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2 years ago

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Three apprentice tailors (X, Y, and Z) are assigned the task of measuring the seam of a pair of trousers. Each one makes three m

easurements. The results in inches are X (31.5, 31.6, 31.4); Y (32.8, 32.3, 32.7); Z (31.9, 32.2, 32.1).The true length is 32.0 in. Comment on the precision and the accuracy of each tailor's measurements.

1 answer:

goldenfox [79]2 years ago

7 0

Answer:

Explanation:

Average of reading of X = 31.5

Average of reading of Y = 32.6

Average of reading of Z = 32.06

True measurement is 32.0 so Z's is closest to the true value so Z is most accurate and Y is least accurate.

Since all of them have reading instrument which can read upto 0.1 so the precision of their measuring instrument is equal . Actually their measuring instrument appears to be same.so precision of their measurement is same.

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A 40.0 mL sample of 0.25 M KOH is added to 60.0 mL of 0.15 M Ba(OH)2. What is the molar concentration of OH-(aq) in the resultin

solmaris [256]

Answer:

C) 0.28 M

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Potassium hydroxide will furnish hydroxide ions as:

$KOH\rightarrow K^{+}+OH^-$

Given :

<u>For Potassium hydroxide : </u>

Molarity = 0.25 M

Volume = 40.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 40.0×10⁻³ L

Thus, moles of hydroxide ions furnished by Potassium hydroxide is same as the moles of Potassium hydroxide as shown below:

$Moles =0.25 \times {40.0\times 10^{-3}}\ moles$

Moles of hydroxide ions by Potassium hydroxide = 0.01 moles

Barium hydroxide will furnish hydroxide ions as:

$Ba(OH)_2\rightarrow Ba^{2+}+2OH^-$

Given :

<u>For Barium hydroxide : </u>

Molarity = 0.15 M

Volume = 60.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 60.0×10⁻³ L

Thus, moles of hydroxide ions furnished by Barium hydroxide is twice the moles of Barium hydroxide as shown below:

$Moles =2\times 0.15 \times {60.0\times 10^{-3}}\ moles$

Moles of hydroxide ions by Barium hydroxide = 0.018 moles

Total moles = 0.01 moles + 0.018 moles = 0.028 moles

Total volume = 40.0×10⁻³ L + 60.0×10⁻³ L = 100×10⁻³ L

Concentration of hydroxide ions is:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity_{OH^-}=\frac{0.028 }{100\times 10^{-3}}$

<u> The final concentration of hydroxide ion = 0.28 M</u>

5 0

2 years ago

Read 2 more answers

Write the balanced Ka and Kb reactions for HSO3- in water. Be sure to include the physical states of each species involved in th

Hunter-Best [27]

Answer:

Ka = [H₃O⁺] [SO₃²⁻] / [HSO₃⁻]

Kb = [OH⁻] [H₂SO₃] / [HSO₃⁻]

Explanation:

An amphoteric substance as HSO₃⁻ is a substance that act as either an acid or a base. When acid:

HSO₃⁻(aq) + H₂O(l) ⇄ H₃O⁺(aq) + SO₃²⁻(aq)

And Ka, the acid dissociation constant is:

<h3>Ka = [H₃O⁺] [SO₃²⁻] / [HSO₃⁻]</h3><h3 />

When base:

HSO₃⁻(aq) + H₂O(l) ⇄ OH⁻(aq) + H₂SO₃(aq)

And kb, base dissociation constant is:

<h3>Kb = [OH⁻] [H₂SO₃] / [HSO₃⁻]</h3>

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1 year ago

How might a collision between Neptune and Halley's Comet<br> affect Neptune's orbit?

Damm [24]

Answer:

It will mess up the orbit around the sun

Explanation:

7 0

2 years ago

When the solutes are evenly distributed throughout a solution, we say the solution has reached _______. when the solutes are eve

german

If the solutes are dispersed evenly in their particular solvent we say that the solution has reached diffusion I believe.

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2 years ago

An evaporation–crystallization process of the type described in Example 4.5-2 is used to obtain solid potassium sulfate from an

timofeeve [1]

Answer:

feed = 220.77 kg/s; maximum production rate of solid crystal = 416 kg/s; the rate of supplying fresh feed to obtain the production rate = 1.6

Explanation:

Material or mass balance can be used to estimate the mass flow rates of all the streams in the diagram shown in the attached file.

Overall balance: $M_{1} = 175 + 11M_{2}$

Water: $0.804M_{1} = 175 + 0.6M_{2}$

Using substitution method, we have:

$M_{1}$ = 220.77 kg/s

$M_{2}$ = 4.16 kg/s

The maximum production rate of solid crystal is $10M_{2}$ = 10*4.16 = 416 kg/s

Around evaporator:

$0.45M_{5} = 175$

$M_{5} = 175/0.45 = 389$ kg/s

Around the mixing point:

$M_{1} + M_{3} = M_{4} + M_{5}$

Solid crystal: $0.196M_{1} + 0.4M_{3} = M_{4}$

Using the last two equations, we can obtain:

$0.804M_{1} + 0.6M_{3} = M_{5}$

$0.6M_{3} = 389 - 0.804*220.77 = 389 - 177.5 = 211.5$

$M_{3} = 211.5/0.6 = 352.5$ kg/s

The rate of supplying fresh feed to obtain the production rate is:

$\frac{M_{3}}{M_{1}}$ = 352.5/220.77 = 1.6

7 0

3 years ago