4. Each time that you prepare a diluted bleach solution you will want it to have a [OH-] = 0.02. Calculate how much 1.00 M NaOH you need to add to d.1. water to get every 20 mL of this concentration of OH- ion. Express your answer in drops of 1.00 M NaOH. (Assume that there are 20 drops in 1 m.)

Answers

Answer 1

Answer:

8 drops of 1.00 M NaOH will be needed.

Explanation:

Concentration of [tex][OH^-][/tex] in bleach solution = 0.02 M

[tex]NaOH\rightarrow OH^-+Na^+[/tex]

[tex]NaOH=[OH^-]=0.02M[/tex]

Concentration of bleach solution we want ,[tex]M_1[/tex] = 0.02 M

Volume of the bleach solution,[tex]V_1[/tex] = 20 ml

Concentration of NaOH solution,[tex]M_2[/tex] = 1.00 M

Volume of the NaOH solution required ,[tex]V_2[/tex] = ?

[tex]M_1V_1=M_2V_2[/tex]

[tex]0.02 M\times 20 mL=1.00 M\times V_2[/tex]

[tex]V_2=\frac{0.02 M\times 20 mL}{1.00 M}=0.4 mL[/tex]

1 mL = 20 drops

0.4 mL = 0.4 × 20 drops = 8 drops

8 drops of 1.00 M NaOH will be needed.


Related Questions

Draw the Lewis Structure for CH4.

Answers

Explanation:

Methane (CH₄)

Valence electrons of carbon = 4

Valence electrons of hydrogen = 1  

The total number of the valence electrons  = 4 + 4(1) = 8

The Lewis structure is drawn in such a way that the octet of each atom and duet for the hydrogen in the molecule is complete. So,  

The Lewis structure is:

                H

                 :

     H    :    C   :    H

                 :

                H

Final answer:

To draw the Lewis Structure for CH4, follow the steps to determine the total number of valence electrons, place the atoms and lone pairs, and draw the final structure. Carbon in CH4 forms 4 single bonds with Hydrogen, resulting in a tetrahedral arrangement of Hydrogen atoms around the Carbon atom.

Explanation:

The Lewis Structure for CH4, which represents the arrangement of electrons in the molecule, can be drawn using the following steps:

Determine the total number of valence electronsDetermine the central atom, which is Carbon (C) in this case, and surround it with the Hydrogen (H) atomsPlace the remaining electrons as lone pairs on the central atom and the outer atomsCheck if the central atom has an octet of electrons. If not, move a lone pair from an outer atom to form a double bond with the central atomDraw the final Lewis structure

For CH4, Carbon has 4 valence electrons and Hydrogen has 1 valence electron each. The total number of valence electrons is 4 + (4 x 1) = 8. Following the steps, Carbon is surrounded by 4 Hydrogen atoms, and each Hydrogen atom shares its electron with Carbon. Carbon ends up with a complete octet by sharing electrons with the Hydrogen atoms, forming 4 single bonds. The final Lewis structure for CH4 is a tetrahedral arrangement of the Hydrogen atoms around the Carbon atom, represented by a molecular formula, structural formula, or other models.

Learn more about Lewis Structure here:

https://brainly.com/question/20300458

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A 0.4657 g sample of a pure soluble bromide compound is dissolved in water, and all of the bromide ion is precipitated as AgBr by the addition of an excess of silver nitrate. The mass of the resulting AgBr is found to be 0.8878 g. What is the mass percentage of bromine in the original compound?

Answers

Answer:

The mass percentage of bromine in the original compound is 81,12%

Explanation:

Step 1: Calculate moles AgBr

moles AgBr = mass AgBr / molar mass  AgBr

= 0.8878 g / 187.77 g/mol

= 0.00472812 moles AgBr

Since 1 mol AgBr contains 1 mol Br-

Then the amount of moles Br- in the original sample must also have  been 0.00472812 moles

Step 2: Calculating mass Br-

mass Br- = molar mass Br x moles  Br-

= 79.904 g/mol x 0.00472812 mol

= 0.377796 g Br-

There were 0.377796 g Br- in the original sample

Step 3: Calculating mass percentage Br-

⇒mass percentage  = actual mass Br- / total mass x 100%

% mass Br = 0.377796 g / 0.4657 g x 100  %

= 81.12%

What is the temperature (°C) of 1.75 g of O2 gas occupying 3.10 L at 1.00 atm? Enter your answer in the box provided.

Answers

Answer:

691.29 K or 418.14 °C

Explanation:

Hello, at first the moles of oxygen gas are required:

[tex]n_{O_2}=1.75 g * \frac{1mol O_2}{32 g O_2} =0.0547 mol[/tex]

Now, based on the ideal gas equation, we solve for the temperature:

[tex]PV=nRT\\T=\frac{PV}{nR}\\T=\frac{1atm * 3.10 L}{0.0547 mol*0.082 \frac{atm*L}{mol*K} }\\T=691.29 K[/tex]

Best regards.

