Light travels three-fourths as fast through water as it does through a vacuum.
What is water's index of refraction?

Answers

Answer 1
1.33 is the answer thank me later
Answer 2
The answer is 1.33hope it helps

Related Questions

consider an exceptionally weak acid, HA, with Ka= 1 x 10-20. you make 0.1M solution of the salt NA. what is the pH.

Answers

Answer:

[tex]pH=10.5[/tex]

Explanation:

Hello,

In this case, the dissociation of the given weak acid is:

[tex]HA\rightleftharpoons H^++A^-[/tex]

Therefore, the law of mass action for it turns out:

[tex]Ka=\frac{[H^+][A^-]}{[HA]}[/tex]

That in terms of the change [tex]x[/tex] due to the reaction extent is:

[tex]1x10^{-23}=\frac{x*x}{0.1-x}[/tex]

Thus, by solving with the quadratic equation or solver, we obtain:

[tex]x=31.6x10^{-12}M[/tex]

Which clearly matches with the hydrogen concentration in the solution, therefore, the pH is:

[tex]pH=-log(-31.6x10^{-12})\\pH=10.5[/tex]

Regards.

Identify the state(s) of matter that each property describes.

Answers

Answer:solid,liquid,gas,plasma

Explanation:

This question seems incomplete. I believe the full question is as followed:

Identify the state(s) of matter that each property describes.

1.) takes the shape of its container:

O gas

O liquid

O solid

2.) fills all available space:

O gas

O liquid

O solid

3.) maintains its shape:

O gas

O liquid

O solid

4.) can be poured:

O gas

O liquid

O solid

5.) is compressible:

O gas

O liquid

O solid

6.) has a fixed volume:

O gas

O liquid

O solid

The answers to the 1st are gas and liquid.

The answer to the 2nd is gas.

The answer to the 3rd is solid.

The answer to the 4th is liquid.

The answer to the 5th is gas.

The answers to the 6th are liquid and solid.

A researcher placed 25.0 g of silver chloride, AgCl, in sunlight and allowed the substance to decompose completely to form silver, Ag, with the release of chlorine gas, Cl2. The gas was collected in a container during the decomposition. The researcher determined that the mass of the silver formed was 18.8 g, and the mass of the chlorine gas formed was 6.2 g. The equation for the reaction is:

Answers

Answer:

A. The law of definite proportions states that all pure samples of a particular chemical compound contain the same elements combines in the same proportion by mass.

B. The law of conservation of mass states that during ordinary chemical reactions, matter can neither be created or destroyed.

Note: The full question is as follows;

A researcher placed 25.0 g of silver chloride, AgCl, in sunlight and allowed the substance to decompose completely to form silver, Ag, with the release of chlorine gas, Cl2. The gas was collected in a container during the decomposition. The researcher determined that the mass of the silver formed was 18.8 g, and the mass of the chlorine gas formed was 6.2 g. The equation for the reaction is:

2AgCl ----> 2Ag + Cl2

a. State the law of definite proportions. Then use the researcher's data to confirm the law of definite proportions. Show your calculations.

b. State the law of conservation of matter. Then use the researcher's data to confirm the conservation of matter. Show your calculations.

Explanation:

A. Mass of silver obtained from AgCl = 18.8g.

Percentage mass of silver in the chloride = (18.8/25.0) * 100 = 75.2 %

Mass of chlorine obtained from AgCl = 6.2

Percentage mass of chlorine = (6.2/25) * 100 = 24.8 %

In one mole of AgCl with a molar mass of 143.3 g/mol; mass of silver = 107.8, mass of Cl = 35.5

Percentage mass of Ag = (107.8/143.3) * 100 = 75.2%

Percentage mass of Cl = (35.5/143.3) * 100 = 24.8%

Since the percentages by mass of Ag and AgCl obtained from the sample is the same to that obtained from a mole of AgCl, the law of definite proportions which states that all pure samples of a particular chemical compound contain the same elements combined in the same proportion by mass is verified.

B. Mass of reactant; AgCl sample = 25.0

Mass of products; At = 18.8 g; Cl = 6.2 g

Sum of products masses = 18.8 + 6.2 = 25.0 g

Therefore mass of reactant = mass of products.

This is in accordance with the law of conservation of mass which states that during ordinary chemical reactions, matter is neither created nor destroyed.

