Boyd takes additional measurements of the mass of product formed in a reaction. He uses a balance that has smaller graduations than the first balance he used. What is Boyd most likely trying to change?
He is trying to increase his accuracy but not his precision.
He is trying to increase his precision but not his accuracy.
He is trying to decrease his precision and increase his accuracy.
He is trying to increase his precision and decrease his accuracy.

apparently not He is trying to increase his accuracy but not his precision.

Answers

Answer 1

Answer:

Boyd is most likely trying to increase his accuracy but not his precision. Therefore, option A is correct.

What are accuracy and precision?

Accuracy and precision can be described as two measures of observational error. Accuracy can be defined as how close a given set of measurements, are to their true value, while precision can be defined as how close the measurements are with respect to each other.

In other words, precision can be considered a description of random errors, as well as a measure of statistical variability. Accuracy is a description of systematic errors while low accuracy can bring a difference between a result and a true value.

Accuracy as describing a combination of both kinds of observational error, so high accuracy needs both high precision and high trueness.

In simpler terms, given a set of data points from repeated measurements of the same amount, the sample can be said to be accurate if its average is close to the true value being measured.

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Answer 2

Answer:

Answer:

The answer is D) He is trying to increase his accuracy and precision.

Explanation:

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Scaled Synthesis of Alum. Show your calculations for:a.the experimental scaling factor giving rise to a 15.0 g theoretical yield;b.the corrected volumes of KOH and H2SO4; andc.the theoretical yield of alum based on the actual amount of Al used.Make sure you carefully show each step for these calculations.

Answers

Answer:

Explanation:

You don't give your experimental data, so I shall assume:

Mass of Al = 1.07 g

20 mL of 3 mol·L⁻¹ KOH

20 mL of 9 mol·L⁻¹ H₂SO₄

The overall equation for the reaction is

Mᵣ: 26.98 474.39

2Al + 2KOH +4H₂SO₄ + 22H₂O ⟶ 2K[Al(SO₄)₂]·12H₂O + 3H₂

m/g: 1.07

(c) Theoretical yield of alum

(i) Moles of Al

$\text{Moles of Al} = \text{1.07 g Al} * \frac{\text{1 mol Al}}{\text{26.98 g Al}} = \text{0.039 66 mol Al}$

(ii) Moles of alum

$\text{Moles of alum} = \text{0.039 66 mol Al} * \frac{\text{2 mol alum }}{\text{2 mol Al}} = \text{0.039 66 mol alum \n}$

(iii) Theoretical yield of alum

$\text{Mass of alum} = \text{0.039 66 mol alum} * \frac{\text{474.39 g alum}}{\text{1 mol alum}} = \textbf{18.8 g alum}$

(a) Scaling factor for 15.0 g alum

You want a theoretical yield of 15.0 g, so you must scale down the reaction.

$\text{Scale factor} = (15.0)/(18.8) = \mathbf{0.798}$

(b) Corrected volumes of NaOH and H₂SO₄

V = 0.798 × 20 mL = 16 mL

Why is it impossible to perform a liquid-liquid extraction using h2o (water)and ch3oh (methanol)?

Answers

Solution : Water and Methanol are easily miscible in any amount. so they are not preferred for the liquid-liquid extraction process.

Liquid-Liquid Extraction is also called as solvent extraction. It is the method of seperation of compound based on their relative solubilities in two different immiscible liquids. Generally we use water (polar) and an organic solvent (non-polar).

It is important that the two solvents should not be mix because it is easy to seperate them.

Water and Methanol are easily miscible in any amount. we can not seperate them easily. So that is why we can not use water and methanol as a solvent in liquid-liquid extraction process.

Answer:

they could be joined due to hydrogen bridge-type intermolecular interactions so no phase splitting will be carried out.

Explanation:

Hello,

Liquid-liquid extraction is a widely used separation operation that is suitable when relative volatilities are so close, so an extra substance is used to modify the equilibrium causing a phase splitting (two liquid immiscible phases) which could be leveraged to mechanically separate the two phases. The basic idea lies on the fact that the extra substance must be largely immiscible with the original solvent, to the solute is selectively separated, nonetheless, in this case, water and methanol are largely soluble to each other since they could be joined due to hydrogen bridge-type intermolecular interactions so no phase splitting will be carried out.

Best regards.

