A student needed to make a 3 g/L NaCl solution. The student weighed 3 g of NaCl in a beaker and measured 1 L of water in a 1L volumetric flask that was labeled TC. The student then added the water to the beaker containing the NaCl. What errors did this student make? Describe how this can be performed properly.

Answers

Answer 1

Answer:

The answer is in the explanation.

Explanation:

A solution is defined as the homogeneous mixture of a solute (In this case, NaCl) and the solvent (water).

To prepare 1L of the solution, the student can weigh the 3g of NaCl in the volumetric flask but need to add slowly water to dissolve the NaCl (That is very soluble in water). When all NaCl is dissolved the student must transfer the solution to the 1L volumetric flask. Then, you must add more water to the beaker until "Clean" all the solute of the beaker to transfer it completely to the volumetric flask.


Related Questions

A sealed vessel initially contains 100 g of chlorine gas and 90 g of hydrogen gas. The two gases undergo reaction to form HCl. Which of the following statements is true?

a. 100 g HCl is produced
b. 190 g HCl is produced
c. less than 90 g HCl is produced
d. between 100 and 190 g of HCl is produced

Answers

Answer:

d.

Explanation:

H2 + Cl2 = 2HCL

From the equation 2g hydrogen combine with 71g of chlorine.

So 35.5 g Cl2 combines with 1g of H2

There are 100g of Cl2 so this will, by proportion, react with 100/35.5 g hydrogen.

This is 2.8 g hydrogen so the mass of HCl formed = 102.8 g.

The true statement is that d. between 100 and 190 g of HCl is produced.

How do find the mass of HCL?

To find mass of HCL:

H2 + Cl2 = 2HCL

From the equation, 2g of hydrogen combines with 71g of chlorine.

So 35.5 g Cl2 combines with 1g of H2

There are 100g of Cl2 so this will, by proportion, react with 100/35.5 g of hydrogen.

This is 2.8 g hydrogen so the mass of HCl formed = 102.8 g.

Hydrogen chloride may be formed by the direct combination of chlorine (Cl2) gas and hydrogen (H2) gas.

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Predict the reactants of this chemical reaction. That is, fill in the left side of the chemical equation. Be sure the equation you submit is balanced. (You can edit both sides of the equation to balance it, if you need to.)

______________ → BaBr2 + H2O

Answers

Answer:

Ba(OH)₂ + 2 HBr ⇒ BaBr₂ + 2 H₂O

Explanation:

We have the products of a reaction and we have to predict the reactants. Since the products are binary salt and water, this must be a neutralization reaction. In neutralizations, acids react with bases. The acid that gives place to Br⁻ is HBr, while the base the gives place to Ba²⁺ is Ba(OH)₂. The balanced chemical equation is:

Ba(OH)₂ + 2 HBr ⇒ BaBr₂ + 2 H₂O

Nitrogen monoxide and ozone react to form nitrogen dioxide and oxygen, like this:
NO(g) + O3 → NO2 + O2

The reaction is exothermic. Suppose a mixture of NO, O3 NO and O2 has come to equilibrium in a closed reaction vessel. Predict the change.

Answers

Answer:

The pressure of O2 will increase. Shift to the right.

The pressure of O3 will increase . Dhift to the left

Explanation:

The perturbations are:

The temperature is lowered. The pressure of O2 will

The temperature is raised. The pressure of O3 will:

We can Apply LeCh's principle and see the heat, ΔH, as a product of the reaction:

NO(g) + O3(g) → NO2(g) + O2(g) + ΔH

If temperature is lowered, the system will shift to the right in order to produce more heat doing:

The pressure of O2 will increase

In the other way, if temperature is raised, the system will shift to the left in order to decrease the amount of heat produced.

The pressure of O3 will increase

What volume (mL) of the sweetened tea described in Example 1 contains the same amount of sugar (mol) as 10 mL of the soft drink in this example

Answers

The question is incomplete, the complete question is:

What volume (mL) of the sweetened tea described in Example 3.14 contains the same amount of sugar (mol) as 10 mL of the soft drink in this example. The example is attached below.