How many significant figures are in the following? a) 0.1111 b) 2000 c) 35.6 d) 180,701

Answers

Answer:

a) 0.1111 : 4 significant figures

b) 2000 : 1 significant figure

c) 35.6 : 3 significant figures

d) 180,701 : 6 significant digits

Explanation:

Significant figures or digits of a number are the digits that carry meaning and contribute to the precision of the number.

a) 0.1111 : 4 significant figures, leading zeros are not significant digits.

b) 2000 : 1 significant figure, trailing zeros are not significant.

c) 35.6 : 3 significant figures, all digits are significant.

d) 180,701 : 6 significant digits, zeros between mom-zero digits are significant.

The number of significant figures in the given examples are:

a) 4 significant figures for 0.1111

b) 1 significant figure for 2000 (assuming no decimal)

c) 3 significant figures for 35.6  

d) 6 significant figures for 180,701

To determine the number of significant figures in a number, certain rules are followed:

Non-zero digits are always significant.Any zeros between significant digits are significant.Leading zeros (zeros before non-zero digits) are not significant.Trailing zeros in a decimal number are significant.In a number without a decimal point, trailing zeros may or may not be significant, depending on whether a decimal point is specified or assumed.

Applying these rules, we have:

0.1111 - All digits are significant because they are non-zero, giving 4 significant figures.2000 - Without additional context or a decimal point, we cannot be certain if the trailing zeros are significant, so this number has 1 significant figure.35.6 - All digits including the zero are significant, giving 3 significant figures.180,701 - All digits are significant, giving 6 significant figures.

If ine mole of pennies were divided amung 250 million peoplein
the US, how many dollars would each person recieve?

Answers

Answer:

[tex]2.4088\times 10^{13}[/tex] dollars each person will receive.

Explanation:

Number of people in which 1 mole of pennies is distributed = 250 million =

[tex]1 million = 10^6 [/tex]

250 million = [tex]2.5\times 10^8 [/tex] persons

Number of pennies in 1 mole = [tex]6.022\times 10^{23}[/tex]

Pennies per person:

[tex]\frac{6.022\times 10^{23} pennies}{2.5\times 10^8 persons}=2.4088\times 10^{15} pennies/person[/tex]

1 penny = 0.01 $

[tex]2.4088\times 10^{15} pennies/person=2.4088\times 10^{15}\times 0.01 \$/person=2.4088\times 10^{13} \$/person[/tex]

[tex]2.4088\times 10^{13}[/tex] dollars each person will receive.

A mixture of Fe2O3 and FeO was found to contain 72.00% Fe
bymass. What is the mass of Fe2O3 in 0.500 g of this mixture...how
doi work this..i dont even know what to look for as the given
asneeded?

Answers

Answer:

The mass of  Fe₂O₃ in 0.500 g of mixture is 0.367 g.

Explanation:

First off, we know that 72% of the mass of the mixture is iron. The information also tells us that the remaining 28% of the mass is oxygen.

Now we calculate the total mass of iron and the total mass of oxygen in the mixture:

0.500 g * 0.72 = 0.360 g of Fe0.500 g * 0.28 = 0.140 g of O

With the mass of each element we can calculate the number of moles of each atom, using the atomic weight:

0.360 g Fe * 1 mol / 55.845 g = 0.00645 moles of Fe

0.140 g O * 1 mol / 16 g = 0.00875 moles of O

The number of moles of Fe in the mixture is equal to the number of moles of FeO plus two times the number of moles of Fe₂O₃:

0.00645 = [tex]2*n_{Fe2O3} +n_{FeO}[/tex]             eq A

The number of moles of O in the mixture is equal to the number of moles of FeO plus three times the number of moles of Fe₂O₃:

0.00875 = [tex]3*n_{Fe2O3} +n_{FeO}[/tex]               eq B

So now we have a system of two equations and two unknowns, we solve for [tex]n_{Fe2O3}[/tex]:

From eq A:

[tex]n_{FeO3}=0.00645-2*n_{Fe2O3}[/tex]

Replacing in  eq B:

[tex]0.00875=3*n_{Fe2O3} + (0.00645-2*n_{Fe2O3})\\0.00230=n_{Fe2O3}[/tex]

Now we just need to convert moles of Fe₂O₃ into grams, using the molecular weight:

0.00230 moles * 159.66 g/mol = 0.367 g Fe₂O₃

a 2.60 g sample of titanium metal chemically combines
withchlorine gas to form 10.31g of a titanium chloride.
a. What is the empirical formula of the titaniumchloride
b. What is the percent by mass of titanium and chlorine in
thesample?

Answers

Answer:

For a: The empirical formula for the given compound is [tex]TiCl_4[/tex]

For b: The percent by mass of titanium and chlorine in the sample is 25.55 % and 74.78 % respectively.

Explanation:

For a:

We are given:

Mass of Titanium = 2.60 g

Mass of sample = 10.31 g

Mass of Chlorine = 10.31 - 2.60 = 7.71 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of titanium =[tex]\frac{\text{Given mass of Titanium}}{\text{Molar mass of Titanium}}=\frac{2.60g}{47.867g/mole}=0.054moles[/tex]

Moles of Chlorine = [tex]\frac{\text{Given mass of Chlorine}}{\text{Molar mass of Chlorine}}=\frac{7.71g}{35.5g/mole}=0.217moles[/tex]

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.054 moles.

