Which chemical can remove color of red/Pink phenol and make it clear like water transparent?

Answers

Answer 1
Lethal of skin by absorption

Related Questions

How is the compound NH3 classified?
A. As a salt
B. As a base
C. As an acid
D. As ionic

Answers

Answer:

B

Explanation:

Ammonia is considered a base as it's pH is 11

Answer from Gauthmath

The  compound NH3 (Ammonia) can be classified as a weak Base. Below you can learn more about Ammonia.

What is Ammonia (NH3)?

Ammonia is a chemical compound which is derived from the combination of Nitrogen and Hydrogen. It is denoted by the chemical formula NH3.

Ammonia is a base and when it reacts with acids to gives out salts. Physically, It is a colorless gas with a distinct characteristic of a pungent smell.

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How many moles of Al are needed to react exactly with 10.00 moles of Fe2O3 according to the following
equation?
Fe2O3 + 2 Al → Al2O3 + 2Fe
A) 15.0 moles
1
B) 20.0 moles
C) 30.0 moles
D) 60.0 moles
E) 35.0 moles

Answers

Answer:

Answer is B) 20.0 moles

Explanation:

From the equation,

1 mole of Fe2O3 = 2 moles of Al

therefore 10.0 moles of Fe2O3 = 10×2

= 20.0 moles.

Suppose you are using distillation to separate cyclohexane and toluene. The boiling point of cyclohexane is Choose... oC and the boiling point of toluene is Choose... oC. Therefore, the liquid collected first should be Choose... .

Answers

Answer:

81°C

111°C

cyclohexane

Explanation:

Distillation is a process of separating two liquids based on differences in Bolling point. For two substances having different boiling points, they are collected as they are converted into vapour, condensed and move down the condenser one after the other.

Since the boiling point of cyclohexane is less than that of toluene, cyclohexane is collected first before toluene.

Determine whether the statement about identifying a halide is true: Regardless of any concentration of ammonium solution, the precipitates in the reaction solution of my unknown halide after 0.1M AgNO3 remain because my unknown halide solution contains Br. Select one: True False

Answers

Answer:

False

Explanation:

The statement ; Regardless of any concentration of ammonium solution the precipitate of unknown halide after 0.1M AgNO3 will remain is FALSE

This is Because the remaining concentration of AgNO3 is dependent on the solubility of Ag⁺

What is Bose Einstein state of matter

Answers

Hffhfudytasujcgigustfihovo

100 mL of 0.2 mol/L sodium carbonate solution and 200 mL of 0.1 mol/L calcium nitrate solution are mixed together. Calculate the mass of calcium carbonate that would precipitate and the concentration of the sodium nitrate solution that will be produced.

Answers

Answer:

Explanation:

Na2CO3+Ca(NO3)2=CaCO3+2NaNO3

nNa2CO3=0.02

nCa(NO3)2=0.02

mCaCO3=0.02*100=2 gram

nNaNo3=0.04

Cm=2/15

From the calculation, the mass of the product is 2 g.

What is a reaction?

A chemical reaction occurs when two more substances are mixed together. In this case, the reaction is shown by; Ca(NO3)2 + Na2CO3 ----> CaCO3(s) + 2NaNO3.

Number of moles of Na2CO3 = 100/1000 L *  0.2 mol/L = 0.02 moles

Number of moles of  Ca(NO3)2 = 200/1000 L *  0.1 mol/L = 0.02 moles

Since the reaction is equimolar, amount of the product = 0.02 moles * 100 g/mol = 2 g

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Given its formula and Avogadro's Number (6.02 x 10^23 molecules/mol), deduce how many molecules are present in 3 x 10^-16 grams of TCDD. Type in only a number without using scientific notation.

Answers

Answer:

5 × 10⁵ molecules (500,000 molecules)

Explanation:

Step 1: Convert 3 × 10⁻¹⁶ g to moles

We will use the molar mass of TCDD (321.97 g/mol).

3 × 10⁻¹⁶ g × 1 mol/321.97 g = 9 × 10⁻¹⁹ mol

Step 2: Convert 9 × 10⁻¹⁹ mol to molecules

The required conversion factor is Avogadro's number (6.02 × 10²³ molecules/mol).