Answer: 75 mL of sweetened tea will contain the same amount of sugar as in 10 mL of soft drink

Explanation:

We first calculate the number of moles of soft drink in a volume of 10 mL

The formula used to calculate molarity:

[tex]\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}}[/tex] .....(1)

Taking the concentration of soft drink from the example be = 0.375 M

Volume of solution = 10 mL

Putting values in equation 1, we get:

[tex]0.375=\frac{\text{Moles of sugar in soft drink}\times 1000}{10}\\\\\text{Moles of sugar in soft drink}=\frac{0.375\times 10}{1000}=0.00375mol[/tex]

Calculating volume of sweetened tea:

Moles of sugar = 0.00375 mol

Molarity of sweetened tea = 0.05 M

Putting values in equation 1, we get:

[tex]0.05=\frac{0.00375\times 1000}{\text{Volume of sweetened tea}}\\\\\text{Volume of sweetened tea}=\frac{0.00375\times 1000}{0.05}=75mL[/tex]

Hence, 75 mL of sweetened tea will contain the same amount of sugar as in 10 mL of soft drink

Draw a Lewis structure for thiocyanic acid, HSCN, adding charges and lone electron pairs to the appropriate atoms.

Answers

Answer:

See explanation and image attached

Explanation:

Thiocyanic acid is made made up of hydrogen, sulphur, carbon and nitrogen atoms. Carbon is the central atom in the molecule.

The molecule has a total of sixteen valence electrons as shown in the image attached. There are no formal charges in the structure of the molecule as shown.

The molecule is linear in shape.

The products obtained from hydroboration-oxidation of cis-2-butene are identical to the products obtained from hydroboration-oxidation of trans-2-butene. Draw the products and explain why the configuration of the starting alkene is not relevant in this case.

Answers

Answer:

a) Attached below

b)  The presence of racemic mixture found as product in both  cases shows that   products are identical ( i.e. they have same configuration

Explanation:

Diagrams of the products obtained from hydroboration-oxidation of cis-2-butene , hydroboration-oxidation of trans-2-butene.

attached below

The presence of racemic mixture found as product in both  cases shows that   products are identical ( i.e. they have same configuration )

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  

What mass of NaOH would need to be dissolved in 500.0 mL of water to produce a solution with a pH of 12.40

Answers

Answer:

0.5024 g

Explanation:

Step 1: Calculate the concentration of H⁺

We will use the definition of pH.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -12.40 = 3.981 × 10⁻¹³ M

Step 2: Calculate the concentration of OH⁻

We will use the ionic product of water expression.

[H⁺] [OH⁻] = 10⁻¹⁴

[OH⁻] = 10⁻¹⁴/[H⁺] = 10⁻¹⁴/3.981 × 10⁻¹³ = 0.02512 M

Step 3: Calculate the initial concentration of NaOH

NaOH is a strong base and the molar ratio of NaOH to OH⁻is 1:1. Thus, the initial concentration of NaOH is 1/1 × 0.02512 M = 0.02512 M.

Step 4: Calculate the moles of NaOH

We will use the definition of molarity.

M = moles of NaOH/liters of solution

moles of NaOH = M × liters of solution

moles of NaOH = 0.02512 mol/L × 0.5000 L = 0.01256 mol

Step 5: Calculate the mass of 0.01256 moles of NaOH

The molar mass of NaOH is 40.00 g/mol.

0.01256 mol × 40.00 g/mol = 0.5024 g

if a=1/2(a+b)h,express a in terms of A,b and h. pls solve with step by step ​

Answers

Answer:

[tex] a = \frac {2A - bh}{h} [/tex]

Explanation:

Given the following mathematical expression;

A = ½(a + b)h

To make a the subject of formula (express a in terms of A, b and h);

First of all, we would cross-multiply;

2A = (a + b)h

Opening the bracket, we have;

2A = ah + bh

Rearranging the mathematical expression, we have;

ah = 2A - bh

[tex] a = \frac {2A - bh}{h} [/tex]

Please help help please

Answers

Answer:  The correct answer is B.