For Titanium = [tex]\frac{0.054}{0.054}=1[/tex]

For Chlorine  = [tex]\frac{0.217}{0.054}=4.01\approx 4[/tex]

Step 3: Taking the mole ratio as their subscripts.

The ratio of Ti : Cl = 1 : 4

Hence, the empirical formula for the given compound is [tex]TiCl_4[/tex]

For b:

To calculate the percentage by mass of substance in sample, we use the equation:

[tex]\%\text{ composition of substance}=\frac{\text{Mass of substance}}{\text{Mass of sample}}\times 100[/tex]       .......(1)

For Titanium:

Mass of sample = 10.31 g

Mass of titanium = 2.60 g

Putting values in above equation, we get:

[tex]\%\text{ composition of titanium}=\frac{2.60g}{10.31g}\times 100=25.22\%[/tex]

For Chlorine:

Mass of sample = 10.31 g

Mass of chlorine = 7.71 g

Putting values in above equation, we get:

[tex]\%\text{ composition of chlorine}=\frac{7.71g}{10.31g}\times 100=74.78\%[/tex]

Hence, the percent by mass of titanium and chlorine in the sample is 25.55 % and 74.78 % respectively.

Calculate the number of moles in 369 grams of CaoCl2? a) 1 b) 2 c) 3 d) 4

Answers

Answer:

First of all, take account the molar mass of the CaOCl2 (Calcium hypochlorite),  which is 126.97 g/mol. If we have 126.97 g in a mol, 369g should be in aproximately 3 moles. Try to the think the rule of 3.

Explanation:

Balance the equation in the box. Click in the answer box to activate the palette. N2(g) + H2(g) → NH3(g)

Answers

Answer:

N2(g) + 3H2(g) → 2 NH3(g)

Explanation:

N2(g) + H2(g) → NH3(g)

We start equaling the number of N atoms in both sides multiplying by 2 the NH3.

N2(g) + H2(g) → 2 NH3(g)

So we equals the H atoms (there are six in products sites)

N2(g) + 3 H2(g) → 2 NH3(g)

The area of a telescope lens is 6.676 x 10 mm. (a) What is the area in square feet (ft)? Enter your answer in scientific notation. x 10ft (b) If it takes a technician 51.7 s to polish 1.46 x 10mm', how long does it take her to polish the entire lens?

Answers

Answer:

a) A = 7.186 E-4 ft²

b) t = 236.403 s

Explanation:

A = 6.676 E1 mm²

a) A = 6.676 E1 mm² * ( ft / 304.8 mm )²

⇒ A = 7.186 E-4 ft²

b) t = 51.7 s → A = 1.46 E1 mm²

⇒ relation (r) = 6.676 E1 mm² / 1.46 E1 mm² = 4.573

⇒ t = 51.7 s * 4.573 = 236.403 s

How many U.S. gallons are there in a cubic mile? The total proven oil reserves of the U.S. are roughly 30 x 10°bbl. How many cubic miles is this?

Answers

Answer:

1 cubic mile = 1.101 * 10^12 US gallons

1 US bbl oil = 42 US gallons = 3.8143*10^ -11 cubic miles

Explanation:

The number of the exponent of the oil reserve is not very well shown in the question so, I provide you the conversion of bbl oil into cubic mile, the only thing you have to do is multiply the number of bbls of the reserve for the conversion in cubic miles and you'll have the answer.

A turbine operating adiabatically is fed with steam at 400 °C and 8.0 MPa at the rate of 1000 kg/h. Process steam saturated at 0.5 MPa is withdrawn from an intermediate location in the turbine at a rate of 300 kg/h, and the remaining steam leaves the turbine saturated at 0.1 MPa. The kinetic energies and differences in potential energies of all streams are negligible. What is the power output of the turbine?

Answers

Explanation:

We select the enthalpy of steam at state 1 at [tex]800^{o}C[/tex] and 8.0 MPa from the steam tables as follows.

                     [tex]h_{1}[/tex] = 3138 kJ/kg

Also, we select the enthalpy of steam at state 2 at 0.5 MPa from the steam tables as follows.

                     [tex]h_{2}[/tex] = 2748.6 kJ/kg

At state 3 also, from the steam tables at state 3 at 0.1 MPa.

                     [tex]h_{3}[/tex] = 2675.1 kJ/kg

Hence, calculate the mass flow rate at state 3 as follows.

                  [tex]m_{3} = m_{1} - m_{2}[/tex]

                              = 1000 kg/h - 300 kg/h

                              = 700 kg/h

Now, we will calculate the power output of the turbine as follows.

                 [tex]W_{r} = m_{1}(h_{1} - h_{2}) + m_{3}(h_{2} - h_{3})[/tex]

                             = 1000 kg/h (3138 kJ/kg - 2748.6 kJ/kg) + 700 kg/h (2784.6 kJ/kg - 2675.1 kJ/kg)

                            = 440850 kJ/h

It is known that 1 kJ/h = 0.000278 kW.

