Which is used to measure conc. HCl for preparation of 0.1M HCl solution?

a. Volumetric flask c. Measuring cylinder
b. Pipette d. Wash bottle​

Answers

Answer 1

Answer:

option a

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Related Questions

Help pls!!!
For groups 13 through 18, the number of valence electrons is equal to the group number

a. minus 10
b. minus the period number
c. plus 1
d. plus the period number ​

Answers

Answer:

a. minus 10

Explanation:

An element in group 13 = Boron ,valence electrons = 3 , therefore, valence electrons in group 13 = group no. -10

An element in group 18 = Neon, valence electrons = 8 , therefore, valence electrons in group 18 = group no. - 10

For groups 13 through 18, the number of valence electrons in an atom is equal to the group number minus 10. Therefore, option (A) is correct.

What is a valence electron?

Valence electrons in an atom can be described as the electrons occupying the outer most electron shell of an atom while the electrons in the inner shell are core electrons. Lewis structures can be helpful to calculate the number of valence electrons.

Valence electrons can be filled in several electron shells as they are caused interaction between atoms and responsible for the formation of chemical bonds. Only valence electrons can contribute to the formation of a chemical bond and decide the reactivity of the element.

The general electronic configuration of group 13 is ns²np¹ has three valence electrons. It can be determined as group number - 10 = 13 - 10 = 3.

The general electronic configuration of group 18 is ns²np⁶ has eight valence electrons. It can be determined as group number - 10 = 18 - 10 = 8.

Learn more about valence electrons, here:

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3)O que são políticas públicas?​

Answers

Answer:

azertyuiopazertyuiiop

The volume of a single tantalum atom is 1.20×10-23 cm3. What is the volume of a tantalum atom in microliters?

Answers

Answer:

1.20x10⁻²⁰μL

Explanation:

1cm³ is equal to 1milliliter. As we must know, 1milliliter = 1000 microliters, 1000μL. To convert the 1.20x10⁻²³mL we need to use the conversion factor: 1mL = 1000μL.

The volume of tantalum in μL is:

1.20x10⁻²³mL * (1000μL /1L) = 1.20x10⁻²⁰μL

How many molecules are present in 1296 g of dinitrogen pentoxide (N2O5)

Answers

Answer:

The molar mass of N2O5 is 108 g/mol. 1296g of N2O5 has 1296 / 108 = 12 moles. 1 mole contains 6.022 x 10^23 molecules

Name two natural sources of esters.​

Answers

Answer:

"Esters are widespread in nature and are widely used in industry. In nature, fats are in general triesters derived from glycerol and fatty acids. Esters are responsible for the aroma of many fruits, including apples, durians, pears, bananas, pineapples, and strawberries."

Explanation:

Name the following aldehyde PLEASE PLEASE HELP

Answers

Answer:

Explanation:

Answer is D 2,5-dimethylheptanal

You should accern the lowest possible number close to the parent name

I don’t want to fail help
I need correct answer if u don’t know I will report

When the researcher compiled information which research method did they most likely utilize?

a) documentary
b) survey
c) participant observation
d) case study

Answers

Answer:

a

Explanation:

documentary is best researcher!.


Consider the synthesis of water as shown in Model 3. A container is filled with 10,0 g of H, and
5.0 g of Oz
Which reactant (hydrogen or oxygen) is the limiting reactant in this case?

Answers

Answer:

Oxygen, O₂ is the limiting reactant

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2H₂ + O₂ —> 2H₂O

Next, we shall determine the masses of H₂ and O₂ that reacted from the balanced equation. This can be obtained as follow:

Molar mass of H₂ = 2 × 1 = 2 g/mol

Mass of H₂ from the balanced equation = 2 × 2 = 4 g

Molar mass of O₂ = 16 × 2 = 32 g/mol

Mass of O₂O from the balanced equation = 1 × 32 = 32 g

SUMMARY:

From the balanced equation above,

4 g of H₂ reacted with 32 g of O₂.

Finally, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

4 g of H₂ reacted with 32 g of O₂.

Therefore, 10 g of H₂ will react with

= (10 × 32)/4 = 80 g of O₂.

From the calculations made above, we can see that a higher mass (i.e 80 g) of O₂ than what was given (i.e 5 g) is required to react completely with 10 g of H₂. Therefore, O₂ is the limiting reactant.

Oxygen has been the limiting reactant in the reaction.

A limiting reactant can be defined as the reactant in the reaction in which the product concentration has been dependent.