PLEASE HELP IMMEDIATELY

Answers

Answer:

Ability to move

Must be able to eat

grow and develop

reproduce

respond to environment

taking in food

Explanation:

Hope this helped! Have a good day!

When radioactive uranium decays to produce thorium, it also emits a particle. As seen in the balanced nuclear equation, this particle can BEST be described asA) a helium atom.
B) an alpha particle or a helium atom.
C) a beta particle or a hydrogen nucleus.
D) an alpha particle or a helium nucleus.

Answers

The radioactive uranium decays to produce thorium and it emits an alpha particle or helium atom. Thus, option A is correct.

What is radioactive decay?

Unstable heavy isotopes of elements undergo nuclear decay to produce stable atoms by the emission of charged particle such as alpha or beta particles.

Based on the emitted particle, there are two types of decay process namely alpha decay and beta decay. In alpha decay atoms emits alpha particles which are helium nuclei and the atom losses its mass number by 4 units and atomic number by two units,

In beta decay, electrons are emitted by the atom, where no change occurs in mass number and atomic number increases by one unit. Uranium undergo alpha decay by emitting alpha particle or helium nuclei.

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You’re answer would be D love!

Calculate the mass of silver chloride required to plate 265 mg of pure silver.

Answers

Given Mass of pure silver (Ag) = 265 mg

Silver chloride AgCl which is used in plating silver contains 75.27 % Ag

This means that:

A 100 mg of silver chloride contains 75.27 mg of silver

Therefore, the amount of silver chloride required to plate a sample containing 265 mg silver would correspond to:

265 mg Ag * 100 mg AgCl/75.27 mg Ag

= 352.1 mg AgCl

Answer:

0.35215 grams of silver chloride required to plate 265 mg of pure silver.

Explanation:

$2AgCl(aq)\rightarrow 2Ag(s)+Cl_2(g)$

Mass of silver = 265 mg = 0.265 g

Moles of silver = $(0.265 g)/(108 g/mol)=0.002454 mol$

According to reaction, 2 moles of silver are obtained from 2 moles of silver chloride.

Then 0.002454 moles of silver will be obtained from :

$(2)/(2)* 0.002454 mol=0.002454 mol$ of silver chloride

Mass of 0.002454 moles of silver chloride:

= 0.002454 mol × 143.5 g/mol = 0.35215 g

0.35215 grams of silver chloride required to plate 265 mg of pure silver.

Consider the titration of 30 mL of 0.030 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added: a) 0 mL; b) 10 mL; c) 20 mL; d)35 mL; e) 36 mL; f) 37 mL.

Answers

The pH of the solution in the titration of 30 mL of 0.030 M NH₃ with 0.025 M HCl, is:

a) pH = 10.86

b) pH = 9.66

c) pH = 9.15

d) pH = 7.70

e) pH = 5.56

f) pH = 3.43

Calculating the pH

a) 0 mL

Initially, the pH of the solution is given by the dissociation of NH₃ in water.

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻ (1)

The constant of the above reaction is:

$Kb = ([NH_(4)^(+)][OH^(-)])/([NH_(3)]) = 1.76\cdot 10^(-5)$ (2)

At the equilibrium, we have:

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻ (3)

0.030 M - x x x

$1.76\cdot 10^(-5)*(0.030 - x) - x^(2) = 0$

After solving for x and taking the positive value:

x = 7.18x10⁻⁴ = [OH⁻]

Now, we can calculate the pH of the solution as follows:

$pH = 14 - pOH = 14 + log(7.18\cdot 10^(-4)) = 10.86$

Hence, the initial pH is 10.86.

b) 10 mL

After the addition of HCl, the following reaction takes place:

NH₃ + HCl ⇄ NH₄⁺ + Cl⁻ (4)

We can calculate the pH of the solution from the equilibrium reaction (3).