Therefore,        [tex]440850 kJ/h \times \frac{0.000278 kW}{1 kJ/hr}[/tex]    

                          = 122.56 kW

Thus, we can conclude that the power output of the turbine is 122.56 kW.

Identify the molecules with a dipole moment: (a) SF (b) CF (c) CCCB (d) CHCI (e) H.CO

Answers

Answer: An atom has either tendency of accepting or losing electron, on the basis of this virtue it is named as

Electronegative: An atom has tendency to attract the shared pair of electron towards itself is called electronegative.Electropositive: An atom that has tendency to give the shared pair of electron towards an electronegative atom is called electropositive.

For the existence of dipole within a molecule there must be a difference in electronegativity of the atoms participating in it. All the molecules here have dipole moment because the atoms participating in them have a difference in electronegative.

Which of the following non-hydrogen atom transitions does the photon have at its long wavelength?

Move from Level n=6 to Level n=1
Move from Level n=4 to Level n=1
Move from Level n=3 to Level n=2
Move from Level n=5 to Level n=4

Please explain thae way, how I can solve such questions and how to calculate (If needed) in future?

Answers

Answer:

The move from Level n=3 to Level n=2 has the long wavelength.

Explanation:

First, due to the selection rules, only transitions between adjacent levels are allowed, thus, only a transition between Level n=3 to Level n=2 or Level n=5 to Level n=4 are allowed. The two first options are wrong.

Second, analyzing the transition between Level n=3 to Level n=2 and the transition between Level n=5 to Level n=4 it is necessary to think in terms of the equation of the difference of energy for these type of transitions:

Δ[tex]E = \frac{h^{2}}{8.m.L}(n_{LUMO}^{2} -n_{HOMO}^{2} )[/tex] (1)

The difference in energy (ΔE) is directly proportional to the quadratic difference between the 'n' levels of transition. Therefore, If the transition occurs between smaller 'n' levels the difference of energy will be smaller too.

Also, the energy (ΔE) is inversely proportional to the wavelength (λ) so a smaller energy means a larger wavelength.

ΔE = c / λ (2)

Hence, the move from Level n=3 to Level n=2 has a long wavelength.

In order to calculate this wavelength is necessary to replace the data on equation (1) and (2).

The specific gravity of gasoline is approximately 0.70.

(a) Estimate the mass (kg) of 50.0 liters of gasoline.

(b) The mass flow rate of gasoline exiting a refinery tank is 1150 kg/min. Estimate the volumetric flow rate in liters/s.
(c) Estimate the average mass flow rate lbm/min delivered by a gasoline pump when the gas pump has a pump rate of 10 gallons per 2 minutes.

Answers

Answer:

a) mass=35 kg

b) volumetric flow rate= 27.37 [tex]\frac{liters}{s}[/tex]

c) average mass flow rate=29.21 [tex]\frac{lbm}{min}[/tex]

Explanation:

Specific gravity is defined as the relation between the density of one substance and the density of another reference substance (it is usual that water is used in this case). In this case the specific gravity is the relationship between the density of gasoline and the density of water. So:

[tex]0.7=\frac{densitygasoline}{densitywater}[/tex]

Now you can know the density of gasoline  with a simple mathematical operation and knowing that densitywater≅ 1000 [tex]\frac{kg}{m^{3} }[/tex]:

densitygasoline=0.7*densitywater

densitygasoline=0.7*1 [tex]\frac{kg}{liters}[/tex]

densitygasoline=0.7 [tex]\frac{kg}{liters}[/tex]

a)

Now that the density is known, you can calculate the mass in 50 liters of gasoline.

By definition of density, you know that in 1 liter there are 700 kg of gasoline. So using The Rule of Three( tool that allows you to solve problems based on proportions), you can calculate the mass in 50 liters:

[tex]\frac{700 kg}{1 liter} =\frac{mass}{50 liters}[/tex]

[tex]mass=\frac{0.7 kg* 50 liters}{1 liter}[/tex]

mass=35 kg

b)

First, it is convenient to change units: from kg / min to kg / seconds. For that you know that 1 min= 60 seconds. So:

[tex]\frac{1150 kg}{min} =\frac{1150 kg}{60 s}[/tex]

[tex]\frac{1150 kg}{min} =\frac{19.16 kg}{s}[/tex]

By definition, density is the relationship between the mass and volume of a substance: This is: [tex]density=\frac{mass}{volume}[/tex]

Applied to this case, it would be [tex]density=\frac{mass flow rate}{volumetric flow rate}[/tex]

The mass flow rate and the density of the gasoline are known, so you can calculate the volumetric flow rate:

[tex]volumetric flow rate= \frac{mass flow rate}{density gasoline}[/tex]

volumetric flow rate= 27.37 [tex]\frac{liters}{s}[/tex]

c)

You know that the pump rate is 10 gallons per 2 minute. This is:

[tex]\frac{10 gallons}{2 minutes} = 5 \frac{gallons}{minute}[/tex]

To calculate the average mass flow, you must relate this data to density and do the necessary unit conversions.  Conversions are done similarly to what was previously applied in the previous exercises.

