in 11.0 mL of 0.30 M HBr solution

= [tex](\frac{0.30}{1000}\times 11.0)[/tex] moles = 0.0033 moles

Let's say V mL of 0.16 M NaOH solution is required to reach equivalence point.

So, number of moles of NaOH in V mL of 0.16 M NaOH solution

= [tex](\frac{0.16}{1000}\times V)[/tex] moles = 0.00016V moles

According to balanced equation-

1 mol of HBr is neutralized by 1 mol of NaOH

So, 0.0033 moles of HBr are neutralized by 0.0033 moles of NaOH

Hence, [tex]0.00016V=0.0033[/tex]

           [tex]\Rightarrow V=\frac{0.0033}{0.00016}=21[/tex]

So, 21 mL of NaOH is required.

Answer:

u

uric acid is a useful diagnostic tool as screening for most of purine metabolic disorders. The importance of uric acid measurement in plasma and urine with respect of metabolic disorders is highlighted. Not only gout and renal stones are indications to send blood to the laboratory for uric acid examination

Answer:

[tex]m=20kg[/tex]

Explanation:

Hello,

In this case, we define the kinnetic energy as:

[tex]K=\frac{1}{2} m*v^2[/tex]

Thus, for finding the mass we simply solve for it on the previous equation given the kinetic energy and the velocity:

[tex]m=\frac{2*K}{v^2}=\frac{500kg*\frac{m^2}{s^2} }{(5\frac{m}{s})^2} =\frac{500kg*\frac{m^2}{s^2} }{25\frac{m^2}{s^2}}\\\\m=20kg[/tex]

Best regards.

Answer:

The answer is 40 kg

Explanation:

You will this formula below:

m=[tex]\frac{2*\\KE}{v^{2} }[/tex]

Now we know our formula, now we plug in the given numbers:

m=[tex]\frac{2(500J)}{(5m/s)^2}[/tex]

Simplify and we get:

m=40 kg

I hope this was helpful.

Answer:

recycled is limited through enviroment and matter on earth

Explanation:

Answer:

E. The ratio of [HNO2] / [NO2-] will increase

D. The pH will decrease.

Explanation:

Nitrous acid ( HNO₂ ) is a weak acid and NaNO₂ is its salt . The mixture makes a buffer solution .

pH = pka + log [ salt] / [ Acid ]

= 3.4 + log .32 / .42

= 3.4 - .118

= 3.282 .

Now .16 moles of nitric acid is added which will react with salt to form acid

HNO₃ + NaNO₂ = HNO₂ + NaNO₃

concentration of nitrous acid will be increased and concentration of sodium nitrite ( salt will decrease )

concentration of nitrous acid = .42 + .16 = .58 M

concentration of salt = .32 - .16 = .16 M

ratio of [HNO₂ ] / NO₂⁻]

= .42 / .32 = 1.3125

ratio of [HNO₂ ] / NO₂⁻] after reaction

= .42 + .16 / .32 - .16

= 58 / 16

= 3.625 .

ratio will increase.

Option E is the answer .

pH after reaction

= 3.4 + log .16 / .58

= 2.84

pH will decrease.

Answer:

See explanation below

Explanation:

This problem is a little long so I'm gonna be as clear as possible.

a) In this case we have two acids, HBr and HCHO2. Between these two acids, the HBr is the strongest, and does not have a Ka value to dissociate, while HCHO2 do.

In order to calculate pH we need the [H₃O⁺], and in this case, as HBr is stronger, the contribution of the weaker acid can be negligible, therefore, the pH of this mixture will be:

pH = -log[H₃O⁺]

pH = -log(0.115)

pH = 0.93

b) In this case it happens the same thing as part a) HNO₃ is the strongest acid, so the contribution of the HNO₂ which is a weak acid is negligible too, therefore the pH of this mixture will be:

pH = -log(0.085)

pH = 1.07

c) Now in this case, HCHO2 and HC2H3O2 are both weak acids, so to determine which is stronger, we need to see their Ka values. In the case of HCHO2 the Ka is 1.8x10⁻⁴ and for the HC2H3O2 the Ka is 1.8x10⁻⁵. Note that the difference between the two values of Ka is just 10¹ order, so, we can neglect the concentration of either the first or the second acid. We need to see the contribution of each acid, let's begin with the stronger acid first, which is the HCHO2, we will write an ICE chart to determine the value of the [H₃O⁺] and then, use this value to determine the same concentration for the second acid and finally the pH:

        HCHO₂ + H₂O <-------> CHO₂⁻ + H₃O⁺     Ka = 1.8*10⁻⁴

i)        0.185                                0          0

c)           -x                                 +x        +x

e)       0.185-x                             x           x

1.8*10⁻⁴ = x² / 0.185-x      

As Ka is small, we can assume that "x is small" too, therefore the (0.185-x) can be rounded to just 0.185 so:

1.8*10⁻⁴ = x²/0.185

1.8*10⁻⁴ * 0.185 = x²

x² = 3.33*10⁻⁵

x = 5.77*10⁻³ M = [H₃O⁺]

Now that we have this concentration, let's write an ICE chart for the other acid, but taking account this concentration of [H₃O⁺] as innitial in the chart, and solve for the new concentration of [H₃O⁺] (In this case i will use "y" instead of "x" to make a difference from the above):

        HC₂H₃O₂ + H₂O <--------> C₂H₃O₂⁻ + H₃O⁺       Ka = 1.8x10⁻⁵

i)          0.225                                  0           5.77x10⁻⁶

c)            -y                                     +y            +y

e)        0.225-y                                y           5-77x10⁻³+y

1.8x10⁻⁵ = y(5.77x10⁻³+y) / 0.225-y   ---> once again, y is small so:

1.8x10⁻⁵ = 5.77x10⁻³y + y² / 0.225

1.8x10⁻⁵ * 0.225 = 5.77x10⁻³y + y²

y² + 5.77x10⁻³y - 4.05x10⁻⁶ = 0

Solving for y:

y = -5.77x10⁻³ ±√(5.77x10⁻³)² + 4*4.05x10⁻⁶ / 2

y = -5.77x10⁻³ ±√4.95x10⁻⁵ / 2

y = -5.77x10⁻³ ± 7.04x10⁻³ / 2

y₁ = 6.35x10⁻⁴ M

y₂ = -6.41x10⁻³ M

We will take y₁ as the value, so the concentration of hydronium will be:

[H₃O⁺] = 5.77x10⁻³ + 6.35x10⁻⁴ = 6.41x10⁻³ M

Finally the pH for this mixture is:

pH = -log(6.41x10⁻³)

pH = 2.19

d) In this case, we have the same as part c, however the Ka values differ this time. The Ka for acetic acid is 1.8x10⁻⁵  while for HCN is 4.9x10⁻¹⁰. In this ocassion, we the difference in their ka is 10⁵ order, so we can neglect the HCN concentration and focus in the acetic acid. Let's do an ICE chart and then, with the hydronium concentration we will calculate pH:

         HC₂H₃O₂ + H₂O <--------> C₂H₃O₂⁻ + H₃O⁺       Ka = 1.8x10⁻⁵

i)          0.050                                  0              0

c)            -y                                     +y            +y

e)        0.050-y                                y              y

1.8*10⁻⁵ = y² / 0.050-y      

As Ka is small, we can assume that "y is small" too

1.8*10⁻⁵ = y²/0.050

1.8*10⁻⁵ * 0.050 = y²

y² = 9*10⁻⁷

y = 9.45*10⁻⁵ M = [H₃O⁺]

Finally the pH:

pH = -log(9.45x10⁻⁵)

pH = 3.02

Answer:

Option C

The specific heat is the heat capacity for one mole of a substance.

Explanation:

The incorrect statement is  The specific heat is the heat capacity for one mole of a substance.

This is because the specific heat capacity is the amount of heat required to raise the temperature of 1 gramme of a substance by 1 degree Celcius.

Note that the unit in question here for the specific heat capacity of the substance is in grammes.

The definition given in the options is actually for the molar heat capacity of the substance, not the specific heat capacity.

Answer:

B.Handling the crucible directly with your hands.

D.Taking the mass of the empty crucible without the lid, but including the lid in all other mass measurements.

E.Taking the mass of all samples with the lid included.

Explanation:

When observed critically , the measures associated with the errors which could cause your percent yield to be falsely high, or even over 100% are those which increase the weight of the substance with the individual neglecting.

Handling the crucible directly with your hands,Taking the mass of the empty crucible without the lid, but including the lid in all other mass measurements and taking the mass of all samples with the lid included will all increase the weight of the substance. Instead the substance should be placed alone without any form of support or contamination.