Why are sediments carried by wind deposited in a sorted manner? ( that is the largest particles at the bottom and the smallest particles on top)

Answers

Answer:

Explanation:

The larger sediment particles are having more weright(mass), hence fall quickly (early) and smaller particles with low mass are carried by wind for longer time and falls slowly ,  hence you observe sorted kind of things. Hope this helps you to understadn this phenomenon.

The aluminum in a package containing 75 ft2 of kitchen foil weighs approximately 12 ounces. Aluminum has a density of 2.70 g/cm3 . What is the approximate thickness of the aluminum foil in millimeters?(1 ounce = 28.4g)

Answers

Answer:

18130 mm

Explanation:

Now we have a lot of unit conversions to do in order to correctly answer this question. We shall do these conversions gradually.

First we convert the weight in ounce to grams.

If 1 ounce = 28.4g

12 ounces = 12×28.4 = 340.8 g

Next we convert the area of aluminum from ft2 to m2

1ft2= 0.0929 m2

75 ft2= 75 × 0.0929= 6.9675m2

Now density of aluminum= 2.70 gcm-3

Density= mass/volume

But volume= area× thickness

Density= mass/ area × thickness

Density × area × thickness= mass

Thickness= mass/ density × area

Thickness= 340.8g / 2.70gcm-3 × 6.9675m2

Thickness= 340.8/18.8

Thickness= 18.13 m

Since 1000 milimeters make 1 metre

Thickness= 18130 mm

A salt solution was found to contain 1.50 g of salt dissolved in 50 mL of water. On evaporation, the recovered salt weighed 1.47 g. What percent of salt was recovered?

A) 20.4%
B) 107%
C) 98%
D) 20.0%

Answers

Answer:

C = 98%

Explanation:

Hello,

To determine the percentage of salt recovered, we'll divide the mass of the salt recovered over by the original mass of the salt.

Mass of salt recovered = 1.47g

Initial mass of salt = 1.50g

Percentage of salt recovered = (mass recovered/ initial mass of salt) × 100

Percentage of salt recovered = (1.47 / 1.50) × 100

Percentage of salt recovered = 0.98 × 100

Percentage of salt recovered = 98%

The percentage of salt recovered is equal to 98%

A galvanic cell at a temperature of 25.0°C is powered by the following redox reaction:

2MnO4^-(aq)+16H+(aq)+5Pb(s)-->2Mn^2+(aq)+8H2O(l)+5Pb^2+(aq)

Suppose the cell is prepared with 1.87 M MnO−4 and 1.37 M H+ in one half-cell and 3.23 M Mn+2 and 6.62 M Pb+2 in the other. Calculate the cell voltage under these conditions. Round your answer to 3 significant digits.

Answers

Answer:

1.63 V

Explanation:

Let us state the reaction equation again for the purpose of clarity;

2MnO4^-(aq)+16H+(aq)+5Pb(s)-->2Mn^2+(aq)+8H2O(l)+5Pb^2+(aq)

The reduction potentials for the two half reaction equations are;

MnO 4 - (aq) + 8H + (aq) + 5e - → Mn2+(aq) + 4H2O(l) Eo=1.51 V

Pb2+(aq) + 2e - → Pb(s) Eo= -0.13 V

E°cell = E°red – E°Ox

E°cell = 1.51 - (-0.13)

E°cell = 1.51 + 0.13

E°cell = 1.64 V

But Q= [Mn^2+]^2 [Pb^2+]^5/[MnO4^-]^2 [H^+]^16

Q= [3.23]^2 [6.62]^5/[1.87]^2 [1.37]^16

Q= 10.43 × 12714.22/3.4969 × 154

Q= 132609.3/538.5226

Q= 246.25

From Nernst equation

E= E° - 0.0592/n log Q

Where n=10

E= 1.64- 0.0592/10 log 246.25

E= 1.64-0.0142

E= 1.63 V

g Provide the complete balanced chemical equation for each reaction. Include the phases (s, l, g, or aq) for each substance. If there is no reaction, write NR. Also, provide the type of reaction (combination, decomposition, combustion, single replacement, double replacement, or neutralization). Gaseous methane (CH4) reacts with gaseous oxygen.

