You have to prepare a pH 3.65 buffer, and you have the following 0.10M solutions available: HCOOH, CH3COOH, H3PO4, HCOONa, CH3COONa, and NaH2PO4. How many mL of HCOOH and HCOONa would you use to make approximately a liter of the buffer?

Answers

Answer 1

Answer:

550mL of HCOOH 0.1M and 450mL of HCOONa 0.1M

Explanation:

It is possible to find the pH of a buffer by using H-H equation:

pH = pKa + log [A⁻]/[HA]

For the formic buffer (HCOOH/HCOONa):

pH = 3.74 + log [HCOONa]/[HCOOH]

As you need a buffer of pH 3.65:

pH = 3.74 + log [HCOONa]/[HCOOH]

3.65 = 3.74 + log [HCOONa]/[HCOOH]

0.81283 = [HCOONa]/[HCOOH] (1)

Where  [HCOONa]/[HCOOH] can be taken as the moles of each specie.

As molarity of both solutions is 0.10M (0.10mol / L) and you need 1L of solution, total moles of the buffer are:

0.10 moles = [HCOONa] + [HCOOH] (2)

Replacing (2) in (1):

0.81283 = 0.10 - [HCOOH]  /[HCOOH]

0.81283[HCOOH] = 0.10 - [HCOOH]

1.81283[HCOOH] = 0.10

[HCOOH] = 0.055 moles

And moles of HCOONa are:

[HCOONa] = 0.1 mol - 0.055mol =

[HCOONa] = 0.045 moles

As concentration of the solutions is 0.1M, the volume you need to add of both solutions is:

HCOOH = 0.055 mol ₓ (1L / 0.1mol) = 0.55L = 550mL of HCOOH 0.1M

HCOONa = 0.045 mol ₓ (1L / 0.1mol) = 0.45L = 450mL of HCOONa 0.1M

Answer 2

The number should be considered like 550mL of HCOOH 0.1M and 450mL of HCOONa 0.1M.

Calculation of mL:

Here we used the H-H equation:

pH = pKa + log [A⁻]/[HA]

Now

For the formic buffer (HCOOH/HCOONa):

So,

pH = 3.74 + log [HCOONa]/[HCOOH]

Now

need a buffer of pH 3.65:

So,

pH = 3.74 + log [HCOONa]/[HCOOH]

3.65 = 3.74 + log [HCOONa]/[HCOOH]

0.81283 = [HCOONa]/[HCOOH] (1)

here  [HCOONa]/[HCOOH] can be considered as the moles of each specie.

Now the total moles should be

0.10 moles = [HCOONa] + [HCOOH] (2)

Now

0.81283 = 0.10 - [HCOOH]  /[HCOOH]

0.81283[HCOOH] = 0.10 - [HCOOH]

1.81283[HCOOH] = 0.10

[HCOOH] = 0.055 moles

And moles of HCOONa should be

[HCOONa] = 0.1 mol - 0.055mol =

[HCOONa] = 0.045 moles

Now

HCOOH = 0.055 mol ₓ (1L / 0.1mol) = 0.55L = 550mL of HCOOH 0.1M

HCOONa = 0.045 mol ₓ (1L / 0.1mol) = 0.45L = 450mL of HCOONa 0.1M

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Related Questions

Isotopes are: A. are only theoretical. B. only formed in laboratories. C. found in nature. D. found in the nuclear reactions in stars but not on Earth.

Answers

Answer:

B. only formed in laboratories

Explanation:

i know

Isotopes are only formed in Laboratories. hence, Option (B) is correct.

What are Isotopes ?

Each of two or more forms of the same element that contain equal numbers of protons but different numbers of neutrons in their nuclei, and hence differ in relative atomic mass but not in chemical properties; in particular, a radioactive form of an element is known as Isotope.

Isotopes are two or more types of atoms that have the same atomic number and position in the periodic table, and that differ in nucleon numbers due to different numbers of neutrons in their nuclei.

Therefore, Isotopes are only formed in Laboratories. hence, Option (B) is correct.

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Which method of separation would be most appropriate for separating a mixture of water and alcohol?

A. evaporation

B. chromatography

C. distillation

D. filtration

Answers

To separate a mixture of alcohol and water, you can use a process known as fractional distillation. This technique relies on the fact that the compounds in the mixture have different boiling points.

