Pls someone help me with this question pls

Answer:

The incorrect or the false options are -

(B) - If NH4NO3 is added, the pH becomes more basic.

(D) - If a small amount of HCl is added, the pH becomes slightly more basic.

 (E) - If NaOH is added, the OH- ions react with the [tex]NH_3[/tex].

Explanation:

The given solution the buffer basic .

Therefore , the options A ( If a small amount of NaOH is added, the pH becomes slightly more basic.) and C ( If HCl is added, the H+ ions react with the NH3.) All other B , D , E are false .

Answer:

85.952 ml [tex]Zn^2^+[/tex]  needed to titrate the excess complexing reagent .

Explanation:

Lets calculate

After addition of 80 ml of EDTA the solution becomes = 20 + 70 = 90 ml

As the number of moles of [tex]CoSO_4[/tex] =[tex]\frac{Given mass }{molar mass}[/tex]

                                                       =[tex]\frac{1.817}{155}[/tex]

                                                          =0.01172

Molarity = [tex]\frac{no. of moles}{volume of solution}[/tex]

           =[tex]\frac{0.01172}{20}[/tex]

        =0.000586 moles

Excess of EDTA = concentration of EDTA - concentration of CoSO4

                            = 0.009005 - 0.000586

                           = 0.008419 M

As M1V1 ( Excess of EDTA ) = M2V2 [tex](Zn^2^+)[/tex]

           [tex]0.008419\times100ml=0.009795\times V2[/tex]

           [tex]V2=\frac{0.008419\times100}{0.009795}[/tex]

             V2 =85.952 ml

Therefore , 85.952 ml [tex]Zn^2^+[/tex] needed to titrate the excess complexing reagent .

Answer:

98.356 ohms is the answer

Answer:

2.04x10⁻⁴ mol

Explanation:

First we convert 3.06x10⁻³ grams of glycine into moles of glycine, using its molar mass:

In order to calculate the number of hydrogen moles, we multiply the number of glycine moles by 5, as there are 5 hydrogen moles per glycine mol:

Answer:

the payloads weight

Explanation:

a p e x on the ground stream

Answer:

The payload's weight

Explanation:

i just took the test on a pex :)

Answer: The formula of chromium fluoride is CFr2.

Answer:

Explanation:

Chemically identical protons in 1HNMR spectra are protons that are found in the same chemical system and produce one signal in the spectra. Non-equivalent protons are protons that reside in various chemical environments and display different signals.

(a) The provided alkyl halide has four carbons, nine hydrogens, and one chlorine atom, according to the molecular formula. To achieve a single NMR peak, all protons (9 hydrogens) must be in the same chemical environment. That is, the carbon skeleton should be made up of three methyl groups, each bound to a single carbon atom.

As a consequence, the tert-butyl chloride structure can be present. That it has three methyl groups that are all the same and are all bound to the same carbon atom. As a result. A single peak is generated by proton NMR.

(b) The other isomeric variant of C4H9Cl, which has several peaks and a doublet at d3.4, could have a proton at a nearby carbon. And then does it produce a doublet. To create the doublet, two methyl groups should be substituted for the carbon. The value of 3.4 indicates that the protons can be exposed to chlorine. As a result, the compound's composition is iso-butyl chloride.

(c) The compound C4H9Cl can have two protons on adjacent carbons since it has multiple peaks and a triplet at d3,5. And then does it generate a triplet. The meaning of 3.5 indicates that the protons should be exposed to chlorine. As a result, the compound's composition is n-butyl chloride.

(d) The compound C4H9Cl has many peaks, including a methyl group triplet at d1.0 and a methyl group doublet after. It's in the same family as the -CH party. Due to the chlorine atom, this doublet occurs at downfield at 1.5. As a result, the compound's composition is sec-butyl chloride.