Answers

Answer:

CH4 (g) + O2 (g) → CO2 (g) + H2O (g)

This is a combustion reaction.

Explanation:

The combination of methane (CH4) and oxygen (O2) yields carbon dioxide (CO2) and water (H2O).

CO2 is typically a gas, and water, in this case, is in a gas form because it evaporated.

The reaction is combustion because the methane reacts with the oxygen to produce carbon dioxide and water. Combustion reactions must involve O2 as one reactant.

. The pI is called ________________. The rule of calculating pI of an amino acid is that first, write the dissociation equation from fully protonated form to fully deprotonated form, label the charge of each form; second, identify the zwitterionic form (zero charge) and find the closest pKs (left and right side in the dissociation equation); third, average these two pKs. Write the dissociation equations for amino acids, glutamate, histidine, and calculate their pIs.

Answers

Answer:

The isoelectric point is that the pH at which the compound is in an electronically neutral form.

For diss equations, please find them in the enclosed file.

The pIs of 2 amino acids:

Glutamate: pI = 3,2Histidine: pI = 7,6

Explanation:

Formula for the pI calculation: pI = (pKa1 + pKa2)/2

Given 3 pKa :

Acid glutamic with an acid sidechain:

Use the lower 2  pKas (corresponding with 2 -COOH groups)

pKa1 = 2,19; pKa2 = 4,25; so pI = 3,2

Histidine with 2 amino groups:

Use the higher 2 pKas ( -COOH group and -NH= group)

pKa1 = 6; pKa2 = 9,17; so pI = 7,6

Consider the following system at equilibrium: P(aq)+Q(aq)⇌3R(aq) Classify each of the following actions by whether it causes a leftward shift, a rightward shift, or no shift in the direction of the net reaction. Drag the appropriate items to their respective bins.
Items:1) Increase [P]2) Increase [Q]3) Increase [R]4) Decrease [P]5) Decrease [Q]6) Decrease [R]7) Triple [P] and reduce [Q] to one third8) Triple both [Q] and [R]

Answers

Explanation:

P(aq)+Q(aq)⇌3R(aq)

This problem involves applying LeChatelier's principle.

LeChatelier's principle states that whenever a system in equilibrium is disturbed, the equilibrium position would change in order to annul that change.

1) Increase [P]

This would cause the equilibrium position to shift to the right. This is because more reactions have been added, to annul that change more products have to be formed.

2) Increase [Q]

This would cause the equilibrium position to shift to the right. This is because more reactions have been added, to annul that change more products have to be formed.

3) Increase [R]

This would cause the equlibrium position to shift to the left. This is because more products have been formed, to annul that change more reactants have to be formed.

4) Decrease [P]

This would cause the equlibrium position to shift to the left. This is because there are now less reactants, to annul that change more reactants have to be formed.

5) Decrease [Q]

This would cause the equilibrium position to shift to the left. This is because there are now less reactants, to annul that change more reactants have to be formed.

6) Decrease [R]

This would cause the equilibrium position to shift to the right. This is because there are now less products, to annul that change more products have to be formed.

7) Triple [P] and reduce [Q] to one third

No shift in the direction of the net reaction because both changes cancels each other.

8) Triple both [Q] and [R]

No shift in the direction of the net reaction because both changes cancels each other.

Consider this reaction:

2Cl2O5 —> 2Cl2 + 5O2

At a certain temperature it obeys this rate law.
rate = (2.7.M^-1•s^-1) [Cl2O5]^2

Suppose a vessel contains Cl2O5 at a concentration of 0.600M. calculate how long it takes for the concentration of Cl2O5 to decrease by 94%. you may assume no other reaction is important. round your answer to two digits

Answers

Answer:

[tex]t=9.7s[/tex]

Explanation:

Hello,

In this case, we have a second order kinetics given the second power of the concentration of chlorine (V) oxide in the rate expression, thus, the integrated equation for the concentration decay is:

[tex]\frac{1}{[Cl_2O_5]}=kt+\frac{1}{[Cl_2O_5]_0}[/tex]

Thus, the final concentration for a 94% decrease is:

[tex][Cl_2O_5]=0.600M-0.600M*0.94=0.036M[/tex]

Therefore, we compute the time for such decrease:

[tex]kt=\frac{1}{[Cl_2O_5]}-\frac{1}{[Cl_2O_5]_0}=\frac{1}{0.036M}-\frac{1}{0.60M} =26.1M^{-1}[/tex]

[tex]t=\frac{26.1M^{-1}}{k}= \frac{26.1M^{-1}}{2.7M^{-1}*s^{-1}}\\\\t=9.7s[/tex]

Regards.