$1.76\cdot 10^(-5)(Cb - x) - (Ca + x)*x = 0$ (5)

Finding the number of moles of NH₃ and NH₄⁺

The number of moles of NH₃ (nb) and NH₄⁺ (na) are given by:

$n_(b) = n_(i) - n_(HCl)$ (6)

$n_(b) = 0.030 mol/L*0.030 L - 0.025 mol/L*0.010 L = 6.5\cdot 10^(-4) moles$

$n_(a) = n_(HCl)$ (7)

$n_(a) = 0.025 mol/L*0.010 L = 2.5 \cdot 10^(-4) moles$

Calculating the concentrations of NH₃ and NH₄⁺

The concentrations are given by:

$Cb = (6.5\cdot 10^(-4) moles)/((0.030 L + 0.010 L)) = 0.0163 M$ (8)

$Ca = (2.5 \cdot 10^(-4) mole)/((0.030 L + 0.010 L)) = 6.25 \cdot 10^(-3) M$ (9)

Calculating the pH

After entering the values of Ca and Cb into equation (5) and solving for x, we have:

$1.76\cdot 10^(-5)(0.0163 - x) - (6.25 \cdot 10^(-3) + x)*x = 0$

x = 4.54x10⁻⁵ = [OH⁻]

Then, the pH is:

$pH = 14 + log(4.54\cdot 10^(-5)) = 9.66$

Hence, the pH is 9.66.

c) 20 mL

We can find the pH of the solution from the reaction of equilibrium (3).

Calculating the concentrations of NH₃ and NH₄⁺

The concentrations are (eq 8 and 9):

$Cb = (0.030 mol/L*0.030 L - 0.025 mol/L*0.020 L)/((0.030 L + 0.020 L)) = 8.0\cdot 10^(-3) M$

$Ca = (0.025 mol/L*0.020 L)/((0.030 L + 0.020 L)) = 0.01 M$

Calculating the pH

After solving the equation (5) for x, we have:

$1.76\cdot 10^(-5)(8.0\cdot 10^(-3) - x) - (0.01 + x)*x = 0$

x = 1.40x10⁻⁵ = [OH⁻]

Then, the pH is:

$pH = 14 + log(1.40\cdot 10^(-5)) = 9.15$

So, the pH is 9.15.

d) 35 mL

We can find the pH of the solution from reaction (3).

Calculating the concentrations of NH₃ and NH₄⁺

$Cb = (0.030 mol/L*0.030 L - 0.025 mol/L*0.035 L)/((0.030 L + 0.035 L)) = 3.85\cdot 10^(-4) M$

$Ca = (0.025 mol/L*0.035 L)/((0.030 L + 0.035 L)) = 0.0135 M$

Calculating the pH

After solving the equation (5) for x, we have:

$1.76\cdot 10^(-5)(3.85\cdot 10^(-4) - x) - (0.0135 + x)*x = 0$

x = 5.013x10⁻⁷ = [OH⁻]

Then, the pH is:

$pH = 14 + log(5.013\cdot 10^(-7)) = 7.70$

So, the pH is 7.70.

e) 36 mL

Finding the number of moles of NH₃ and NH₄⁺

$n_(b) = 0.030 mol/L*0.030 L - 0.025 mol/L*0.036 L = 0$

$n_(a) = 0.025 mol/L*0.036 L = 9.0 \cdot 10^(-4) moles$

Since all the NH₃ reacts with the HCl added, the pH of the solution is given by the dissociation reaction of the NH₄⁺ produced in water.

At the equilibrium, we have:

NH₄⁺ + H₂O ⇄ NH₃ + H₃O⁺

Ca - x x x

$Ka = (x^(2))/(Ca - x)$

$Ka(Ca - x) - x^(2) = 0$ (10)

Calculating the acid constant of NH₄⁺

We can find the acid constant as follows:

$Kw = Ka*Kb$

Where Kw is the constant of water = 10⁻¹⁴

$Ka = (1\cdot 10^(-14))/(1.76 \cdot 10^(-5)) = 5.68 \cdot 10^(-10)$

Calculating the pH

The concentration of NH₄⁺ is:

$Ca = (9.0 \cdot 10^(-4) moles)/((0.030 L + 0.036 L)) = 0.0136 M$

After solving the equation (10) for x, we have:

x = 2.78x10⁻⁶ = [H₃O⁺]

Then, the pH is:

$pH = -log(H_(3)O^(+)) = -log(2.78\cdot 10^(-6)) = 5.56$

Hence, the pH is 5.56.

f) 37 mL

Now, the pH is given by the concentration of HCl that remain in solution after reacting with NH₃ (HCl is in excess).

Calculating the concentration of HCl

$C_(HCl) = (0.025 mol/L*0.037 L - 0.030 mol/L*0.030 L)/((0.030 L + 0.037 L)) = 3.73 \cdot 10^(-4) M = [H_(3)O^(+)]$

Calculating the pH

$pH = -log(H_(3)O^(+)) = -log(3.73 \cdot 10^(-4)) = 3.43$

Therefore, the pH is 3.43.