Answer:

Because thats where the animal passed away

Explanation:

Answer:

25 minutes

Explanation:

V = Volume = [tex]90\ \text{m}^3[/tex]

[tex]C_0[/tex] = Radon concentration under steady state = 1.5 Bq/L

k = Radon decay rate = [tex]2.09\times 10^{-6}\ \text{s}^{-1}[/tex]

[tex]Q[/tex] = Venting rate = [tex]0.14\ \text{m}^3/\text{s}[/tex]

[tex]C_f[/tex] = Final concentration of radon = 0.15 Bq/L

Theoretical detention time is given by

[tex]\theta=\dfrac{V}{Q}\\\Rightarrow \theta=\dfrac{90}{0.14}\\\Rightarrow \theta=642.86\ \text{s}[/tex]

We have the relation

[tex]C_f=C_0e^{-(\dfrac{1}{\theta}+k)t}\\\Rightarrow t=\dfrac{\ln\dfrac{C_f}{C_0}}{-(\dfrac{1}{\theta}+k)}\\\Rightarrow t=\dfrac{\ln\dfrac{0.15}{1.5}}{-(\dfrac{1}{642.86}+2.09\times 10^{-6})}\\\Rightarrow t=1478.25\ \text{s}=\dfrac{1478.25}{60}=24.63\approx 25\text{minutes}[/tex]

The time taken to reach the acceptable level of concentration is 25 minutes.

Answer:

For a): The mass percent of naphthalene in the original sample is 31.00 %.

For b): The mass percent of 3-nitroaniline in the original sample is 32.03 %.

For c): The mass percent of benzoic acid in the original sample is 29.97 %.

For d): The total percent recovered is 93.00 %.

Explanation:

Percentage by mass is defined as the ratio of the mass of a substance to the mass of the solution multiplied by 100. The formula used for this is:  

[tex]\text{Percent by mass}=\frac{\text{Mass of substance}}{\text{Mass of a solution}} \times 100[/tex]               ......(1)

a):

Mass of naphthalene = 0.483 g

Mass of the sample = 1.558 g

Plugging values in equation 1:

[tex]\%\text{ mass of naphthalene}=\frac{0.483 g}{1.558}\times 100\\\\\%\text{ mass of naphthalene}=31.00 \%[/tex]

b):

Mass of 3-nitroaniline = 0.499 g

Mass of the sample = 1.558 g

Plugging values in equation 1:

[tex]\%\text{ mass of 3-nitroaniline}=\frac{0.499 g}{1.558}\times 100\\\\\%\text{ mass of 3-nitroaniline}=32.03 \%[/tex]

c):

Mass of benzoic acid = 0.467 g

Mass of the sample = 1.558 g

Plugging values in equation 1:

[tex]\%\text{ mass of benzoic acid}=\frac{0.467 g}{1.558}\times 100\\\\\%\text{ mass of benozic acid}=29.97 \%[/tex]

d):

Total mass recovered = [0.483 + 0.499 + 0.467] = 1.449 g

Mass of the sample = 1.558 g

Plugging values in equation 1:

[tex]\text{Total percent recovered}=\frac{1.449 g}{1.558}\times 100\\\\\text{Total percent recovered}=93.00\%[/tex]

Answer:

c. uracil

Explanation:

uracil is not found in DNA. the missing base would be thymine

Answer:

Alkane

Explanation:

Alkenes have CH in their condensed formulas and alkynes have C in their condensed formulas. There are only CH3 and CH2 so it's an alkane.

Answer:

27.27%

Explanation:

Answer:

0.3832

Explanation:

Givens

m = 437.2 grams

c = ?

delta t = 69.8 - 19.3

H = 8460 J

Formula

H = m*c*Δt

Solution

8460 = 437.2 * c * (69.8 - 19.3)

8460 = 437.2 * c * 50.5

8460 = 22078.6 * c

c = 8460 / 22078,6

c = .3832 J / (oC * gr)

Explanation:

copper (II) nitrate

Illustrate variety of substances of which an element can be part:

metal --> blue solution --> blue solid --> black solid --> blue solution (again) --> metal (again).When solid copper reacts with aqueous silver nitrate, the products are copper (II) nitrate and solid silver. and carbon monoxide gas produces solid iron and carbon dioxide gas.