Answer:C. distillation

Explanation:

Compounds having  different boiling points when  mixed can be separated by Fractional distillation.

Now when there is a mixture of  water and alcohol, Upon heat, we would see that the alcohol will boil at a lower temperature than water  for eg, ethanol will boil at 78.5 degrees Celsius, vaporizes and  distills into the fractionating columns leaving the  water which we know boils at 100 degrees  Celsius.

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The rate constant for the decay of a radioactive element is 3.68 × 10⁻³ day⁻¹. What is the half-life of this element?

Answers

Answer:

half life=0.693/rate constant =188.3

The half-life of this element is 188.32 years

The formula for calculating the half-life of an element is expressed according to the equation:

[tex]t_{1/2}=\frac{ln 2}{\lambda}[/tex]

[tex]\lambda[/tex] is the decay constantt1/2 is the half-life

Given the following parameters:

The rate constant for the decay = 3.68 × 10⁻³ day⁻¹.

Substitute into the formula for calculating the half-life as shown:

[tex]3.68\times 10^{-3}=\frac{0.693}{\lambda}\\ 0.00368=\frac{0.693}{\lambda}\\\lambda=\frac{0.693}{0.00368}\\\lambda = 188.32 years[/tex]

Hence the half-life of this element is 188.32 years

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Consider the following practical aspects of titration.
(a) how can you tell when nearing the end point in titration?
(b) What volume of NaOH is required to permanently change the indicator at the end point?
(c) If KHP sample #1 requires 19.90 mL of NaOH solution to reach an end point, what volume is required for samples #2 and #3?
(d) if vinegar sample #1 requires 29.05 mL of NaOH solution to reach an endpoint, what volume is required for samole #2 and #3?

Answers

Answer:

A) when the titration is nearing the end point in titration the color of the solution starts to change and the change in color does not disappear as fast as it was during the beginning of the titration  

B) The volume of NaOH required to permanently change the indicator at the end point is  a  drop of NaOH

c) The volume required by samples #2 and #3 will be the same i.e 19.90 mL of NaOH, if the concentration of KHP base used for the samples are the same

D) The volume required by samples #2 and #3 will be the same i.e 29.05 mL OF NaOH, if the concentration of Vinegar and base for the samples are the same

Explanation:

A) when the titration is nearing the end point in titration the color of the solution starts to change and the change in color does not disappear as fast as it was during the beginning of the titration  

B) The volume of NaOH required to permanently change the indicator at the end point is  a  drop of NaOH

c) The volume required by samples #2 and #3 will be the same i.e 19.90 mL of NaOH, if the concentration of KHP base used for the samples are the same

D) The volume required by samples #2 and #3 will be the same i.e 29.05 mL OF NaOH, if the concentration of Vinegar and base for the samples are the same

Consider the compound hydrazine N2H4 (MW = 32.0 amu). It can react with I2 (MW = 253.8 amu) by the following reaction 2 I2 + N2H4 ------------- 4 HI + N2 (a) How many grams of I2 are needed to react with 36.7 g of N2H4? (b) How many grams of HI (MW = 127.9 amu) are produced from the reaction of 115.7 g of N2H4 with excess iodine?

Answers

Answer:Cobb

Explanation:What y'all

For which of the following elements (in their normal, stable, forms) would it be correct to describe the bonding as involving "a sea of electrons"?

a. hydrogen
b. nellum
c. sulfur
d. Iodine
e. Ethium

Answers

Answer:

e. Lithium

Explanation:

Correct list of options!

a. hydrogen  b. Helium  c. sulfur  d. Iodine  e. Lithium

Sea of electrons generally refers to metal atoms. This  is because of the delocalized nature of the electrons compared to non metals where the electrons are localized (fixed to a specific atoms).

Among all the elements in the options, the metal is option e. Lithium

The surface of silver metal, Ag(s), became tarnished when it was exposed to oxygen, producing Ag2O. In Ag2O, the oxidation state of silver is 1. According to this information, silver metal was _____. Please choose the correct answer from the following choices, and then select the submit answer button. Answer choices reduced

Answers

Answer: The silver metal was OXIDIZED.

Explanation:

OXIDATION is defined as loss of electrons and increase in oxidation number of an atom.