You have to know that:

1 gallon = 3.785 liters1 kg= 2.205 kg

So:

[tex]\frac{5 gallons*3.785 liters}{1gallons*min} =\frac{18.925 liters}{min}[/tex]

You know that the density is [tex]density=\frac{mass flow rate}{volumetric flow rate}[/tex]

The volumetric flow rate (pump rate) and the density of the gasoline are known, so you can calculate the average mass flow rate:

average mass flow rate=density*volumetric flow rate

[tex]average mass flow rate=\frac{0.7 kg}{liters} *\frac{18.925 liters}{min}[/tex]

average mass flow rate=13.2475 [tex]\frac{kg}{min}[/tex]

Converting the units:

[tex]average mass flow rate=\frac{13.2475 kg*2.205 lbm}{1 kg*min}[/tex]

average mass flow rate=29.21 [tex]\frac{lbm}{min}[/tex]

What are the major causes of environmental problems?

Answers

Answer: This world is rapidly moving towards to some serious enviornmental problems, the cause of this problems are mentioned below;

(1) The increase in average temperature of Earth.

(2) Deforestation

(3) Over Population

(4) Mining

Because of these reasons mentioned above there are many enviornmental issue occurring, and many problems we will face in future for example, water crisis, lack of resources.

Menthol (molar mass = 156.3 g/mol), the strong-smelling substance in many cough drops, is a compound of carbon, hydrogen and oxygen. When 0.1595g of menthol was burned in a combustion apparatus, 0.449g of CO2 and 0.184g of H2O formed. What is menthol’s molecular formula?

Answers

Answer: The molecular formula for the menthol is [tex]C_{10}H_{20}O[/tex]

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

[tex]C_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of [tex]CO_2=0.449g[/tex]

Mass of [tex]H_2O=0.184g[/tex]

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.449 g of carbon dioxide, [tex]\frac{12}{44}\times 0.449=0.122g[/tex] of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.184 g of water, [tex]\frac{2}{18}\times 0.184=0.0204g[/tex] of hydrogen will be contained.

Mass of oxygen in the compound = (0.1595) - (0.122 + 0.0204) = 0.0171 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = [tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.122g}{12g/mole}=0.0102moles[/tex]

Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.0204g}{1g/mole}=0.0204moles[/tex]

Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0171g}{16g/mole}=0.00107moles[/tex]

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00107 moles.

For Carbon = [tex]\frac{0.0102}{0.00107}=9.53\approx 10[/tex]

For Hydrogen  = [tex]\frac{0.0204}{0.00107}=19.54\approx 20[/tex]

For Oxygen  = [tex]\frac{0.00107}{0.00107}=1[/tex]

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 10 : 20 : 1

Hence, the empirical formula for the given compound is [tex]C_{10}H_{20}O_1=C_{10}H_{20}O[/tex]

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

[tex]n=\frac{\text{molecular mass}}{\text{empirical mass}}[/tex]

We are given:

Mass of molecular formula = 156.3 g/mol

Mass of empirical formula = 156 g/mol

Putting values in above equation, we get:

[tex]n=\frac{156.3g/mol}{156g/mol}=1[/tex]

Multiplying this valency by the subscript of every element of empirical formula, we get:

[tex]C_{(1\times 10)}H_{(1\times 20)}O_{(1\times 1)}=C_{10}H_{20}O[/tex]

Thus, the molecular formula for the menthol is [tex]C_{10}H_{20}O[/tex]

You are given a protein solution with a concentration of 0.15 mg/ml.

v. Suppose that we want to prepare 100 microliters of 10 micrograms/microliters solution. How much of H2O and protein stock do we need to add to obtain the target concentration and volume? (THE CONCENTRATION IS NOT GREATER IN THE QUESTION, YOU NEED TO CONVERT IT TO MICROGRAMS/MICROLITERS.... THEN IT SHOULD MAKE SENSE)

Answers

Answer:

6,666.66 micro liter of water and protein stock we will need to add to obtain the target concentration and volume.

Explanation:

Concentration of given solution[tex]C_1 = 0.15 mg/mL[/tex]

1 mL = 1000 μL , 1 mg = 1000 μg

[tex]C_1=0.15 mg/mL=\frac{0.15\times 1000 \mu g}{1\times 1000 \mu L}=0.15 \mu g/\mu L[/tex]

The volume of the given solution =[tex]V_1= V[/tex]

Concentration of required solution = [tex]C_2=10 \mu g/\mu L[/tex]

Volume of required solution = [tex]V_2=100 \mu L[/tex]

[tex]C_1V_1=C_2V_2[/tex]

[tex]V=\frac{C_2V_2}{C_1}=\frac{10 \mu g/\mu L\times 100 \mu L}{0.15 \mu g/\mu L}=6,666.66 \mu L[/tex]

6,666.66 micro liter of water and protein stock we will need to add to obtain the target concentration and volume.