Each of the insoluble salts below are put into 0.10 M hydrobromic acid solution. Do you expect their solubility to be more, less, or about the same as in a pure water solution ?

a. Calcium sulfite
b. Calcium fluoride
c. Silver bromide

Answers

Answer:

A. Solubility of calcium sulfite increases

B. Solubility of calcium fluoride increases

C. Solubility of Silver bromide decreases

Explanation:

The solubility factor is proportional to ions' concentration. The solubility of a solution can be predicted from Le Chatelier's principle which states that if an external constraint is imposed on a system in equilibrium, the equilibrium position will shift in order to annul the effect of the external constraint. So, If the reactant's concentration increases, the equilibrium shifts to the right indicating a higher solubility of the solution and if the product's concentration increases, the equilibrium shifts to the left indicating a lesser solubility of the solution.

Case 1. Calcium sulfite

The dissociation reaction of CaSO3 is given below:

CaSO3 ----> Ca²+ + SO3²-

SO3²- is the conjugate base of the weak acid, H2SO3. Therefore, on the addition of hydrobromic acid, some of the sulfite ion is removed from the solution by the following reaction;

H+ + SO3²- ---> HSO3-

This shifts the equilibrium to the right, more dissociation, thereby resulting in more solubility of the solute.

Case 2. Calcium fluoride

The dissociation reaction of calcium fluoride (CaF2) is shown below.

CaF2 ----> Ca²+ + 2F-

Fluoride ion (F-) is a strong conjugate base of the weak acid. Therefore, some of fluoride ions is removed by the addition of hydrobromic acid as shown below:

H+ + F- ---->. HF

Hence, the concentration of fluoride ions reduces, shifting equilibrium in the forward direction. Therefore, the solubility will be more than in pure water solution.

Case 3: Silver bromide

The dissociation reaction of AgBr is as follows:

AgBr ----> Ag+ + Br-

The addition of HBr will increase the concentration of bromide ions. Hence, equilibrium will shift in backward direction resulting in a lesser solubility than in water.

The solubility of calcium sulfite and calcium fluoride is greater in 0.10 M hydrobromic acid solution than in pure water while the solubility of silver bromide is lesser in 0.10 M hydrobromic acid solution than in pure water.

Common ion effect refers to the decrease in solubility of a substance in a solution that contains another solute with which it has a common ion. If a substance is dissolved in a solution that contains a solute with which it has a common ion, the solubility of the substance in that solution is less than its solubility in pure water.

Considering the substances given, the solubility of calcium sulfite and calcium fluoride in 0.10 M hydrobromic acid solution is more than their solubility in pure water the equilibrium position is shifted in the forward direction.

However, solubility of silver bromide in 0.10 M hydrobromic acid solution is less than its solubility in pure water due to common ion effect.

Learn more: https://brainly.com/question/6505878

When 25ml of sulphuric acid, was titrated with 0.0820 mol/L sodium hydroxide solution the end point was detected (with phenolphthalein) at 22.5ml . Calculate the concentration of sulphuric
acid in mol/L.​

Answers

Answer:

the concentration of sulphuric acid is 14mol/l

The following data show the rate constant of a reaction measured at several different temperatures. Temperature (K) Rate Constant (1/s) 310 0.194 320 0.554 330 1.48 340 3.74 350 8.97 Part APart complete Use an Arrhenius plot to determine the activation barrier for the reaction. Express your answer using three significant figures. Ea

Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

Part A

    activation barrier for the reaction [tex]E_a = 84 .0 \ KJ/mol[/tex]

Part B

    The frequency plot is  [tex]A = 2.4*10^{13} s^{-1}[/tex]    

Explanation:

From the question we are told that

     at  [tex]T_1 = 300 \ K[/tex]   [tex]k_1 = 5.70 *10^{-2}[/tex]

and  at  [tex]T_2 = 310 \ K[/tex]   [tex]k_2 = 0.169[/tex]