Oxidation number is the charge on an ion in an ionic compound or the charge that an atom in a covalent compound would have it it were ionic. There are basic rules governing oxidation number, they include:

--> the oxidation number of elements in their free states is ZERO. Example O2, Cl2,Na, Al, Ag.

-->The oxidation number for an ion is the same as the size and sign of the charge on the ion. For example the oxidation number of Zn2+ is +2.

--> the sum of all the oxidation numbers of the elements in a compound is zero.

--> oxidation numbers are always written with either a positive or a negative sign.

On the other hand, reduction is the opposite of oxidation. As oxidation is taking place, reduction is also taking place. Reduction involves:

--> Gain of electron and

--> decrease in oxidation number

From the question, the silver metal had ZERO as it's oxidation number because it's in free state. After being exposed to oxygen to form silver oxide, the oxidation number became +1.

Since there is an increase in the oxidation number, the silver metal was OXIDIZED

How many grams is 5.8 moles of hydrochloric acid (HCI)?
Answer to the nearest 0.01 g.

Answers

Answer:

211.47 grams

Explanation:

We need to set up a dimensional analysis to solve this problem by converting from moles to grams.

First, find the molar mass of HCl. Since the molar mass of H (hydrogen) is 1.01 g/mol and the molar mass of Cl (chlorine) is 35.45 g/mol, then the molar mass of HCl is:

1.01 + 35.45 = 36.46 g/mol

We have 5.8 moles of HCl, so multiply by its molar mass:

(5.8 mol) * (36.46 g/mol) = 211.468 ≈ 211.47 g

The answer is thus 211.47 grams.

~ an aesthetics over

Answer:

[tex]\large\boxed{211.47}\\[/tex] grams

Explanation:

First, you need to gather the atomic masses of the elements involved in the compound - hydrogen and chlorine. Referencing a modern periodic table will give you this information.

Hydrogen has an atomic weight of 1.00784 and Chlorine has an atomic mass of 35.453.Add those two values together - 1.00784 + 35.453 = 36.46084Multiply this value by 5.8 (one mole is equivalent to the atomic mass of the compound) - 5.8 x 36.46084 = 211.472872Round to the nearest 0.01 gram - 211.47

[tex]\large\boxed{211.47}[/tex] is the final answer.

Explain the general properties of aqueous solution based on the following support your answer with examples for each case
1. Electrolytes versus non-electrolyte
2. Precipitation reaction
3. Solubility

Answers

An aqueous solution, based on its name, is a water based solution, such that the solvent is water. In such solution, ionic compounds when dissolved, tend to dissociate into the constituent ions, for example, sodium chloride, NaCl forms an aqueous solution of sodium, Na⁺ and chloride, Cl⁻

1) Therefore, aqueous solutions are good electrolyte when ionic, and are therefore, good electrolytes which conduct electricity compared to solids that form non-electrolyte

2) A precipitation reaction is the insoluble product formed by the combination of cations and anions to form ionic solids that are insoluble

3) Aqueous solutions are made with substance that are soluble in water

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Choose the correct statement
a) The maximum value of principal quantum number (n) is 7
b) The angular quantum number (l) can receive value from 1 to (n-1)
c) The magnetic quantum number (ml) shows the energy of electron
d) The magnetic quantum number (ml) show how many orbitals in each subshell

Answers

Answer:

maybe number b is correct. ...thank you

If you dissolve 0.1 mol of formic acid in 1 L of water, ther esulting solution contains 0.004 mol of H₃O+. Based on this information, is formic acid a strong acid, or is it a weak acid?

Please explain!

Answers

In an ideal case, the acid would completely protonate. such an acid would be a strong acid.  Dissociation of Formic acid looks like:

HCOOH + H₂O → COOH⁻ + H₃O⁺

one mole of formic acid should give one mole of Hydronium(H₃O⁺) ions

similarly, 0.1 moles of formic acid should produce 0.1 moles of hydronium ions. but we know that it is not the case, a much lesser amount is actually formed.

Which means that Formic acid did not completely dissociate into COOH⁻ and H⁺ ions.

Hence, Formic Acid is a weak acid

We know

Any acid

Containing carbon is weak Containing no carbon is strong

Formic acid stands for HCOOH

It contains C means dissociation is less .

The dissociation of H+ is less means acidity is less.