Express each of the following values in the given base unit. 1st attempt Part 1 (2.5 points) m = 2.54 cm Part 2 (2.5 points) S = 316 MS

Answers

Answer:

1) 2.54 cm = 0.0254 m

2) 316 Ms = [tex]3.16\times 10^{8} s[/tex]

Explanation:

1) 2.54 cm

In 1 centimeter is there are 0.01 meters.

1 cm = 0.01 m

[tex]2.54 cm = 2.54\times 0.01 m = 0.0254 m[/tex]

2) 316 Ms

In 1 mega second  is there are 1 million seconds.

[tex]1 Ms = 10^{6} s[/tex]

[tex]316 Ms= 316\times 10^{6} s= 3.16\times 10^{8} s[/tex]

How many atoms of Phosphorus (symbol P) are in 45.67g of P? (Your answer should just be a number. Do not include atoms as the unit. For scientific notation use the "^" symbol before the exponent. For example 2.54 x 10^5)

Answers

Find the Ar of P, then apply the moles = mass divided by Mr. Then do mr x 45.67 x 6.02 x 10^23 and that should give you the answer.

Answer:

8.882 × 10^23

Explanation:

The molar mass of phosphorus is 30.97 g/mol. The moles of P corresponding to 45.67 g are:

45.67 g × (1 mol/30.97 g) = 1.475 mol

In 1 mole of P, there are 6.022 × 10²³ atoms of P (Avogadro's number). The number of atoms of P in 1.475 moles are:

1.475 mol × (6.022 × 10²³ atoms/1 mol) = 8.882 × 10²³ atoms

Suppose the gas above the soda in a bottle of soft drink
ispure CO2 at a pressure of 2atm. Calculate [CO2] at 25
degreeC.
Henry's law at 25C of CO2 is 0.033363 (mol/L*atm).

Answers

Answer: The concentration of carbon dioxide in bottle of soft drink is [tex]6.6\times 10^{-2}mol/L[/tex]

Explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

[tex]C_{CO_2}=K_H\times p_{liquid}[/tex]

where,

[tex]K_H[/tex] = Henry's constant = [tex]0.033363mol/L.atm[/tex]

[tex]p_{CO_2}[/tex] = partial pressure of gas in a bottle of soft drink = 2 atm

Putting values in above equation, we get:

[tex]C_{CO_2}=0.033363mol/L.atm\times 2atm\\\\C_{CO_2}=0.066726mol/L=6.6\times 10^{-2}[/tex]

Hence, the concentration of carbon dioxide in bottle of soft drink is [tex]6.6\times 10^{-2}mol/L[/tex]

You have a sample of water that contains the organic compound
C7H12ON2 at a concentration of 50
mg/L. The compound can be oxidized by bacteria to form carbon
dioxide, water, and ammonia. How many mg/L of oxygen is needed to
biodegrade the compound? Note: Determine only carbonaceous
demand.

Answers

Answer:

131.4 mg/L of oxygen is needed to  biodegrade the organic compound.

Explanation:

The chemical reaction will be written as:

[tex]2C_7H_{12}ON_2+23O_2\rightarrow 14CO_2+12H_2O+4NO_2[/tex]

Concentration of the organic compound = 50 mg/L

This means that 50 milligrams of organic compound in present in 1 L of the solution.

50 mg = 0.050 g

1 mg = 0.001 g

Moles of organic compound = [tex]\frac{0.050 g}{140 g/mol}=0.0003571 mol[/tex]

According to reaction, 2 moles of organic compound reacts with 23 moles of oxygen gas.

Then 0.0003571 moles of an organic compound will react with:

[tex]\frac{23}{2}\times 0.0003571 mol=0.004107 mol[/tex] oxygen gas.

Mass of 0.004107 moles of oxygen gas:

0.004107 mol × 32 g/mol = 0.1314 g = 131.4 mg

131.4 mg/L of oxygen is needed to  biodegrade the organic compound.

How many mL of concentrated hydrochloric acid should be added to 48.3 g of sodium nitrite to prepare 2.50 L of a buffer solution with a pH of 2.60?

Answers

Answer:

47 mL

Explanation:

The equilibrium of nitrous acid is:

HNO₂ ⇄ NO₂⁻ + H⁺ Ka = 4,5x10⁻⁴

If desire pH is 2,60 the [H⁺] concentration in equilibrium is:

[H⁺] = [tex]10^{-2,6}[/tex] =2,51x10⁻³ M

Initial molarity of sodium nitrite is:

43,8g × [tex]\frac{1mol}{68,9953 g}[/tex]÷ 2,5 = 0,254 M

Thus, in equilibrium the concentration of chemicals is:

[NO₂⁻] = 0,254 - x

[HNO₂] = x

[H⁺] =  2,51x10⁻³ = Y-x Where Y is initial concentration.

Equilibrium formula is:

4,5x10⁻⁴ =  [tex]\frac{[2,51x10^{-3}[0,254 - x] ]}{[x]}[/tex]

Solving, x = 0,215

Thus, initial [H⁺] concentration is:

0,215 + 2,51x10⁻³ = 0,2175 M

If total volume is 2,50 L:

2,50L ×[tex]\frac{0,2175 mol}{L}[/tex] = 0,5438 mol of HCl

As molarity of concentrated hydrochloric acid is 11,65 mol per liter:

0,5438 mol HCl ×[tex]\frac{1L}{11,65}[/tex] = 0,047 L ≡ 47 mL

I hope it helps!