The balanced equation for the formation of water has been:

[tex]\rm 2\;H_2\;+\;O_2\;\rightarrow\;2\;H_2O[/tex]

For the formation of reaction to form 2 moles of water, 2 moles of hydrogen reacts with 1 mole of oxygen.

The moles can be calculated as:

Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]

The moles of Hydrogen in 10 g [tex]\rm H_2[/tex]:

Moles = [tex]\rm \dfrac{10}{2}[/tex]

Moles of hydrogen = 5 mol.

Moles of Oxygen in 5 grams Oxygen:

Moles = [tex]\rm \dfrac{5}{32}[/tex]

Moles of oxygen = 0.156 mol.

For the reaction with 2 moles of Hydrogen 1 mole of Oxygen has been required.

For reacting with 5 mol of Hydrogen, moles of oxygen required are:

Moles of oxygen = [tex]\rm \dfrac{1}{2}\;\times\;5[/tex]

Moles of oxygen required = 2.5 moles.

The available oxygen = 0.156 moles.

Since the moles of oxygen available is lesser than required, the formation of the product has been dependent on the concentration of the oxygen.

Thus, oxygen has been the limiting reactant in the reaction.

For more information about the limiting reactant, refer to the link:

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We can use bond-line formulas to represent alkenes in much the same way that we use them to represent alkanes. Consider the following alkene: h5ch5e4 How many carbon atoms are sp2−hybridized in this alkene?

Answers

Answer:

2

Explanation:

The number of carbon atoms that are sp²-hybridized in this alkene is 2

Because all the single bonded carbon atoms in the alkene are  sp²-hybridized

There are three(3) single formed via sp² orbitals and one ( 1 ) PI bond formed via Pure-P-orbital

attached below is the some part of the solution

g Arrange the following compounds in order of acidity (highest to lowest): H2O, H3O , HCl A. CH3COOH > HCl > H2O B. H2O > CH3COOH > HCl C. HCl > H2O > CH3COOH D. HCl > CH3COOH > H2O

Answers

Answer:

Arrange the following compounds in order of acidity (highest to lowest): H2O, CH3COOH , HCl

A. CH3COOH > HCl > H2O

B. H2O > CH3COOH > HCl

C. HCl > H2O > CH3COOH

D. HCl > CH3COOH > H2O

Explanation:

The given substances are acetic acid, hydrochloric acid, and water.

Since HCl is a strong acid and it undergoes complete ionization.

CH3COOH acetic acid is a weak acid and it undergoes partial dissociation in water.

Pure water is a neutral substance.

Hence, the order of acidity is shown below:

HCl > CH3COOH > H2O.

Among the given options, option D is the correct answer.

15. You are interested in separating 4-methylbenzoic acid from 1,4-dimethoxybenzene using a procedure similar to the extraction procedure we used in lab. You plan to use sodium bicarbonate instead of sodium hydroxide. a) Show the reaction between salicylic acid and sodium bicarbonate. Label the acid, base, conjugate acid, conjugate base. b) Give the pKa values of the acid and conjugate acid. c) Which base will work better, sodium hydroxide or sodium bicarbonate

Answers

Solution :

a). The separation of 4-methylbenzoic acid from 1,4-dimethoxybenzene will work but it will result in lower recovery.

In the reaction of acid-base to form a sodium 4 - methoxy benzoate, that is soluble in the water, 4-methoxy benzoic acid reacts with the sodium bicarbonate to give sodium 4-methoxybenzoate as well as carbonic acid.

b). The pKa for the 4-methoxybenzoic acid is [tex]4.46[/tex], and that of carbonic acid is [tex]6.37[/tex]

c). The Keq for the reaction is [tex]10(6.37 - 4.46) = 101.91[/tex]

Therefore, the equilibrium lies to the right  and also the reaction favors the products and the separation works.

But the recovery will be low when compared to the extraction with Sodium hydroxide as the strong base will drive the equilibrium further to the right position, thus neutralizing all the acids virtually. And the weak base will partially neutralize the acid.

Under certain conditions, the substance mercury(II) oxide can be broken down to form mercury and oxygen. If 32.2 grams of mercury(II) oxide react to form 29.8 grams of mercury, how many grams of oxygen must simultaneously be formed

Answers

Explanation:

This is a decomposition reaction. Firstly, you will want to write the chemical equation out and balance it.

[tex]2Hg_2O->4Hg+O_2[/tex] (The -> is supposed to be an arrow, sorry!)

We see that there's only 1mol of Oxygen made in the products, we can do some simple math to solve for the amount of grams of Oxygen produced according to the amount of the reactant (Hg2O).