The  Arrhenius plot is mathematically represented as

      [tex]ln [\frac{k_2}{k_1} ] = \frac{E_a}{R} [\frac{1}{T_1} - \frac{1}{T_2} ][/tex]

Where [tex]E_a[/tex] is the activation barrier for the reaction

         R is the gas constant with a value of  [tex]R = 8.314*10^{-3} KJ/mol \cdot K[/tex]

Substituting values

          [tex]ln [\frac{0.169}{6*10^-2{}} ] = \frac{E_a}{8.314*10^{-3}} [\frac{1}{300} - \frac{1}{310} ][/tex]

=>       [tex]E_a = 84 .0 \ KJ/mol[/tex]

The  Arrhenius plot can also be  mathematically represented as

      [tex]k = A * e^{-\frac{E_a}{RT} }[/tex]

Here we can use any value of k from the data table with there corresponding temperature let take  [tex]k_2 \ and \ T_2[/tex]

So substituting values

        [tex]0.169 = A e ^{- \frac{84.0}{8.314*10^{-3} * 310} }[/tex]

=>       [tex]A = 2.4*10^{13} s^{-1}[/tex]    

Suppose you were preparing 1.0 L of a bleaching solution in a volumetric flask, and it calls for 0.21 mol of NaOCl. If all you had available was a jug of bleach that contained 0.78 M NaOCl, what volume of bleach would you need to add to the volumetric flask before you added enough water to reach the 1.0 L line

Answers

Answer: brainliesss plssssssss

0.256 L  

Explanation:

We should use the following formula:

concentration (1) × volume (1) =  concentration (2) × volume (2)

concentration (1) = 0.82 M NaOCl

volume (1) = ?

concentration (2) = 0.21 M NaOCl

volume (2) = 1 L

volume (1) = [concentration (2) × volume (2)] / concentration (1)

volume (1) = [0.21 / 1] / 0.82 = 0.256 L

What occurs when potassium reacts with bromine to form potassium bromide?

1) Electrons are shared and the bonding is ionic.

2) Electrons are shared and the bonding is covalent.

3) Electrons are transferred and the bonding is ionic

4) Electrons are transferred and the bonding is covalent.

Answers

2) Electrons are shared meaning the bond is covalent.

Two scientists study data collected during an experiment and reach different conclusions. How would the scientific community address their disagreement?

Please

Answers

Answer: D. They would device an experiment that could test the two scientists conclusions.

Explanation:

The results of the scientific study must be verified by peer scientists or members of the scientific community to proof whether the research has been conducted produce a valid evidence.

In the given situation, the two scientists had developed different conclusion for the same experiment. This may mean either of the two may have put up an incorrect conclusion.

The scientific community may address this issue by performing the experiment. Every scientific conclusion is based upon the results of the experimental approach.

Answer:d

Explanation:

What is the equilibrium constant for the following reaction:HCO2H(aq) + CN–(aq) HCO2–(aq) + HCN(aq)Does the reaction favor the formation of reactants or products? The acid dissociation constant, Ka, for HCO2H is 1.8 x 10–4and the acid dissociation constant for HCN is 4.0 x 10–10.(A) K = 1.00. The reaction favors neither the formation of reactants nor products.(B) K = 2.2 x 10–6. The reaction favors the formation of products.(C) K = 2.2 x 10–6. The reaction favors the formation of reactants.(D) K = 4.5 x 105. The reaction favors the formation of products.(E) K = 4.5 x 105. The reaction favors the formation of reactants.

Answers

Answer:

(D) K = 4.5 x 10⁵. The reaction favors the formation of products

Explanation:

HCOOH + CN⁻ ⇆ HCOO⁻ + HCN

K = [HCOO⁻] [ HCN ] / [ HCOOH] [ CN⁻]

HCOOH ⇄ H ⁺ + COO⁻

K₁ = [ H⁺] [ COO⁻ ] / [HCOOH ]

HCN ⇆ H⁺ + CN⁻

K₂ = [ H⁺] [ CN⁻] / [ HCN ]

K₁ / K₂

= [ H⁺] [ COO⁻ ] / [HCOOH ]  X  [ HCN ] / [ H⁺] [ CN⁻]

= [ COO⁻ ][ HCN ] / [HCOOH ]  [ CN⁻]

= K

K = K₁ / K₂

= 1.8  x 10⁻⁴ / 4 x 10⁻¹⁰

= 4.5 x 10⁵

So equilibrium constant of the reaction

HCOOH + CN⁻ ⇆ HCOO⁻ + HCN

is very high . Hence reaction favours the formation of product.

option (D) is correct.