Which is more stable, a trans-1,4-disubstitutedcyclohexane or
its cis isomer?

Answers

Answer:

If the substituents are the same, the cis distribution is more stable than trans distribution.

Explanation:

A cis cyclohexane is one in which both substituents are oriented towards the same face of the ring regardless of the conformation.

A trans cyclohexane has substituents on opposite sides of the ring.

In trans-1,4-disubstituted cyclohexane, the conformation with the two substituent groups in equatorial is more stable than the constitution that has the axial groups.

In cis-1,4-disubstituted cyclohexane, both conformations have the same stability as they have an axial and an equatorial substituent, so they have the same energy and there is no equilibrium shift.

How many degrees of freedom does each of the following systems have? (Answer as a number, i.e., 1, 2, 3, etc.)

1. Liquid water in equilibrium with its vapor?

2. Liquid water in equilibrium with a mixture of water vapor and nitrogen?

3. A liquid solution of alcohol in water in equilibrium with its vapor?

Answers

Answer:

Explanation:

Hello, since the Gibbs' phase rule states the following equation:

[tex]F=C-P+2[/tex]

Whereas C is the number of components and P the present phases, you answers are:

1. F=1-2+2=1.

2. F=2-2+2=2.

3. F=2-2+2=2.

Best regards.

A sample of nitric acid has a mass of 8.2g. It is dissolved in 1L of water. A 25mL aliquot of this acid is titrated with NaOH. The concentration of the NaOH is 0.18M. What titre volume was added to the aliquot to achieve neutralisation?

Answers

Answer:

18.075 mL of NaOH solution was added to achieve neutralization

Explanation:

First, let's formulate the chemical reaction between nitric acid and sodium hydroxide:

NaOH + HNO3 → NaNO3 + H2O

From this balanced equation we know that 1 mole of NaOH reacts with 1 mole of HNO3 to achieve neutralization. Let's calculate how many moles we have in the 25 mL aliquot to be titrated:

63.01 g of HNO3 ----- 1 mole

8.2 g of HNO3 ----- x = (8.2 g × 1 mole)/63.01 g = 0.13014 moles of HNO3

So far we added 8.2 grams of nitric acid (0.13014 moles) in 1 L of water.

1000 mL solution ---- 0.13014 moles of HNO3

25 mL (aliquot) ---- x = (25 mL× 0.13014 moles)/1000 mL = 0.0032535 moles

So, we now know that in the 25 mL aliquot to be titrated we have 0.0032535 moles of HNO3. As we stated before, 1 mole of NaOH will react with 1 mole of HNO3, hence 0.0032535 moles of HNO3 have to react with 0.0032535 moles of NaOH to achieve neutralization. Let's calculate then, in which volume of the given NaOH solution we have 0.0032535 moles:

0.18 moles of NaOH ----- 1000 mL Solution

0.0032535 moles---- x=(0.0032535moles×1000 mL)/0.18 moles = 18.075mL

As we can see, we need 18.075 mL of a 0.18 M NaOH solution to titrate a 25 mL aliquot of the prepared HNO3 solution.

The half-life of a pesticide determines its persistence in the environment. A common pesticide degrades in a first-order process with a half-life of 2 days. What fraction (in decimal notation) of the pesticide remains in the environment after 18 days? Enter to 4 decimal places.

Answers

Answer:

0.0020 fraction of the pesticide remains in the environment after 18 days

Explanation:

For a first order reaction, [tex]\frac{N}{N_{0}}=(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}}[/tex]

where N is remaining mass after "t" time , [tex]N_{0}[/tex] is initial mass,  [tex]\frac{N}{N_{0}}[/tex] represents fraction of mass remains after "t" time and [tex]t_{\frac{1}{2}}[/tex] is half-life

Here t is 18 days and [tex]t_{\frac{1}{2}}[/tex] is 2 days

So, [tex]\frac{N}{N_{0}}=(\frac{1}{2})^{\frac{18}{2}}=(\frac{1}{2})^{9}=0.0020[/tex]

Hence 0.0020 fraction of the pesticide remains in the environment after 18 days

The fraction (in decimal notation) of the pesticide that remains in the environment after 18 days is 0.0020

We'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:

Half-life (t½) = 2 days

Time (t) = 18 days

Number of half-lives (n) =?

n = t / t½

n = 18 / 2

n = 9

Thus, 9 half-lives has elapsed.

Finally, we shall determine the fraction of the pesticide that remains in the environment. This can be obtained as follow:

Original amount (N₀) = 1

Number of half-lives (n) = 9

Fraction remaining (N / N₀) =?