[tex]32.2gHg_2O*\frac{1molHg_2O}{417.18gHg_2O}*\frac{1molO_2}{2molHg_2O}*\frac{32gO_2}{1molO_2}[/tex]

I want to break this down, just in case:

The 417.18gHg2O is the molecular mass of the molecule (so I doubled Hg and added 16 to it to get this number).

As we can see in the chemical equation, 1mol Hg2O produces 2mol O because Oxygen is a diatomic molecule (so there will always be two of it when it's by itself).

And finally, in 1mol O2 there are 32g of O2.

** When you do math like this, always make sure that all of your units cancel out except for the units you're looking for. For example, here we're looking for the grams of Oxygen, so after everything else cancels out, we should only have grams O2.

So, 1.23gO2 should be your answer.

Calculate the mass of water produced when 1.57g of butane reacts with excess oxygen

Answers

Explanation:

So, first you will want to write the balanced chemical equation for this reaction.

Butane = [tex]C_4H_{10}[/tex]

[tex]2C_4H_{10}+13O_2=>10H_2O+8CO_2[/tex]

^ This ends up being your balanced chemical equation. Now, you can do the math!

[tex]1.57gC_4H_{10}*\frac{1molC_4H_{10}}{58.12gC_4H_{10}}*\frac{10molH_2O}{2molC_4H_{10}}*\frac{18gH_2O}{1molH_2O}[/tex]

After plugging this into a calculator, your final mass of water should be:

2.43gH2O

Suppose you analyze a 30.3 g sample of bleach and determine that there are 2.61 g of sodium hypochlorite present. What is the percent of sodium hypochlorite in the bleach sample

Answers

Answer:

8.61 %

Explanation:

The percent of sodium hypochlorite in the bleach sample can be calculated using the following formula:

% of sodium hypochlorite = Mass of sodium hypochlorite / mass of sample * 100%

We input the data given by the problem:

% of sodium hypochlorite = 2.61 g / 30.3 g * 100 % = 8.61 %

atomaticity of chlorine 1) 2, 2)1, 3) 32 , 4) 4.​

Answers

Answer:

ATOMICITY OF CHLORINE IS 2

Explanation:

Atomicity is defined as the total number of atoms present in a molecule.

A gas bottle contains 0.650 mol of gas at 730. mmHg pressure. If the final pressure is 1.15 atm, how many moles of gas were added to the bottle

Answers

Answer: There are 0.779 moles of gas were added to the bottle.

Explanation:

Given: [tex]n_{1}[/tex] = 0.650 mol,     [tex]P_{1}[/tex] = 730 mm Hg (1 mm Hg = 0.00131579 atm) = 0.96 atm

[tex]n_{2}[/tex] = ?,           [tex]P_{2}[/tex] = 1.15 atm

Formula used is as follows.

[tex]\frac{P_{1}}{n_{1}} = \frac{P_{2}}{n_{2}}[/tex]

Substitute the values into above formula as follows.

[tex]\frac{P_{1}}{n_{1}} = \frac{P_{2}}{n_{2}}\\\frac{0.96 atm}{0.650 mol} = \frac{1.15 atm}{n_{2}}\\n_{2} = 0.779 mol[/tex]

Thus, we can conclude that there are 0.779 moles of gas were added to the bottle.

A sample of calcium fluoride was decomposed into the constituent elements. Write a balanced chemical equation for the decomposition reaction. If the sample produced 294 mg of calcium, how many g of fluorine were formed

Answers

Answer:

A sample of calcium fluoride was decomposed into the constituent elements. Write a balanced chemical equation for the decomposition reaction. If the sample produced 294 mg of calcium, how many g of fluorine was formed

Explanation:

The balanced chemical equation for the decomposition of calcium fluoride is shown below:

[tex]CaF_2(s)->Ca(s)+F_2(g)[/tex]

The sample produced 294 g of calcium then, how many grams of fluorine is formed?

From the balanced chemical equation,

1 mol of CaF2 forms 1mol of calcium and 1 mol of fluorine.

That is:

40g of calcium and 38.0 g of fluorine are formed.

then,

If 294 g of calcium is formed then how many grams of fluorine is formed?

[tex]294g Ca * 38g F2 / 40g Ca\\=279.3 g F_2[/tex]

Hence, 279.3 g of fluorine will be formed.

If a student drops 2.3g pieces of magnesium into a flask of hydrochloric acid, this reaction occurs: Mg + 2HCl= MgCl2 + H2

How many liters of hydrogen can be produced at a pressure of 2 atm and a temperature of 298 K

Answers

Answer:

1.2 L

Explanation:

Step 1: Write the balanced equation

Mg + 2 HCl ⇒ MgCl₂ + H₂

Step 2: Calculate the moles corresponding to 2.3 g of Mg

The molar mass of Mg is 24.31 g/mol.