What happens in a neutralization reaction?
a
The hydrogen (H+) ions from the base and the hydroxide (OH-) ions from the acld come together to form water.
The hydrogen (H+) ions from the acid and the hydroxide (OH) ions from the base come together to form water.
A substance's pH is increased to a value greater than 7.
A solution of a known concentration and volume is added until the reaction is completed.

Answers

Answer:

Answer is letter B

Explanation:

The first one is wrong because acids release H+, not bases.

The third one is wrong because the pH is exactly 7, not greater.

The last one is wrong because it is vague and does not fit a neutralization reaction.

3. Scientific methods may include three steps of study as listed below. Explain each step in detail with a complete content related sentence. (ref: p.12-16) a. Hypothesis b. Theory c. Scientific law

Answers

Answer:

Hypothesis is an assumption or idea about a particular topic or argument. An hypothesis should be one which is able to be tested and measurable to determine its authenticity.

A theory is an explanation of a scientific observation which has undergone series of experiments and is reproducible in any part of the world.

A law is simply a rule which gives an in depth explanation of a scientific finding. If new findings emerge the law could be changed or modified.

what are mineralocorticoids

Answers

Explanation:

it is used to describe those action of adrenal corticosteroids that produce sodium

it is a corticosteroid that is involved with maintaining the salt balance in the body.

Calculate the number of moles of magnesium, chlorine, and oxygen atoms in 4.90 moles of magnesium perchlorate, Mg(ClO4)2.]\

Answers

Answer:

 4.90  moles of  [tex]Mg(ClO_4)_2[/tex]   will produce  (9.8) moles of  [tex]Cl^{-}[/tex] ,

                                                    (4.90) moles of [tex]Mg^{2+}[/tex] and

                                                    (39.2) moles of  [tex]O^{2-}[/tex]

Explanation:

From the question we are told that

  The number of moles of  is  [tex]n = 4.90 \ mols[/tex]

The formation reaction of [tex]Mg(ClO_4)_2[/tex]  is

             [tex]Mg^{2+} + 2 Cl^{-} + 8O^{2+} \to Mg(ClO_4)_2[/tex]  

From the reaction we see that

   1 mole of  [tex]Mg(ClO_4)_2[/tex]  is formed by 2 moles of [tex]Cl^{-}[/tex] 1 mole of  [tex]Mg^{2+}[/tex] and 4  [tex]O^{2-}[/tex]

 This implies that

   4.90  moles of  [tex]Mg(ClO_4)_2[/tex]   will produce  (2 * 4.90) moles of  [tex]Cl^{-}[/tex] ,

                                                    (1 * 4.90) moles of [tex]Mg^{2+}[/tex] and

                                                    (8 * 4.90) moles of  [tex]O^{2-}[/tex]

So

  4.90  moles of  [tex]Mg(ClO_4)_2[/tex]   will produce  (9.8) moles of  [tex]Cl^{-}[/tex] ,

                                                    (4.90) moles of [tex]Mg^{2+}[/tex] and

                                                    (39.2) moles of  [tex]O^{2-}[/tex]

                           

Answer:

- [tex]n_{Mg}=4.90molMg[/tex]

- [tex]n_{Cl}=9.6molCl[/tex]

- [tex]n_{O}=38.4molO[/tex]

Explanation:

Hello,

In this case, for the given 4.90 moles of magnesium perchlorate, we can compute the moles of each atom by identifying the subscript each atom has in the molecule as shown below:

- Moles of magnesium atoms: here, one mole of magnesium perchlorate has only one mole of magnesium atom (subscript is one), this the moles of magnesium atoms are also 4.90 moles.