N = 1/2ⁿ × N₀

Divide both side by N₀

N / N₀ = 1/2ⁿ

N / N₀ = 1/2⁹

N / N₀ = 0.0020

Thus, the fraction of the pesticide that remains in the environment is 0.0020

Learn more: https://brainly.com/question/20516965

On a website devoted to answering engineering questions, viewers were invited to determine how much power a 100-MW power plant generates annually. The answer declared to be best was submitted by a civil engineering student, who stated, "It produces 100 MW/hr so over the year that's 100*24*365.25 & do the math." a. Carry out the calculation, showing all the units. Answer b. What is wrong with the statement of the question? c. Why was the student wrong in saying that the plant produces 100 MW/hr?

Answers

Answer:

(a) The plant generates 3,153,600,000 MJ a year.

(b) The problem with the question is that "power" is not "generated". What is generated is Energy, and Power is the rate of generation of Energy.

(c) The student is wrong because the plants generates 100 MJ of energy a second. "MW/hr" is not a unit of energy or power, so it has no sense.

Maybe he get confused with MW-h, which is a unit of energy, used to measure electrical consumption.

Explanation:

(a) The power of the plant (100 MW) is the rate at which electrical energy is produced. It has units of [energy]/[time].

In this case, the plant produces 100 MJ/s. The energy produced can also be expressed in other units, like MJh.

To calculate the energy generated in one year, we have

[tex]Energy = Power * Time  = \\\\Energy = 100 MW*1year\\\\Energy = (100 \frac{MJ}{s})*(1 year*\frac{365 days}{1year}*   \frac{24hours}{1day}*\frac{3600s}{1hour})\\\\Energy=(100 \frac{MJ}{s})*(31,536,000s)=3,153,600,000MJ[/tex]

The plant generates 3,153,600,000 MJ a year.

(b) The problem with the question is that "power" is not "generated". What is generated is Energy, and Power is the rate of generation of Energy.

(c) The student is wrong because the plants generates 100 MJ of energy a second. "MW/hr" is not a unit of energy or power, so it has no sense.

Maybe he get confused with MW-h, which is a unit of energy, used to measure electrical consumption. That unit represents the amount of energy consumed or generated by a 1 MW unit in one hour. It is equivalent to 3600 MJ (1 MW-h = 3600 MJ).

In this unit, the 100 MW plant generates 876,000 MWh.

How many electrons are there in the valence shell of the oxygen atom of water? O a. 2 O 6.4 OC.6 O d. 8

Answers

Answer:

d. 8

Explanation:

Valence electrons of oxygen = 6

Valence electrons of hydrogen = 1  

The total number of the valence electrons  = 6 + 2(1) = 8

The Lewis structure is drawn in such a way that the octet of each atom and duet for the hydrogen in the molecule is complete.

In the water molecule,

Oxygen has 2 bond pairs and two lone pairs which means that the total electrons in the valence shell is 8.

A 1.00 g sample of a metal X (that is known to form X ions in solution) was added to 127.9 mL of 0.5000 M sulfuric acid. After all the metal had reacted, the remaining acid required 0.03340 L of 0.5000 M NaOH solution for complete neutralization. Calculate the molar mass of the metal and identify the element.

Answers

Answer: The metal having molar mass equal to 26.95 g/mol is Aluminium

Explanation:

To calculate the number of moles for given molarity, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]     .....(1)

Molarity of NaOH solution = 0.5000 M

Volume of solution = 0.03340 L

Putting values in equation 1, we get:

[tex]0.5000M=\frac{\text{Moles of NaOH}}{0.03340L}\\\\\text{Moles of NaOH}=(0.5000mol/L\times 0.03340L)=0.01670mol[/tex]

The chemical equation for the reaction of NaOH and sulfuric acid follows:

[tex]2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O[/tex]

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, 0.01670 moles of NaOH will react with = [tex]\frac{1}{2}\times 0.01670=0.00835mol[/tex] of sulfuric acid

Excess moles of sulfuric acid = 0.00835 moles

Calculating the moles of sulfuric acid by using equation 1, we get:

Molarity of sulfuric acid solution = 0.5000 M

Volume of solution = 127.9 mL = 0.1279 L    (Conversion factor:  1 L = 1000 mL)

Putting values in equation 1, we get:

[tex]0.5000M=\frac{\text{Moles of }H_2SO_4}{0.1279L}\\\\\text{Moles of }H_2SO_4=(0.5000mol/L\times 0.1279L)=0.06395mol[/tex]

Number of moles of sulfuric acid reacted = 0.06395 - 0.00835 = 0.0556 moles

The chemical equation for the reaction of metal (forming [tex]M^{3+}[/tex] ion) and sulfuric acid follows:

[tex]2X+3H_2SO_4\rightarrow X_2(SO_4)_3+3H_2[/tex]

By Stoichiometry of the reaction:

3 moles of sulfuric acid reacts with 2 moles of metal

So, 0.0556 moles of sulfuric acid will react with = [tex]\frac{2}{3}\times 0.0556=0.0371mol[/tex] of metal

To calculate the molar mass of metal for given number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Mass of metal = 1.00 g

Moles of metal = 0.0371 moles

Putting values in above equation, we get:

[tex]0.0371mol=\frac{1.00g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{1.00g}{0.0371mol}=26.95g/mol[/tex]

Hence, the metal having molar mass equal to 26.95 g/mol is Aluminium

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