Weak acid

By December 31, 2003, concerns over arsenic contamination had prompted the manufacturers of pressure-treated lumber to voluntarily cease producing lumber treated with CCA (chromated copper arsenate) for residential use. CCA-treated lumber has a light greenish color and was widely used to build decks, sand boxes, and playground structures.

Required:
Draw the Lewis structure of the arsenate ion (ASO4^3-) that yields the most favorable formal charges.

Answers

Answer:

Explanation:

lewis structure can be defined as a process of how the valence shell electrons of a molecule is being arranged, the pattern of it arrangement and the relationship between the bonding atoms and the lone pairs present in the molecule.

In order to draw the Lewis structure for Arsenate ion  [tex]\mathsf{AsO \ _4^{3-}}[/tex], first thing is to count the valence electrons in the molecule. Once we determine the valence electrons, then we distribute them around the central atom. The Arsenate ion structure is tetrahedral in nature with a bond angle of 109.5° and it is sp³ hybridized.

How many dozen (dz) eggs are needed to make 12 muffins? What about 15.5
muffins? (hint cross out units first) *

Answers

Answer:

I think its 1.2 cause I divided 15.5 with 12 and got 1.2 as an answer

Should be 1.2 I divided and got 1.2

An oxidation-reduction reaction in which 3 electrons are Transferred has ∆G° = +18.55 kJ at 25°C? What is the value of E°?

Answers

Answer:

The correct answer is: 0.064 V

Explanation:

In a oxidation-reduction reaction, the relation between Gibbs free energy (ΔGº) and cell potential (Eºcell) is given by:

[tex]\Delta G^{0} = -nFE^{0} _{cell}[/tex]

where n is the number of electrons that are transferred in the reaction and F is the Faraday constant (96,500 C/mol e-). Given: ∆G° = +18.55 kJ and n= 3 mol e-, we calculate Eºcell as follows:

+18.55 kJ = (-3 mol e-) x (96500 C/mol e-) x Eºcell

Eºcell= (+18.55 kJ)/(-3 mol e-) x (96500 C/mol e-)

Eºcell= (18550 J)/ (289500 C)

Eºcell= 0.064 J/C

Since 1 Volt= 1 Joule/1 Coulomb, thus:

Eºcell= 0.064 J/C = 0.064 V

Both chlorine and fluorine are represented by a green modeling piece that has 4 holes. Is using the same piece for two different atoms acceptable? Why or why not

Answers

Answer:

Yes, same piece can be used.

Explanation:

The same piece can be used for two different atoms are acceptable because both atoms has 7 electrons in their outermost valance shell. Both atoms belong to same group i. e. halogens so same piece can be used for both atoms. If the atoms belong to different groups and they have different number of electrons in their outermost shell so using same piece will be a problem so it is recommended to use different pieces for different atoms.

The use of the same modeling piece for chlorine and fluorine has been accepted as it has consisted of the same properties and belongs to the same group.

Chlorine and fluorine have been the elements of group 17. The elements are halogens with the presence of 7 valence electrons.

The elements have been belonging to the same group and have the same number of valence electrons thus resembling each other in the chemical properties.

Since both the elements are similar to each other, the use of the same piece for two different atoms has been acceptable.

For more information about the modeling piece, refer to the link:

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Without doing any calculations, determine the sign of ΔSsys for each of the following chemical reactions. Drag the appropriate items to their respective bins.
1. 2H30' (aq) + CO23- (aq) - CO2(g) +3H2O(1)
2. CH4(g) + 202,(g) - CO2(g) + 2H2O(l)
3. Mg (s) + Cl2(g) - MgCǐ2(s)
4. SO3(g) + H2O(I) - H2SO4(I)
A. ΔSsys greater than
B. ΔSsys smaller than

Answers

Answer:

Answers are in the explanation.

Explanation:

In a chemical reaction we can determine the sign of ΔSsys based on the states of products and reactants knowing that:

Entropy of gases >>> entropy of liquid > entropy of solids.

The entropy of solids is lower than entropy of liquids that is lower than entropy of gases.

In the reactions:

1. 2H₃O⁺(aq) + CO₃²⁻(aq) → CO₂(g) +3H₂O(l)

As 1 gas is produced, entropy of products is higher than entropy of reactants. That means  ΔSsys > 0 (That because ΔSsys is ΔSProducts - ΔSReactants)

2. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

3 moles of gas are converted in 1 mole of gas in products. Entropy of reactants is higher than entropy of products, ΔSsys < 0.