- Moles of chlorine atoms: here, one mole of magnesium perchlorate has two moles of chlorine atoms as it has a two out of the parenthesis enclosing the perchlorate anion, thus, we have:

[tex]n_{Cl}=4.80molMg(ClO_4)_2*\frac{2molCl}{1molMg(ClO_4)_2} =9.6molCl[/tex]

- Moles of oxygen atoms: here, one mole of magnesium perchlorate has eight moles of oxygen atoms as it has a four in the oxygen subscript and a two out of the parenthesis enclosing the perchlorate anion, thus, we have:

[tex]n_{O}=4.80molMg(ClO_4)_2*\frac{8molO}{1molMg(ClO_4)_2} =38.4molO[/tex]

Best regards.

What is the name of this molecule? (will give BRAINLIEST)


A straight chain of four carbons. There is a triple bond between the second and third carbons when counting from left to right or right to left.

Answers

Answer:

2 - Butyne

Explanation:

The name of the molecule with a carbon atoms arranged in a straight chain with a triple bond between the second and third carbons is 2 - Butyne.

2- Butyne is an alkyne with structural formula given below. Some of the properties of Butyne include it is a produced artificially, it is volatile and colorless in nature.

Hence, the given molecules described is 2 - Butyne.

An ideal gaseous reaction occurs at a constant pressure of 35.0 atm and releases 66.8 kJ of heat. Before the reaction, the volume of the system was 8.20 L. After the reaction, the volume of the system was 2.21 L. Calculate the total change in internal energy for the system. Enter your answer numerically in units of kJ.

Answers

Answer:

U = -45.557kj

Explanation:

Before we can calculate the totally internal energy change in kilojoules firstly we need to calculate W

U=q + w .

We know that

w = PΔ V

where P is the pressure of

and V is the volume

then we can calculate the work

w = 35 atm * ( 8.20L - 2.21L)

W=35atm* 5.99L

W=209.65atmJ

But 1 atm = 101.325J

then ,

w = 209.65* 101.325 J = 21242.79 J

let us convert it to Kj

But we know that 1kJ = 10^3 J .

Then w = 21.243 kJ .

Then we can now calculate the internal energy as

U = 21.243- 66.8 kJ = -45.557kj

But we know that heat was released. Theeefore, the total internal energy change was -45.557kj

How many kg of gas fill a 11.6 gal gas tank

Answers

Answer:

43.964

Explanation:

i think i used a calculator so let me know if its wrong

Answer:

39.49 kg

Explanation:

:)

An ideal gas sealed in a rigid 4.86-L cylinder, initially at pressure Pi=10.90 atm, is cooled until the pressure in the cylinder is Pf=1.24 atm. What is the enthalpy change for this process? ΔH =

Answers

Answer:

[tex]\Delta H=-11897J[/tex]

Explanation:

Hello,

In this case, it is widely known that for isochoric processes, the change in the enthalpy is computed by:

[tex]\Delta H=\Delta U+V\Delta P[/tex]

Whereas the change in the internal energy is computed by:

[tex]\Delta U=nCv\Delta T[/tex]

So we compute the initial and final temperatures for one mole of the ideal gas:

[tex]T_1= \frac{P_1V}{nR}=\frac{10.90atm*4.86L}{0.082*n}=\frac{646.02K }{n} \\\\T_2= \frac{P_2V}{nR}=\frac{1.24atm*4.86L}{0.082*n}=\frac{73.49K }{n}[/tex]

Next, the change in the internal energy, since the volume-constant specific heat could be assumed as ³/₂R:

[tex]\Delta U=1mol*\frac{3}{2} (8.314\frac{J}{mol*K} )*(73.49K-646.02K )=-7140J[/tex]

Then, the volume-pressure product in Joules:

[tex]V\Delta P=4.86L*\frac{1m^3}{1000L} *(1.24atm-10.90atm)*\frac{101325Pa}{1atm} \\\\V\Delta P=-4756.96J[/tex]

Finally, the change in the enthalpy for the process:

[tex]\Delta H=-7140J-4757J\\\\\Delta H=-11897J[/tex]

Best regards.

The change in enthalpy is 70.42J

Data;

Volume of the gas = 4.86LInitial Pressure = 10.90 atmFinal Pressure = 1.24 atmChange in Enthalpy = ?