Explanation:  Segregate most organic acids from oxidizing mineral acids. Keep oxidizers away from other chemicals, especially flammables.

Answer:

Segregate most organic acids from oxidizing mineral acids. Keep oxidizers away from other chemicals, especially flammables, combustibles, and toxic materials. Keep corrosives away from substances that they may react with and release corrosive, toxic, or flammable vapors.

Calculate the volume in liters of a 1.60 mol/L sodium nitrate solution that contains of sodium nitrate . Round your answer to significant digits.

Answers

Answer:

1.5L of NaNO3 must be present

Explanation:

That contains 200g of sodium nitrate. Round to 2 significant digits

To solve this question we need to convert the mass of NaNO3 to moles using its molar mass (85g/mol). With the moles and the molar concentration we can find the volume in liters of the solution:

Moles NaNO3:

200g * (1mol / 85g) = 2.353 moles NaNO3

Volume:

2.353 moles NaNO3 * (1L / 1.60moles) =

1.5L of NaNO3 must be present

There are four containers: a 100-mL beaker, 250-mL Erlenmeyer flask, a 500-mL beaker, and a 1-L Florence flask. They contain coffee, tea, water, and milk, although not in that order. Use the following facts to identify the beverage in each container.

a. the 500-mL container has a beverage commonly associated with breakfast.
b. the largest container has a colorless liquid (i.e. neither yellow nor orange).
c. the beverage in the smallest container is opaque. (you cannot see through it).
d. One clear liquid is in a container half the volume of a colored liquid.
e. The only combustible liquid has exactly twice the volume of an opaque liquid.

Answers

Answer:

There are four containers: a 100-mL beaker, 250-mL Erlenmeyer flask, a 500-mL beaker, and a 1-L Florence flask. They contain coffee, tea, water, and milk, although not in that order. Use the following facts to identify the beverage in each container.

a. the 500-mL container has a beverage commonly associated with breakfast.

b. the largest container has a colorless liquid (i.e. neither yellow nor orange).

c. the beverage in the smallest container is opaque. (you cannot see through it).

d. One clear liquid is in a container half the volume of a colored liquid.

e. The only combustible liquid has exactly twice the volume of an opaque liquid.

Explanation:

a. The 500-mL container has a beverage commonly associated with breakfast is coffee.

b. The largest container has a colorless liquid (i.e. neither yellow nor orange) water.

c. The beverage in the smallest container is opaque. (you cannot see through it) milk.

d. One clear liquid is in a container half the volume of a colored liquid tea.

The 500-mL container has a beverage commonly associated with breakfast is coffee. (rest answers are as follows)

How to indentify beverages ?

The Indentification of the beverages can be done by knowing the content and optical activity that uniquely identify the container.

The 500-mL container has a beverage commonly associated with breakfast is coffee.The largest container has a colorless liquid (i.e. neither yellow nor orange) water.The beverage in the smallest container is opaque. (you cannot see through it) milk.One clear liquid is in a container half the volume of a colored liquid tea.

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Give the number of lone pairs around the central atom and the molecular geometry of SCl2. Multiple Choice 3 lone pairs, linear 1 lone pair, bent 3 lone pairs, bent 0 lone pairs, linear 2 lone pairs, bent

Answers

Answer:

2 lone pairs, bent

Explanation:

According to the Valence Shell Electron Pair Repulsion Theory, the number of electron pairs on the valence shell of the central atom in a molecule influences the shape of the molecule.

The presence of lone pairs on the valence shell of the central atom causes the observed molecular geometry to deviate from the ideal geometry predicted on the basis of the valence shell electron pair repulsion theory.

SCl2 has four regions of electron density. This means that its electron domain geometry is tetrahedral. However, there are two lone pairs on the valence shell of the central atom hence the observed molecular geometry is bent.

what following oxide react with both acids and bases to form salts.​

Answers

Answer:

mainly metal oxide use to react with both acid and bases to form salts such as zinc, aluminum etc.