3. Mg(s) + Cl₂(g) → MgCǐ₂(s)

You have 1 mole of gas in reactants and 1 mole of solid in products. ΔSProducts <<< ΔSReactants. ΔSsys < 0.

4. SO₃(g) + H₂O(I) → H₂SO₄(I)

1 mole of gas in reactants, a liquid in products. ΔSProducts <<< ΔSReactants. ΔSsys < 0.

Determining the sign of ΔSsys for each given chemical reaction, the following can be obtained without calculations:

1. ΔSsys > 0.

2. ΔSsys < 0

3. ΔSsys < 0

4. ΔSsys < 0

Recall:

The states of the reactants and the products in a chemical reaction determines the sign of ΔSsys of the reaction.The entropy of gasses is greater than the entropy of liquid and solids.The entropy of solids is less than the entropy of liquid and gasses.Gasses have the highest entropy, while solids have the least.

Thus:

In the first chemical reaction, 1 mole of gas is produced, therefore: ΔSsys > 0.

In the second chemical reaction, 3 moles of gasses gives a products of 1 mole of gas, therefore: ΔSsys < 0.

In the third chemical reaction, 1 mole of gas gives 1 mole of solid as product, therefore: ΔSsys < 0.

In the fourth chemical reaction, 1 mole of gas gives 1 mole of liquid as product, therefore: ΔSsys < 0.

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Consider the reaction: NaNO3(s) + H2SO4(l) NaHSO4(s) + HNO3(g) ΔH° = 21.2 kJ

How much heat must absorbed by the reaction system to convert 100g of NaNO3 into NaHSO4(s)?
Please anyone help me out

Answers

The system will absorb 24.9 KJ of heat energy.  

Step 1:

Data obtained from the question

NaNO₃ + H₂SO₄ —> NaHSO₄(s) + HNO₃ ΔH° = 21.2 KJ

Heat absorbed by converting 100 g of NaNO₃ =?  

Step 2

Determination of the mass of NaNO₃ that was converted from the balanced equation.

Molar mass of NaNO₃ = 23 + 14 + (16×3)

= 23 + 14 + 48

= 85 g/mol

Mass of NaNO₃ from the balanced equation = 1 × 85 = 85 g

Thus, 85 g of NaNO₃ were converted from the balanced equation.

Step 3

Determination of the heat absorbed by converting 100 g of NaNO₃.

From the balanced equation above,

When 85 g of NaNO₃ were converted, 21.2 KJ were absorbed by the system.

Thus, the conversion of 100 g of NaNO₃ will cause the system to absorb = [tex]\frac{100 * 21.2}{85} =[/tex] 24.9 KJ of heat energy.

Therefore, the system will absorb 24.9 KJ of heat energy.

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A substance, W has a concentration of 0.02mol when its molar mass was found to be

74.0 gmol-1 . Another substance V contains 1.00x1023 atoms and has molar mass of

40.0gmol-1. Which of the two substances has the greater mass (in grams)? [L =

6.02x1023]

ii) A 250 cm3

solution contains 14.63g of sodium chloride (NaCl). Calculate the

concentration of the solution in moldm-3

[Na= 23, Cl = 35.5]​

Answers

Answer:

Explanation:

mass of W in gram  = mole x molecular weight

= .02 x 74 = 1.48 gm

mass of V in gram

first of all we shall calculate the no of moles of V

1 mole = 6.0 x  10²³ atoms

1 x 10²³ atoms = 1 / 6 moles

mass of V in grams

= 40 / 6

= 6.67 grams .

So V has greater mass .

ii )

molecular weight of sodium chloride

= 58.5 gm

14.63 gram of sodium chloride

= 14.63 / 58.5 = .25 moles

250 cm³ = 250 x 10⁻³ dm³

So 250 x 10⁻³ dm³ of solution contains .25 moles of salt

1 dm³ of solution will contain .25 / 250 x 10⁻³ mole

= 1 mole

so concentration of solution is 1 mole per dm³

Which element always exists in monatomic form? silicon, a metalloid oxygen, a nonmetal calcium, a metal neon, a noble gas

Answers

Answer:

[tex]\huge\boxed{A \ noble \ gas}[/tex]

Explanation:

A noble gas always exists in a mono-atomic form. This is because the valency of noble gas is zero and thus they don't tend to lose, gain or share electrons. They are inert gases and do not react.