Answer:

THE INITIAL TEMPERATURE OF THE SECOND SAMPLE IS 4.93 C OR 277.93 K

Explanation:

Mass of first sample of water = 106 g

Initial temp of first sample = 21.4  °C = 21.4 + 273 K = 294.4 K

Mass of second sample = 64.3 g

Final temp of theresulting mixture = 46.8  °C = 46.8 + 273 K = 319.8 K

Specific heat capacity of water = 4.184 J/g K

It is worthy to note that;

Heat gained by the first sample = Heat lost by the second sample

Since heat = mass * specific heat capacity * change in temperature, we have

Mass * specific heat * change in temp of the first sample  = Mass * specific heat * change in temp. of the second sample

MC (T2 - T1) = MC (T2-T1)

106 * 4.184 * ( 319.8 - 294.4) = 64.3 * 4.184 * ( 319.8 - T1)

106 * 4.184 * 25.4 = 269.0312 ( 319.8 - T1)

11 265.0016 = 269.0312 (319.8 - T1)

Since the change in temperature = 319.8 -T1

Change in temperature =11265.0016 / 269.0312

Change in temperature =  41.87

Change in temperature = 319.8 -T1

41.87 = 319.8 - T1

T1 = 319.8 - 41.87

T1 = 277.93 K

T1 = 4.93  °C

So therefore, the initial temperature of the sacond sample is 4.73  °C or 277.93 K

Answer:

A. 0.75 moles NO2 are required

B. 82.2 gnof HNO3 are produced

C. 205.3 g of HNO3 are produced

Explanation:

Check attachment below for explanation and calculations

Answer:

7 mol

Explanation:

Caffeine molecular formula C8H10N4O2. It has 2 atoms of oxygen.

                          C8H10N4O2         - 2O

                           1 mol                        2 mol

                           3.5 mol                    x mol

x = 3.5*2/1 = 7 mol

Answer:

HBr + H2O = H3O+ + Br-

So our conjugate acid is the H3O+ to H2O

Explanation:

A conjugate acid of a base results when the base accepts a proton.

Consider ammonia reacting with water to form an equilibrium with ammonium ions and hydroxide ions:

NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)

Ammonium, NH4+, acts as a conjugate acid to ammonia, NH3.

Answer:

24 carbons

48 hydrogens

24 oxygen

Explanation:

Answer:

D

Explanation:

Logitudinal waves also known as compression waves.

It involves displacing the medium perpendicular to the motion of the wave is not a characteristic of a transverse mechanical wave. Option B is correct.

A transverse mechanical wave is a disturbance created by it to transfer energy from one point to another. while the proposition happens the particle present within the medium get vibrates.

in a transverse wave, the particle present will vibrate up and down and are perpendicular to the wave's propagation direction. The particles shake in a directional wave in the longitudinal wave propagation.

Therefore, is not a characteristic of a transverse mechanical wave. Option B is correct. It involves displacing the medium perpendicular to the motion of the wave.

Learn more about transverse mechanical waves, here:

https://brainly.com/question/23374194

#SPJ6

Answer:

0.077M is the concentration of the hydroxyl ion

Explanation:

If 7.35 cm3 of this base is take and diluted to 147 cm3, then what is the concentration of the hydroxyl ion?

Use the dilution equation:

M1V1 = M2V2

M1 * 147cm³ = 1.5 M * 7.35 cm³

M1 = 1.5 M * 7.35 cm³ / 147 cm³

M1 = 0.077 M

How many moles of hydroxyl ion are there in the 7.35 cm3?

1000 cm³ contains 1.5 mol OH- ions

7.35 cm³ contains : 7.35 cm³ / 1000 cm³ *1.5 mol

= 0.011025 mol

Answer:

D

Explanation:

trust me its correct i think

Answer:

Explanation:

The equation for the react between Acetic acid and ethanol  to form ethyl acetate and water is :

[tex]HCH_3CO_2_{(aq)}+C_2H_5OH_{(aq)} \to C_2H_5CO_2CH_3_{(aq)} + H_2O_{(l)}[/tex]

Imagine if 94.0 mmol of [tex]C_2H_5CO_2CH_3[/tex] are removed from a flask; Then:

We are to answer the following questions:

1. What is the rate of the reverse reaction before any [tex]C_2H_5CO_2CH_3[/tex]  has been removed from the flask?

The reaction above is called an esterification reaction;

So the rate of reverse reaction before any [tex]C_2H_5CO_2CH_3[/tex]  is removed is  greater than zero and equal to forward reaction rate.

2. What is the rate of the reverse reaction just after the [tex]C_2H_5CO_2CH_3[/tex]  has been removed from the flask?

Just after the [tex]C_2H_5CO_2CH_3[/tex]  has been removed from the flask, the rate of the reverse reaction is  greater than zero but less than forward reaction rate.

3. What is the rate of the reverse reaction when the system has again achieved equilibrium?

When the system has again achieved equilibrium, the rate of the reverse reaction is greater than zero and equal to forward reaction rate because we it has achieved the equilibrium, hence, the reaction tends to proceed in the forward direction.

4. How much less [tex]C_2H_5CO_2CH_3[/tex]  is in the flask when the system has again reached equilibrium?

The [tex]C_2H_5CO_2CH_3[/tex] in the flask when the system has again reached equilibrium is lesser by 94.0 mmol as given right from the question