Answer:

hydro, water

Explanation:

Answer:

35.8g of Cl₂ is the yield

Explanation:

Based on the reaction:

MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂

1 mole of MnO₂ and 4 moles of HCl react producing 1 mole of Cl₂

To solve this question we must find limiting reactant. with limiting reactant we can find the theoretical yield of Cl₂. As the actual yield is the 62.7% we can find actual yield of Cl₂ in grams:

Moles MnO₂ -Molar mass: 86.9368g/mol-:

70.0g * (1mol / 86.9368g) = 0.805 moles

Moles HCl -Molar mass: 36.46g/mol-:

128.0g * (1mol / 36.46g) = 3.51 moles

For a complete reaction of 3.51 moles of HCl are required:

3.51 moles HCl * (1mol MnO₂ / 4mol HCl) = 0.878 moles MnO₂.

As there are just 0.805 moles of MnO₂, MnO₂ is limiting reactant.

1 mole of MnO₂ produce 1 mole of Cl₂. The theoretical moles of Cl₂ produced are 0.805 moles.

As the yield is of 62.7%, the yield of Cl₂ is:

0.805 moles * (62.7 / 100) = 0.505 moles Cl₂. In grams:

0.505 moles Cl₂ * (70.906g / mol) =

Answer:

a PE teacher would be in both catagories  of heath and fitness

Explanation:

potassium chloride from potassium hydroxide and hydrochloric acid

there are 6 O atoms suvdjwndje

Answer:

NO is the limiting reagent.  

In this reaction 0.886 mole of NO2 is produced

Explanation:

The chemical equation for this reaction is  

2NO(g) + O2(g) → 2NO2(g)

In this limiting reagent reaction, 2 moles of NO reacts with one mole of O2  to produce 2 mole of 2NO2

0.886 mole of NO * (2 mole of NO2/2 mole of NO) = 0.886 mole of NO2

0.503 mole of O2 * (1 mole of NO2/1 mole of O2) = 1.01 mole of NO2

Hence, NO is the limiting reagent.  

In this reaction 0.886 mole of NO2 is produced

Answer:

1.5x1024

Explanation:

Answer:

5.86mol 2Fe(OH)3 * [3 mols Na SO4 / 2 mols Fe(OH)3] = 8.79 mols sodium sulfate

Explanation:

To solve this question you need to look at the ratio of sodium sulfate to iron(III) hydroxide. The ratio is 3 to 2, meaning that for every 3 moles of Na SO4 there are also 2 moles of Fe(OH)3. So, you can multiply 5.86 by (3/2) to find the mols of sodium sulfate.

Answer: pH of buffer solution is 8.1

Explanation:

The formula for the Henderson–Hasselbalch equation is:

[tex]pH=pK_a+\log\frac{[A^-]}{[HA]}[/tex]

[tex]pH[/tex] is the concentration of [tex][H^+][/tex]

[tex]pK_a[/tex] is the acid dissociation constant,

[tex]A^-[/tex] and [tex]HA[/tex] are concentrations of the conjugate base and starting acid.

Putting in the values we get:

[tex]pH=7.5+\log\frac{x}{\frac{x}{4}}[/tex]

[tex]pH=8.1[/tex]

Thus pH of buffer solution is 8.1

Answer:

The answer is agriculture.

Explanation:

Answer: Agriculture

Explanation:

I got it right on my exam

Cyclohexene is a cyclic, six-membered hydrocarbon that contains one double bond. The types of reactions that can occur in cyclohexene would be those that are typical with alkenes generally.

The pi-bonded electrons in the double bond are nucleophilic. So, electrophilic addition reactions could occur with cyclohexene. For example,

cyclohexene + HBr → bromocyclohexane

cyclohexene + H2O/H+ → cyclohexanol

cyclohexene + Br2 → trans-1,2-dibromocyclohexane (racemic)

The latter is a common test for alkenes where one adds bromine to a sample to see if there is decolorization, which would indicate the presence of nucleophilic pi bonds. Bromine, which is dark reddish-brown, will become clear as it reacts with an alkene to form a colorless haloalkane.

Cyclohexene can also be converted to the fully saturated cyclohexane by hydrogenation: cyclohexene + H2/Pd → cyclohexane.