2.3 g × 1 mol/24.31 g = 0.095 mol

Step 3: Calculate the moles of H₂ produced

0.095 mol Mg × 1 mol H₂/1 mol Mg = 0.095 mol H₂

Step 4: Calculate the volume occupied by the hydrogen

We will use the ideal gas equation.

P × V = n × R × T

V = n × R × T/P

V = 0.095 mol × (0.0821 atm.L/mol.K) × 298 K/2 atm = 1.2 L

What quantity of sodium azide in grams is required to fill a 56.0 liters air bag with nitrogen gas at 1.00 atm and exactly 0 °C:
2 NaN3 is) 2Na (s) + 3N2 (8)

Answers

Answer:

108.6 g

Explanation:

2NaN₃(s) → 2Na(s) + 3N₂(g)

First we use the PV=nRT formula to calculate the number of nitrogen moles:

P = 1.00 atmV = 56.0 Ln = ?R = 0.082 atm·L·mol⁻¹·K⁻¹T = 0 °C ⇒ 0 + 273.2 = 273.2 K

Inputting the data:

1.00 atm * 56.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.2 Kn = 2.5 mol

Then we convert 2.5 moles of N₂ into moles of NaN₃, using the stoichiometric coefficients of the balanced reaction:

2.5 mol N₂ * [tex]\frac{2molNaN_3}{3molN_2}[/tex] = 1.67 mol NaN₃

Finally we convert 1.67 moles of NaN₃ into grams, using its molar mass:

1.67 mol * 65 g/mol = 108.6 g

Wine goes bad soon after opening because the ethanol dissolved in it reacts with oxygen gas to form water and aqueous acetic acid , the main ingredient in vinegar. Calculate the moles of acetic acid produced by the reaction of of ethanol. Be sure your answer has a unit symbol, if necessary, and round it to significant digits.

Answers

Answer:

The answer is "It takes 1,70 mol of ethanol".

Explanation:

To make acetic acid, we must first write the balanced reaction that occurs of ethanol with oxygen

The response is balanced:

[tex]CH_3CH_2OH+O_2\to CH_3COOH+H_2O[/tex]

1 mol of ethanol creates 1 mol of According the equilibrium Ethanol moles, therefore, required 1.70 mol of water = 1.70 mol

Calculate the mass of butane needed to produce 80.0g of carbon dioxide

Answers

Answer:

Multiply the number of moles of butane by its molar mass, 58.12g/mol, to produce the mass of butane. Mass of butane = 18.8g.

What effect does a high carbon level have on a deep ocean

Answers

Explanation:

High carbon concentration in the deep ocean means increased absorption of carbon to the atmosphere resulting to even greater and harmful amounts of carbon in the atmosphere. Therefore we need to keep a close eye of the deep ocean in the quest to monitor and pump out excess carbon from this part of marine life.

Calculate the Experimental Molar Volume in L/mol of the Hydrogen gas, H2, if the volume of H2 at STP is 52.8 mL and the mass of Magnesium metal, Mg, used in the experiment is 0.055 g.

Answers

Answer:

The Experimental Molar Volume in L/mol of the Hydrogen gas=23.36L/mol

Explanation:

We are given that

Volume of H2 at STP=52.8mL

Mass of magnesium metal ,M(Mg)=0.055g

We have to find the Experimental Molar Volume in L/mol of the Hydrogen gas.

Molar mass of Mg=24.305 g/mol

Number of moles=[tex]\frac{given\;mass}{molar\;mass}[/tex]

Using the formula

Number of moles of Mg=[tex]\frac{0.055}{24.305}[/tex]moles

Number of moles of Mg=0.00226moles

Number of moles of Mg=Number of moles of H2

Number of moles of H2=0.00226moles

Molar volume of Hydrogen gas (H2)=[tex]\frac{volume\;at\;STP}{No\;of\;moles\;H_2}[/tex]

Molar volume of Hydrogen gas (H2)=[tex]\frac{52.8}{0.00226}mL/mol[/tex]

Molar volume of Hydrogen gas (H2)=[tex]\frac{52.8}{0.00226}\times 10^{-3}L/mol[/tex]

[tex]1L=1000mL[/tex]

Molar volume of Hydrogen gas (H2)=23.36L/mol

Hence, the Experimental Molar Volume in L/mol of the Hydrogen gas=23.36L/mol

g 0.500 L of a solution with a concentration of 0.25 M is needed. To prepare this solution, a stock solution with a concentration of 1.25 M is prepared. What volume of the stock solution is needed to create the desired solution

Answers

Answer:

0.1 L

Explanation:

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

Concentration of stock solution (C₁) = 1.25 M

Volume of diluted solution (V₂) = 0.5 L

Concentration of diluted solution (C₂) = 0.25 M

Volume of stock solution needed solution (V₁) =?