9 × 10⁻¹⁹ mol × 6.02 × 10²³ molecules/1 mol = 5 × 10⁵ molecules

When the equation,
O2 + __C 10H 22 →
CO2 +
H2O is balanced, the coefficient of O2 is?

Please help!

Answers

(10)O_2+C_10 H_22-->(10)CO_2+(11)H_2 O

This is the correct balanced equation

The coefficient would be (20) when you multiply 10 by 2

9. Consider a magnesium atom with charge +2. How many overall electrons are on this particle?
Hint: Magnesium's atomic number is 12.
10
12
14

Answers

14 try that one bro and let me know
10 when it’s the + sign you subtract and when it’s the - sign you add

By process of incineration, a mystery substance is empirically determined to contain 40.00% carbon by weight, 6.67% hydrogen, and 53.33% oxygen. Its molecular weight ranges between 55 and 62 g/mole. a. (6 points) Determine the chemical formula of this substance

Answers

Answer:

C₂H₄O₂

Explanation:

Step 1: Divide each percentage by the atomic mass of the element

C: 40.00/12.01 = 3.331

H: 6.67/1.01 = 6.60

O: 53.33/16.00 = 3.333

Step 2: Divide all the numbers by the smallest one

C: 3.331/3.331 = 1

H: 6.60/3.331 ≈ 2

O: 3.333/3.331 ≈ 1

The empirical formula is CH₂O, with a molecular weight of 12 g/mol + 2 × 1 g/mol + 16 g/mol = 30 g/mol. The molecular weight of the compound must be a product of 30, such as 60 (between 55 and 62 g/mol). Since we have to multiply by 2 (30 to 60) to get to the molecular weight of the compound, we also have to multiply the empirical formula by 2 to get the chemical formula of the compound.

CH₂O × 2 = C₂H₄O₂

A sample of helium gas at a pressure of 0.778 atm and a temperature of 20.2 °C, occupies a volume of 574 mL. If the gas is allowed to expand at constant temperature until its pressure is 0.373 atm, the volume of the gas sample will be mL.

Answers

Answer:

I don't know ❤️

Karl-Anthony is trying to plate gold onto his silver ring. He constructs an electrolytic cell using his ring as one of the electrodes. He runs this cell for 94.3 minutes at 205.3 mA. How many moles of electrons were transferred in this process

Answers

Answer:

0.012 moles of electrons were transferred

Explanation:

We can find the number of electrons transferred from the time in seconds and the current in Amperes using the equation:

n = I*t / F

Where n are moles of electrons transferred

I is current in Amperes = 0.2053A

t is time in seconds:

94.3min*(60s/1min) = 5658s

F is faraday constant 96485A*s/mol

Replacing:

n = 0.2053A*5658s / 96485A*s/mol

n = 0.012 moles of electrons were transferred

Moles are the mass per unit molar mass of compound. The number of moles of electrons that are transferred is 0.012 moles.

What is an electrolytic cell?

An electrolytic cell is a type of electrochemical cell that uses electrical energy from external sources to conduct the chemical energy in a cell.  

The moles transferred in electrolytic cells are measured as:

[tex]\rm n = \rm \dfrac{I\times t}{F}[/tex]

Given,

Current (I) in amperes = 0.2053 A

Time (t) in seconds = 5658 sec

Faraday constant (F) = 96485 A -s/mol

Substituting values in the equation above moles (n) can be calculated as:

[tex]\begin{aligned} \rm n &= \dfrac{ 0.2053 \times 5658 }{96485}\\\\&= 0.012 \;\rm mol\end{aligned}[/tex]

Therefore, 0.012 moles are transferred in this process.

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……….is strong due to the ……………..between positive ions and negative delocalized electrons

Answers

Answer:

atom &bond

Explanation:

atom is strong due to the bond

b. Sodium has an emission spectrum with two visible wavelengths, both very close to 590 nm. If you had a light source that contained a mixture of sodium and hydrogen, what color filter would you use to measure only the hydrogen spectrum

Answers

Answer:

Blue or Purple color filter

Explanation:

Given that Sodium has an emission spectrum with wavelength ≈ 590nm and for a wavelength of 590nm  the color is yellow.