Answer:

a noble gas

Explanation:

the the noble gases exist as mono atomic elements that is helium, neon ,argon, Krypton,etc.

mono atomic elements exist as stable single atoms

11mg of cyanide per kilogram of body weight is lethal for 50% of domestic chickens. How many grams per kilogram of body weight is a lethal dose for 50% of domestic chickens?

Answers

Answer:

[tex]0.033g[/tex]

Explanation:

Hello,

In this case, since 11 mg per kilogram of body weight has the given lethality, the mg that turn out lethal for a chicken weighting 3 kg is computed by using a rule of three:

[tex]11mg\longrightarrow 1kg\\\\x\ \ \ \ \ \ \longrightarrow 3kg[/tex]

Thus, we obtain:

[tex]x=\frac{3kg*11mg}{1kg}\\ \\x=33mg[/tex]

That in grams is:

[tex]=33mg*\frac{1g}{1000mg} \\\\=0.033g[/tex]

Regards.

At a constant temperature, a sample of gas occupies 1.5 L at a pressure of 2.8 ATM. What will be the pressure of this sample, in atmospheres, if the new volume is 0.92 L?

Answers

V1=1.5LV2=0.92LP1=2.8atmP2=?

Using boyles law

[tex]\boxed{\sf v\propto \dfrac{1}{p}}[/tex]

[tex]\\ \sf\longmapsto P_1V_1=P_2V_2[/tex]

[tex]\\ \sf\longmapsto P_2=\dfrac{P_1V_1}{V_2}[/tex]

[tex]\\ \sf\longmapsto P_2=\dfrac{2.8\times 1.5}{0.92}[/tex]

[tex]\\ \sf\longmapsto P_2=\dfrac{4.2}{0.92}[/tex]

[tex]\\ \sf\longmapsto P_2=4.56atm[/tex]

[tex]\\ \sf\longmapsto P_2\approx 4.6atm[/tex]

Answer:

[tex]\boxed {\boxed {\sf 4.6 \ atm}}[/tex]

Explanation:

We are asked to find the new pressure given a change in volume. We will use Boyle's Law, which states the volume of a gas is inversely proportional to the pressure. The formula for this law is:

[tex]P_1V_1= P_2V_2[/tex]

Initially, the gas occupies 1.5 liters at a pressure of 2.8 atmospheres.

[tex]1.5 \ L * 2.8 \ atm = P_2V_2[/tex]

The volume is changed to 0.92 liters, but the pressure is unknown.

[tex]1.5 \ L * 2.8 \ atm = P_2* 0.92 \ L[/tex]

We are solving for the final pressure, so we must isolate the variable P₂. It is being multiplied by 0.92 liters. The inverse operation of multiplication is division, so we divide both sides by 0.92 L.

[tex]\frac {1.5 \ L * 2.8 \ atm}{0.92 \ L} = \frac{P_2* 0.92 \ L}{0.92 \ L}[/tex]

[tex]\frac {1.5 \ L * 2.8 \ atm}{0.92 \ L}= P_2[/tex]

The units of liters cancel each other out.

[tex]\frac {1.5 * 2.8 \ atm}{0.92 }=P_2[/tex]

[tex]\frac {4.2}{0.92} \ atm= P_2[/tex]

[tex]4.565217391 \ atm = P_2[/tex]

The original measurements of pressure and volume have 2 significant figures, so our answer must have the same. For the number we calculated, that is the tenths place. The 6 in the hundredth place tells us to round the 5 up to a 6.

[tex]4.6 \ atm \approx P_2[/tex]

The pressure is approximately 4.6 atmospheres.

A piece of plastic sinks in oil but floats in water. Place these three substances in order from lowest density to greatest density.

Answers

Answer:

[tex]\rho _{oil}<\rho _{plastic}<\rho _{water}[/tex]

Explanation:

Hello,

In this case, since water and oil are immiscible due to the oil's nonpolarity and water's polarity, when mixed, the oil remains on the water since it is less dense than water. In such a way, for a plastic sunk in the oil and floating on the water (in middle of them) we can conclude that the plastic have a mid density, therefore, the required organization is:

[tex]\rho _{oil}<\rho _{plastic}<\rho _{water}[/tex]

Best regards.

Please help, Which type of molecule is shown below?