The volume of the stock solution needed can be obtained as follow:

C₁V₁ = C₂V₂

1.25 × V₁ = 0.25 × 0.5

1.25 × V₁ = 0.125

Divide both side by 1.25

V₁ = 0.125 / 1.25

V₁ = 0.1 L

Therefore, the volume of the stock solution needed is 0.1 L

Given the amount of camphor (200mg) we are using in this experiment, please determine how many mg of sodium borohydride to use in this reaction. We would like you to use 5.2 molar equivalents of this reagent. This means 5.2 times the mmol of camphor we are using. As an example: for 110.0 mg of camphor,142 mg of NaBH4 would be used (see if you can confirm this result). For complete credit, your work needs to be clearly drawn out!

Answers

Answer:

Explanation:

From the given information:

Camphor may be reduced as readily in the presence of sodium borohydride(NaHB4). The resulting compound which is stereoselective requires 1 mole of sodium borohydride (NaHB4) to reduce 1 mole of camphor in this reaction. The reaction is shown below.

Through the reduction process of camphor, the reducing agent can reach the carbonyl face with a one-carbon linkage. The product stereoisomer is known as borneol.

If the molecular weight of camphor = 152.24 g/mol

and it mass = 200 mg

The its no of moles = 200 mg/ 152.24 g/mol

= 1.3137 mmol

Now the amount of the required mmol for NaBH4 to be consumed in the reaction = 5.2 × 1.3137 mmol

= 6.831 mmol

since the molar mass of NaBH4 = 37.83 g/mol

Then, using the same formula:

No of moles = mass/molar mass

mass = No of moles × molar mass

mass = 6.831 mmol × 37.83 g/mol

mass of NaBH4 used = 258.42 mg  

A sample of 10.6 g of KNO3 was dissolved in 251.0 g of water at 25 oC in a calorimeter. The final temperature of the solution was 21.5 oC. What is the molar heat of solution of KNO3

Answers

Answer:

36.55kJ/mol

Explanation:

The heat of solution is the change in heat when the KNO3 dissolves in water:

KNO3(aq) → K+(aq) + NO3-(aq)

As the temperature decreases, the reaction is endothermic and the molar heat of solution is positive.

To solve the molar heat we need to find the moles of KNO3 dissolved and the change in heat as follows:

Moles KNO3 -Molar mass: 101.1032g/mol-

10.6g * (1mol/101.1032g) = 0.1048 moles KNO3

Change in heat:

q = m*S*ΔT

Where q is heat in J,

m is the mass of the solution: 10.6g + 251.0g = 261.6g

S is specififc heat of solution: 4.184J/g°C -Assuming is the same than pure water-

And ΔT is change in temperature: 25°C - 21.5°C = 3.5°C

q = 261.6g*4.184J/g°C*3.5°C

q = 3830.87J

Molar heat of solution:

3830.87J/0.1048 moles KNO3 =

36554J/mol =

36.55kJ/mol

compound of aspartame is a dipeptide that is often used as a sugar substitute which functional groups are present

Answers

Answer:

Carboxyl, primary amine, amide, ester, and phenyl.

Explanation:

The functional groups present in the compound of aspartame are carboxyl, primary amine, amide, ester, and phenyl. Aspartame is an artificial non-saccharide sweetener which is 200 times sweeter than sucrose. This aspartame is commonly used as a sugar substitute in many foods and beverages. It has the trade names such as NutraSweet, Equal, and Canderel.

g Calculate the percent yield when you start with 0.50 grams of salicylic acidand end with 0.33 grams of aspirin. The molecular weight of salicylic acid is 138.12 g/mol, and the molecular weight of aspirin is 180.157 g/mol.

Answers

Explanation:

hope the picture above helps you to understand:)

If the volume of the gas is increased to 9.6 L , what will the pressure be?

Answers

The pressure will be 438 mm Hg

How many molecules of C 2H 5Br will be present if you had 4.52 g of this compound?

Answers

okay ty what what happened to me lol lolhope this helps i lol lol oh yeah that’s okay lol lol yeah lol lol !!! hope it helps :)
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