Hence To filter out the color ( yellow ) to enable the measurement of the Balmer series of hydrogen spectrum, we have to use a filter that possess the complementary color of yellow ( i.e. purple(RYB color model) or blue (RGB additive color model )

therefore color filter to be used = Blue or Purple

The sample concentration was measured at 50mg/ml. The loading concentration needs to be 10mg/ml. The final volume needs to be 25ul. What is the volume of sample needed and the amount of buffer needed to reach 25ul

Answers

Answer:

a)  [tex]V_1=5ul[/tex]

b)  [tex]v=20ul[/tex]

Explanation:

From the question we are told that:

initial Concentration [tex]C_1=50mg/ml[/tex]

Final Concentration [tex]C_2=10mg/ml[/tex]

Final volume needs [tex]V_2 =25ul[/tex]

Generally the equation for Volume is mathematically given by

[tex]C_1V_1=C_2V_2[/tex]

[tex]V_1=\frac{C_1V_1}{C_2}[/tex]

[tex]V_1=\frac{10*25}{50}[/tex]

[tex]V_1=5ul[/tex]

Therefore

The volume of buffer needed is

[tex]v=V_2-V_1\\\\v=25-5[/tex]

[tex]v=20ul[/tex]

How many nitrate ions are present in the following aqueous solution?
5.08 L of a solution containing 2.68 x 1021 formula units of lithium nitrate per liter.
When you have the number, determine its log (base 10) and enter that value with 3 decimal places

Answers

Answer:

22.1

Explanation:

Step 1: Calculate the number of formula units of lithium nitrate

A 5.08 L solution contains 2.68 × 10²¹ formula units per liter.

5.08 L × 2.68 × 10²¹ formula units/1 L = 1.36 × 10²² formula units

Step 2: Calculate the number of nitrate ions

Lithium nitrate dissociates completely in water according to the following equation.

LiNO₃(aq) ⇒ Li⁺(aq) + NO₃⁻(aq)

The molar ratio of LiNO₃ to NO₃⁻ is 1:1. The number of nitrate ions is 1/1 × 1.36 × 10²² = 1.36 × 10²².

Then,

log (1.36 × 10²²) = 22.1

The number of nitrate ions are present in the following aqueous solution is 13.6x10²¹, and log(base 10) value of this is 22.2.

What are strong electrolytes?

Those salts which are completely dissociate into their ions in the solution form will known as strong electrolyte.

Lithium nitrate is a strong electrolyte and it shows complete dissociation as:

LiNO₃ → Li⁺ + NO₃⁻

Given that,

1 liter of solution containing = 2.68 x 10²¹ formula unit

5.08 L of solution containing = 5.08 x 2.68 x 10²¹ = 13.6x10²¹ formula unit

So, number of nitrate ion in given solution = 13.6x10²¹ formula unit

Log(13.6x10²¹) = 22.1

Hence, required values are 13.6x10²¹ and 22.1.

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what is the IUPAC name of 2NaOH(s)​

Answers

Answer:

NaoH= sodium hydroxide

please see attachment

Answers

Answer:

I'll see it

Explanation:

lution: What is the molarity of 245 g of H, SO4 dissolved in 1.00 L of solution?

Answers

Answer:

Cm = n/V

n(H2SO4) = 245/98 = 2.5 mol

Cm(H2SO4) = 2.5/1 = 2.5 M

Explanation:

20. An oxide of osmium (symbol Os) is a pale yellow solid. If 2.89 g of the compound contains 2.16 g of osmium, what is its empirical formula?​

Answers

The empirical formula is OsO₄ :

Explanation:

Osmium oxide contains osmium and oxygen only.

Thus, we shall determine the mass of oxygen in osmium oxide. This can be obtained as follow:

Mass of compound = 2.89 g

Mass of Os = 2.16 g

Mass of O =?

Mass of O = (Mass of compound) – (Mass of Os)

Mass of O = 2.89 – 2.16

Mass of O = 0.73 g

Finally, we shall determine the empirical formula of the compound. This can be obtained as follow:

Mass of Os = 2.16 g

Mass of O = 0.73 g

Empirical formula =..?