What effect does hybridization have on chemical bond​

Answers

Effect of Hybridisation on Single, Double and Triple Bond Lengths of Carbon. Since the sp hybrid orbital contains more s-character (50%), it is closer to its nucleus; therefore, it forms shorter bonds. Because of the same reason sp2 hybrid orbital forms shorter bonds than sp3 hybrid orbitals.

What is the
energy
2) The energy transition from n = 1 →n= 3 in hydrogen is 12.09 eV (1.6022 X 10J= 1 eV)-
of light emitted from n = 3 →n=1?
E e fiz

Answers

Answer:

energy is the ability to do work

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:

option a

hope helps you

have a great day

A gas mixture is made by combining 8.7 g each of Ar, Ne, and an unknown diatomic gas. At STP, the mixture occupies a volume of 17.28 L. What is the molar mass of the unknown gas

Answers

Answer: Molar mass of the unknown gas is 73.153 g/mol.

Explanation:

Given: Mass of each gas = 8.7 g

Volume = 17.28 L

Let us assume that the molar mass of gas is m g/mol.

Molar mass of Ar is 40 g/mol and Ne is 20 g/mol.

Hence, total moles of each gas are as follows.

[tex](\frac{8.7}{40} + \frac{8.7}{20} + \frac{8.7}{m}) mol[/tex]

At STP, the total volume of these gases is as follows.

[tex](\frac{8.7}{40} + \frac{8.7}{20} + \frac{8.7}{m}) mol \times 22.4 L = 17.28 L\\(\frac{8.7}{40} + \frac{8.7}{20})22.4 L + \frac{8.7}{m} \times 22.4 L = 17.28 L\\14.616 + \frac{8.7}{m} \times 22.4 L = 17.28 L\\\frac{8.7}{m} \times 22.4 L = (17.28 L - 14.616)\\\frac{8.7}{m} \times 22.4 L = 2.664 \\m = 73.153 g/mol[/tex]

Thus, we can conclude that molar mass of the unknown gas is 73.153 g/mol.

If mercury barometer is replaced by water barometer, height of water column

i. will be less than that of Hg Column
ii. will be greater than that of Hg column iii. will be equal to that of Hg column
iv. will be none of these​

Answers

Answer:

answer is first one 1 will be less then that of hg coloumn

Calculate the average atomic mass for X

Answers

Answer:

39.0229 amu

Explanation:

Hello there!

In this case, according to given information, the idea here is to multiply the percent abundance by the mass number of each isotope and then add them all together as shown below:

[tex]=0.0967*38+0.7868*39+0.1134*40+0.0031*41\\\\=3.6746+30.6852+4.536+0.1271\\\\=39.0229amu[/tex]

Regards!

Question 16(Multiple Choice Worth 5 points)

(04.01 LC) Which statement is true about the total mass of the reactants during a chemical change?

O It is destroyed during chemical reaction.
O It is less than the total mass of the products. O It is equal to the total mass of the products.
O It is greater than the total mass of the products.​

Answers

Answer:

It is equal to the total mass of the products.

Explanation:

Hope this helps :)

If mass of an empty 9.4 mL pycnometer is 10.3 and the mass of the same pycnometer with an unknown liquid is 20.1. Determine the density of the unknown liquid to the correct number of significant figures in g/mL

Answers

Answer:

1.04 g/mL

Explanation:

Applying,

D = (m-m')/V................. Equation 1

Where D = Density of the unknown liquid, m = mass of the pycnometer when filled with unkwon liquid, m' = mass of the empty pycnometer, V = volume of the empty pycnometer

From the question,

Assuming the mass are in grams

Given: m = 20.1 g, m' = 10.3 g, V = 9.4 mL

Substitute these values into equation 1

D = (20.1-10.3)/9.4

D = 9.8/9.4

D = 1.04 g/mL

The time required to pass one Faraday of electricity through a solution with a current of 0.3A is​

Answers

Answer:

89.35 hour

Explanation:

Recall :

Charge on 1 electron = 1.6 × 10^-19 C

1 mole contains = 6.023 × 10^23

Therefore, the charge on 1 mole of electron will be :

Charge per electron × 1 mole :

(1.6 × 10^-19) * (6.023 * 10^23) = 96500 C = 1 Farad

1 Farad = 96500 C

Using the formula :

Q = Current(I) * time(t)

Q = I*t

t = Q/I

Current, I = 0.3 A

t = 96500 / 0.3

t = 321666.66 second

t = 321666.66 / 3600 = 89.35 hour

What should be done if a spectrophotometer reports an absorbance that is too high? Select one: Reposition the cuvette in the spectrophotometer. Pour out half the volume of the sample. Restart the spectrophotometer and try again. Dilute the sample. g

Answers

Answer:

The sample should be diluted

Explanation:

According to Beer Lambert's law, the absorbance of a sample depends on the concentration of the sample.

Hence, if the concentration of the sample is very high, the spectrophotometer will also report a very high value of absorbance.

When this is the case, the sample should simply be diluted and the readings are taken again using the spectrophotometer.

how many lone pair electrons are on the central oxygen atom in the Lewis structure for dinitrogen pentoxide 9

Answers

Answer:

Two

Explanation:

Lone pairs are electron pairs on an atom that resides only with one of the atoms in a molecule.

Dinitrogen pentaoxide is shown in the image attached. There are five oxygen atoms and two nitrogen atoms in the molecule. The molecule has a total of 40 valence electrons.

There are two electrons present on the central oxygen atom in the Lewis structure of dinitrogen pentaoxide as shown in the image attached.

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

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 mixture is -
A. a pure substance made through chemical processes
B. a pure substance made through physical processes
C.
an impure substance made through chemical processes
D.
an impure substance made through physical processes

Answers

Answer:

C

Explanation:

an impure substance made through chemical process

A chemist combines 33 g of methane with 289 g of oxygen to from 189 g of carbon dioxide and 30 g of water. Determine if the results of the following word problem adheres to the Law of Conservation of Mass.

Answers

Answer:

The correct answer is - no not adhere to the law of mass conservation.

Explanation:

According to the law of mass conservation in an isolated system, the mass can not be created or destroyed and in a chemical or physical change, the mass of products should be always equal to the mass of reactants.

On the basis of the law the mass of the chemical reaction-

Mass of products = mass of reactants

33 g of methane + 289g of oxygen = 189g of carbon dioxide + 30g of water

322g ≠ 219 g

which means this reaction does not adhere to the law of conservation.

A cylinder contains 3.1 L of oxygen at 300 K and 2.7 atm. The gas is heated, causing a piston in the cylinder to move outward. The heating causes the temperature to rise to 610 K and the volume of the cylinder to increase to 9.4 L.
How many moles of gas are in the cylinder?
Express your answer using two significant figures.

Answers

Answer: The moles of gas present in the cylinder is 0.34 moles.

Explanation:

Given: [tex]P_{1}[/tex] = 2.7 atm,   [tex]V_{1}[/tex] = 3.1 L,     [tex]T_{1}[/tex] = 300 K

[tex]P_{2}[/tex] = ?,      [tex]V_{2}[/tex] = 9.4 L,       [tex]T_{2}[/tex] = 610 K

Formula used to calculate the final temperature is as follows.

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

Substitute the values into above formula as follows.

[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{2.7 atm \times 3.1 L}{300 K} = \frac{P_{2} \times 9.4 L}{610 K}\\P_{2} = \frac{5105.7}{2820} atm\\= 1.81 atm[/tex]

Now, moles present upon heating the cylinder are as follows.

[tex]P_{2}V_{2} = n_{2}RT_{2}\\1.81 atm \times 9.4 L = n_{2} \times 0.0821 L atm/mol K \times 610 K\\n_{2} = \frac{17.014}{50.081} mol\\= 0.34 mol[/tex]

Thus, we can conclude that moles of gas present in the cylinder is 0.34 moles.

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