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Answer:

a)10.87

b)9.66

c)9.15

d)7.71

e) 5.56

f) 3.43

Explanation:

tep 1: Data given

Volume of 0.030 M NH3 solution = 30 mL = 0.030 L

Molarity of the HCl solution = 0.025 M

Step 2: Adding 0 mL of HCl

The reaction: NH3 + H2O ⇔ NH4+ + OH-

The initial concentration:

[NH3] = 0.030M [NH4+] = 0M [OH-] = OM

The concentration at the equilibrium:

[NH3] = 0.030 - XM

[NH4+] = [OH-] = XM

Kb = ([NH4+][OH-])/[NH3]

1.8*10^-5 = x² / 0.030-x

1.8*10^-5 = x² / 0.030

x = 7.35 * 10^-4 = [OH-]

pOH = -log [7.35 * 10^-4]

pOH = 3.13

pH = 14-3.13 = 10.87

Step 3: After adding 10 mL of HCl

The reaction:

NH3 + HCl ⇔ NH4+ + Cl-

NH3 + H3O+ ⇔ NH4+ + H2O

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.010 L = 0.00025 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.00025 =0.00065 moles

Moles HCl = 0

Moles NH4+ = 0.00025 moles

Concentration at the equilibrium:

[NH3]= 0.00065 moles / 0.040 L = 0.01625M

[NH4+] = 0.00625 M

pOH = pKb + log [NH4+]/[NH3]

pOH = 4.75 + log (0.00625/0.01625)

pOH = 4.34

pH = 9.66

Step 3: Adding 20 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.020 L = 0.00050 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.00050 =0.00040 moles

Moles HCl = 0

Moles NH4+ = 0.00050 moles

Concentration at the equilibrium:

[NH3]= 0.00040 moles / 0.050 L = 0.008M

[NH4+] = 0.01 M

pOH = pKb + log [NH4+]/[NH3]

pOH = 4.75 + log (0.01/0.008)

pOH = 4.85

pH = 14 - 4.85 = 9.15

Step 4: Adding 35 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.035 L = 0.000875 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.000875 =0.000025 moles

Moles HCl = 0

Moles NH4+ = 0.000875 moles

Concentration at the equilibrium:

[NH3]= 0.000025 moles / 0.065 L = 3.85*10^-4M

[NH4+] = 0.000875 M / 0.065 L = 0.0135 M

pOH = pKb + log [NH4+]/[NH3]

pOH = 4.75 + log (0.0135/3.85*10^-4)

pOH = 6.29

pH = 14 - 6.29 = 7.71

Step 5: adding 36 mL HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.036 L = 0.0009 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.0009 =0 moles

Moles HCl = 0

Moles NH4+ = 0.0009 moles

[NH4+] = 0.0009 moles / 0.066 L = 0.0136 M

Kw = Ka * Kb

Ka = 10^-14 / 1.8*10^-5

Ka = 5.6 * 10^-10

Ka = [NH3][H3O+] / [NH4+]

Ka =5.6 * 10^-10 = x² / 0.0136

x = 2.76 * 10^-6 = [H3O+]

pH = -log(2.76 * 10^-6)

pH = 5.56

Step 6: Adding 37 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.037 L = 0.000925 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.000925 =0 moles

Moles HCl = 0.000025 moles

Concentration of HCl = 0.000025 moles / 0.067 L = 3.73 * 10^-4 M

pH = -log 3.73*10^-4= 3.43

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In addition to displacing halide ions, the acetylide ion also adds to carbonyl groups. 2-Methyl-3-butyn-2-ol (MBI) is an acetylenic alcohol used in the manufacture of products for the agrochemical and specialty chemical industry. It can be synthesized by the addition of acetylene to acetone to form the alkoxide ion and, as a second step, protonation of the alkoxide ion to produce the alcohol. Complete the mechanism for 2-methyl-3-butyn-2-ol production by drawing in the products of each step and the missing curved arrows. Sodium amide deprotonates the terminal alkyne to form sodium ethynide. Draw all missing reactants and/or products in the appropriate boxes by placing atoms on the canvas and connecting them with bonds. Add charges where needed. Electron flow arrows should start on an atom or a bond and should end on an atom, bond, or location where a new bond should be create