Os = 2.16 g

O = 0.73 g

Divide by their molar mass of

Os = 2.16 / 190 = 0.011

O = 0.73 / 16 = 0.046

Divide by the smallest

Os = 0.011 / 0.011 = 1

O = 0.046 / 0.011 = 4

Empirical formula = OsO₄

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What volume of 1.50 mol/L stock solution is needed to make 125 mL of 0.60 mol/L solution?

Answers

Chemistry 11 Solutions

978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 8 Solutions and Their Properties • MHR | 85

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

NaCl

2.5 g

m

n

M

58.44 g

2

4.2778 10 m l

ol

o

/m

u

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

–2 4.2778 × 10 mol

0.100

0.42778 mol/L

0.43 mol

L

/L

n

c

V

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

Check Your Solution

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

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

seems reasonable.

Section 8.4 Preparing Solutions in the Laboratory

Solutions for Practice Problems

Student Edition page 386

51. Practice Problem (page 386)

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

(NH4)2SO4(aq).

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

following solutions?

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

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

c. 250 mL of 0.300 mol/L NH4

+

(aq)

What Is Required?

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

needed to prepare each given dilute solution.

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

What is dilution?

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

Given,

Initial volume = V₁

Initial molar concentration (M₁) = 1.50 mol/L

Final volume (V₂) = 125 mL = 0.125 L

Final molar concentration (M₂)= 0.60 mol/L

The dilution is calculated as:

M₁V₁ = M₂V₂

V₁ = M₂V₂ ÷ M₁

Substituting the values in the above formula as

V₁ = M₂V₂ ÷ M₁

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

V₁ = 0.05 L

= 50 mL

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

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Based on the standard EMF series and your knowledge of half-reactions, determine the cell potential and spontanei ty of a cell that consists of a pure cobalt electrode in a solution of Co^2+ ions; the other half is a lead electrode immersed in a Pb^2+ solution.
Pb +2e- Pb Sn +2e Sn Ni 2e Ni Co 2e -0.126 -0.136 -0.250 -0.277 Co
a. +0.403, spontaneous
b. -0.403, nonspontaneous
c. +0.151, spontaneous
d. -0.151, nonspontaneous

Answers

Answer:

+0.151, spontaneous

Explanation:

Given that;

Co^2+(aq) + 2e ---->Co(s) -0.28 V

Pb^2+(aq) + 2e ---->Pb(s). -0.13 V

Hence Co is the anode and Pb is the cathode

E°cell = E°cathode - E°anode

So;

E°cell = -0.13 V - (-0.28 V)

E°cell = 0.15 V

The cell reaction is spontaneous since E°cell is positive.

Determine whether each of the examples represents a colligative property or a non-colligative property. boiling point elevation Choose... color Choose... freezing point depression Choose... vapor pressure lowering Choose... density Choose...

Answers

Answer:

boiling point elevation - colligative property

color - non-colligative property

freezing point depression - colligative property

vapor pressure lowering - colligative property

density - non-colligative property

Explanation:

A colligative property is a property that depends on the number of particles present in the system.

Freezing point depression, boiling point elevation and vapour pressure lowering are all colligative properties of solutions.

Colour and density do not depend on the number of particles present hence they are not colligative properties.

The boiling point elevation, freezing point depression, and vapor pressure lowering are colligative properties. And color and density are non-colligative properties.

Explanation:

The colligative properties are the properties depending upon the number of particles of solute not on the nature of the solute.Example of colligative properties:Vapor pressure loweringElevation boiling pointDepression in freezing pointOsmotic pressureThe non-colligative properties are the properties depending upon the nature of solute and solvent.Example of non-colligative properties :ViscositySurface tensionDensitySolubility

So, from this, we can conclude that boiling point elevation, freezing point depression, and vapor pressure lowering are colligative properties. And color and density are non-colligative properties.