Change in Enthalpy

The change of enthalpy is calculated as

[tex]\delta H = \delta V + \delta nRT\\\delta n = 0\\\delta H = \delta U \\[/tex]

The volume change is negligible

The change in enthalpy here is equal to change in internal energy over ΔE

[tex]\delta H = \delta U = nCv\delta T\\\delta H = \frac{3}{2}(nR\delta T)\\\delta H = \frac{3}{2}\{\delta PV)\\ \delta H = \frac{3}{2}[(10.90-1.24)*4.86] \\\delta H = 70.42J[/tex]

The change in enthalpy is 70.42J

Learn more on change in enthalpy here;

https://brainly.com/question/14047927

When 1.550 gg of liquid hexane (C6H14)(C6H14) undergoes combustion in a bomb calorimeter, the temperature rises from 25.87 ∘C∘C to 38.13 ∘C∘C. Find ΔErxnΔErxn for the reaction in kJ/molkJ/mol hexane. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.73 kJ/∘CkJ/∘C.

Answers

Answer:

ΔErxn[tex]= -3.90*10^3KJ[/tex]

Explanation:

Given from the question

T1 = 25.87∘C

T2= 38.13∘C.

C= 5.73Kj/C

CHECK THE ATTACHMENT FOR DETAILED EXPLATION

The boiling point of diethyl ether, CH3CH2OCH2CH3, is 34.500 °C at 1 atmosphere. Kb(diethyl ether) = 2.02 °C/m In a laboratory experiment, students synthesized a new compound and found that when 14.94 grams of the compound were dissolved in 279.5 grams of diethyl ether, the solution began to boil at 35.100 °C. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound ?

Answers

Answer:

The correct answer is 179.94 g/mol.

Explanation:

Based on the given question, the boiling point of diethyl ether us 34.500 degree C at 1 atm pressure. The boiling point of the solution given is 35.100 degree C. The Kb of diethyl ether given is 2.02 degree C/m. The weight of the compound given is 14.94 grams, the weight of the solvent (diethyl ether) is 279.5 grams.

The molecular weight of the compound can be determined by using the formula,

deltaTb = Kb * molality

Tb-To = Kb * molality

Tb-To = Kb*wt/mol.wt*1000/w (solvent)

35.100 - 34.500 = 2.02 * 14.94 / mol. wt * 1000 g / 279.5 g

0.6 = 2.02 * 53.45/ mol.wt

mol. wt = 2.02*53.45/0.6

mol. wt = 179.94 g/mol

Hence, the molecular weight of the compound is 179.94 gram per mol.

A 25.0 mLsample of an acetic acid solution is titrated with a 0.175 M NaOH solution. The equivalence point is reached when 37.5 mL of the base is added. The concentration of acetic acid in the sample was:_______,
A) 0.263
B) 0.365
C) 0.175
D) 1.83×10−4
E) 0.119

Answers

Answer:

0.263M of CH₃COOH is the concentration of the solution.

Explanation:

The reaction of acetic acid (CH₃COOH) with NaOH is:

CH₃COOH + NaOH → CH₃COO⁻Na⁺ + H₂O

1 mole of acetic acid reacts per mole of NaOH to produce sodium acetate and water.

In the equivalence point, moles of acetic acid are equal to moles of NaOH and moles of NaOH are:

0.0375L × (0.175 moles / L) = 6.56x10⁻³ moles of NaOH = moles of CH₃COOH.

As the sample of acetic acid had a volume of 25.0mL = 0.025L:

6.56x10⁻³ moles of CH₃COOH / 0.0250L =

0.263M of CH₃COOH is the concentration of the solution

Given the information you now know, what is the effect of hyperventilation on blood pH?pH? During hyperventilation, the rapid in the blood CO2CO2 concentration shifts the equilibrium to the which the concentration of H+,H+, thereby the blood pH.

Answers

Answer:

When hypercapnia processes occur, where the concentration of carbon dioxide gas increases in the blood, the protonization of the blood increases, this means that the H + ions increase in concentration, thus generating metabolic acidosis.

This metabolic acidosis is regulated by various systems, but the respiratory system collaborates by generating hyperventilation, to increase blood oxygen pressures, decrease CO2 emissions, and indirectly decrease acidity.

Explanation:

This method of regulating the body is crucial, since the proteins in our body will not be altered if they do not happen.

The enzymes, the red globules, and many more fundamental things for life ARE PROTEINS, that in front of acidic media these modify their structure by denaturing themselves and ceasing to fulfill their functions. This is the reason why it seeks to neutralize the blood pH when it comes to an increase in CO2.

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