Answers

Answer:

Carbohydrate

Explanation:

A.pex

Name the following alkane molecule:
CH3
|
CH3CHCH2CHCH2CHCH3
CH
CH2CH2CH3
A. 2,4,6-trimethylheptane
B. 4-ethyl-2,6-dimethylheptane
C. 2,6-dimethyl-4-propylheptane

Answers

Answer:

2,6dimethyle-4-propylheptane

I guess!

The correct name of the compound is 4-ethyl-2,6-dimethylheptane.

What is IUPAC nomenclature?

The IUPAC nomenclature shows the correct way to name a compound this rule is established by the International Union of Pure and Applied Chemistry.

Looking at the structure of the compound as shown, the correct name of the compound is 4-ethyl-2,6-dimethylheptane.

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MnO₄- chemical name for ALEKS

Answers

Mno4- chemical name is permanganate

If an individual proton has mass 1.007825 amu, and an individual neutron has mass 1.008665 amu, what's the calculated mass of a neptunium-236 nucleus? options: A) 237.92482 amu B) 236.99873 amu C) 237.96682 amu D) 237.04817 amu

Answers

Answer:

C) 237.96682 amu

Explanation:

The symbol for neptunium-236 is given as;

²³⁶₉₃Np

This element has 93 protons and (236 - 93 = 143) neutrons.

Mass Number =Total mass of Protons + Total mass of neutrons

Total Mass pf protons = 93 * 1.007825 amu, = 93.727725 amu

Total mass of Neutrons = 143 * 1.008665 amu = 144.239095 amu

Mass = 144.239095 + 93.727725  = 237.96682 amu

Correct option is option C.

A star is observed from two positions of Earth in its orbit, in summer and winter. Which of these is the best method to calculate the approximate distance of the star from Earth? measure the parallax and use it in calculations measure the red shift of emitted light and use it in calculations use doppler effect to calculate the shift in light traveling from star to Earth in winter use doppler effect to calculate the shift in light traveling from star to Earth in summer

Answers

Answer:

measure the parallax and use it in calculations

Explanation:

got it right on test

There are many more stars at different distances from the earth. The distance to the stars calculated in light years and it is measured using parallax method.Thus option a is correct.

What is parallax method?

Parallax method is used to measure the approximate distance of stars from earth. It uses the position of nearby star from two points opposite to earth and the small angular displacement observed from the remote stars are noted.

The orbit radius of earth and distance to the stars can be calculated from the parallactic angle p, that is one second of arc. Thus the distance is described in the units parsec.

The distance to the stars are usually calculated in light years. One parsec equals 3.26 light years. The nearest star to earth is named as proxima century having the distance parallax 0.76813'' which equals 4.24 light years. Thus, parallax is inversely proportional to the distance.

To find more on parallax method, refer here:

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A sample of a hydrocarbon is found to contain 7.99g carbon and 2.01g hydrogen. What is the empirical formula for this compound

Answers

Answer:

The empirical formulae for the compound is CH3.

The decrease in the water table due to overuse of water.

Answers

Answer:  Groundwater and surface water are connected. When groundwater is overused, the lakes, streams, and rivers connected to groundwater can also have their supply diminished. Land subsidence occurs when there is a loss of support below ground. This is most often caused by human activities, mainly from the overuse of groundwater, when the soil collapses, compacts, and drops.

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Give the formula of each coordination compound. Include square brackets around the coordination complex. Do not include the oxidation state on the metal. Use parentheses only around polyatomic ligands.
For ethylenediamine, use (en) in the formula.
a) sodium hexachloroplatinate(IV)
b) dibromobis(ethylenediamine)cobalt(III) bromide
c) pentaamminechlorochromium(III) chloride

Answers

Answer:

sodium hexachloroplatinate(IV)- Na2[PtCl6]

dibromobis(ethylenediamine)cobalt(III) bromide- [Co(en)2Br2]Br

pentaamminechlorochromium(III) chloride-[Cr(NH3)5Cl]Cl2

Explanation:

The formulas of the various coordination compounds can be written from their names taking cognisance of the metal oxidation state as shown above. The oxidation state of the metal will determine the number of counter ions present in the coordination compound.

The number ligands are shown by subscripts attached to the ligand symbols. Remember that bidentate ligands such as ethylenediamine bonds to the central metal ion via two donors.

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