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Determine the number of water molecules in 0.2830g Na. ​

Answers

Answer:

7.38*10^21

Explanation:

2Na+2H20=2NaOH+H2

nNa=0.0123

number of water moles: 0.012*6*10^23=7.38*10^21

What mass of precipitate (in g) is formed when 250.0 mL of 0.150 M CuCl₂ is mixed with excess KOH in the following chemical reaction?
CuCl₂(aq) + 2 KOH(aq) → Cu(OH)₂(s) + 2 KCl(aq)

Answers

Answer:

3.6487g

CuCl2 moles reacted = (0.15×250)/1000

according to balanced chemical equation

precipitated Cu(OH)2 moles = Reacted CuCl2 moles

molar mass of Cu(OH)2 = 63.5+ (17+1)×2 = 97.5

mass of precipitate = (97.5 × 0.15×250)/1000

= 3.648g

Give four examples illustrating each of the following terms. a. homogeneous mixture b. heterogeneous mixture c. compound e. physical change d. element f. chemical change

Answers

1. homogenous: sugar solution

2. heterogeneous: sand solution

3. compound: water

4. physical change: ice melting

5. element: hydrogen

6. chemical change: burning fire

Suppose a 250.mL flask is filled with 1.7mol of H2 and 0.90mol of I2. The following reaction becomes possible:
+H2gI2g 2HIg
The equilibrium constant K for this reaction is 5.51 at the temperature of the flask.
Calculate the equilibrium molarity of I2. Round your answer to two decimal places.

Answers

Explanation:

here's the answer. I just plug the expression into my calculator and find the intercept to avoid the quadratic formula

Complete (predict the products, write correct formulas) and balance the following reaction:

Gaseous hydrochloric acid reacts with solid calcium hydroxide. Include states of matter and use subscripts as needed to receive full credit for a correct answer.

Answers

Explanation:

since hydrochloric acid is an acid and calcium hydroxide is a base, we know that this is an acid base reaction. The ions will then dissociatiate and bond with one another.

Valproic acid, used to treat seizures and bipolar disorder, is composed of C, H, and O. A 0.165-g sample is combusted to produce 0.166 g of water and 0.403 g of carbon dioxide. What is the empirical formula for valproic acid

Answers

Answer:

The empirical formula is C4H8O

Explanation:

Step 1: Data given

Valproic acid is composed of C, H, and O

Mass of the sample = 0.165 grams

Mass of water = 0.166 grams

Mass of CO2 = 0.403 grams

Molar mass of water ( H2O) = 18.02 g/mol

Molar mass of CO2 = 44.01 g/mol

Atomic mass of C = 12.01 g/mol

Atomic mass of O = 16.0 g/mol

Atomic mass H = 1.01 g/mol (H2 = 2.02 g/mol)

Step 2: The equation

CxHyOz + O2 → CO2 + H2O

Step 3: Calculate the number of carbon in the sample

The carbon comes from CO2

Mass C = (12.01 g/mol/44.01 g/mol) * 0.403 grams

Mass C = 0.110 grams

Step 4: Calculate mass of hydrogen in the sample

The hydrogen comes from H2O

Mass H = (2.02/18.02) * 0.166 grams

MAss H = 0.0186 grams

Step 5: Calculate mass of O

The mass of O in the sample = Mass of sample - mass of H - mass of C

The mass of O = 0.165 grams - 0.110 grams - 0.0186 grams

The mass of O = 0.0364 grams

Step 6: Calculate moles

Moles C = 0.110 grams / 12.01 g/mol = 0.00916 moles

Moles H = 0.0186 / 1.01 = 0.0184 moles

Moles O = 0.0364/16.0 = 0.00228 moles

Step 7: Calculate empirical formula

We divide by the smallest amount of moles

C: 0.00916/ 0.00228 = 4

H: 0.0184/0.00228 = 8

O: 0.00228/0.00228 = 1

The empirical formula is C4H8O

State whether the error introduced by each of the following problems would result in a high or a low value for the Cu recovery or would not affect the results. Explain. a. Some of the copper nitrate solution splashes out of the beaker in step 1. _____________

Answers

Answer:

Low value for copper recovery

Explanation:

The percentage recovery is obtained from;

Percent recovery = amount of substance you actually collected / amount of substance you were supposed to collect × 100

Note that the fact that some of the copper nitrate solution splashed out of the beaker means that some amount copper has been lost from the system. This loss of copper leads to a lower value of copper recovered from solution.

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