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Amiraneli [1.4K]
3 years ago
7

A 50 mL beaker has an uncertainty of ± 10%. What is the absolute uncertainty for this beaker?

Chemistry
2 answers:
Pani-rosa [81]3 years ago
5 0

Answer:

The absolute uncertainty is 10%.

Explanation:

The absolute uncertainty, sometimes referred to as absolute error, is the size of the possible range of values where by the actual or true value of a measurement probably lies.

The uncertainty of the 50 mL beaker is given as  ± 10%. This implies that the volume recorded using the beaker will either be 10 % above or below the actual volume.

The absolute uncertainty ignores the direction of the error, whether above or below, and is mostly concerned with the magnitude of the error. We simply ignore the signs of the uncertainty and this will give the value of the absolute uncertainty.

kipiarov [429]3 years ago
5 0

The absolute uncertainty for this beaker is ± 5 mL.

Explanation

The uncertainty is the measure of error an instrument can exhibit during measurement.

It is the error range in any kind of measurement.

There are two kinds of uncertainty: relative and absolute.

Relative uncertainty is the percentage of error occurred during measurement while absolute uncertainty quantifies the perfect value of error range with the measuring unit.

In this problem, the relative uncertainty was given as ±10 % and the estimated value is 50 mL.

So the conversion from relative uncertainty to absolute uncertainty can be done by the following formula

Absolute uncertainty=Relative uncertainty*Estimated value

Absolute uncertainty=10/100*50=5 mL

So the value can be written as 50 mL±5 mL with the absolute uncertainty as 5 mL.

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10. 17.5 g of NaOH were dissolved in 350 mL of water. The solution was titrated with HNO3 to a methyl orange endpoint (methyl or
Tomtit [17]

1747.2 ml of HNO3 is required to reach methyl orange endpoint.

Explanation:

Number of moles of NaOH solution having 17.5 gram NaOH in 350ml of water.

Atomic weight of NaOH = 40 g/mol

thus number of moles= 17.5/40

                                      =  0.437 moles

Now the molarity of the solution is calculated as

M=n/C

  = 0.437/O.35 LITRES

  = 1.248 M solution of the base

the molarity of acid is given as 0.25 M of HNO3

From the formula

M1V1=M2V2  we can calculate the volume of HNO3 required to make solution acidic.

1.248*350= 0.25*v2

v2=  1.248*350/0.25

     = 1747.2 ml

Thus 1747.2 ml of HNO3 is required.

3 0
3 years ago
How many moles of ethane (C2H6) would be needed to react with 62.3 grams of oxygen gas?
Black_prince [1.1K]
The combustion of ethane is expressed in the balanced reaction C2H6 + 3.5O2= 2CO2 + 3 H2O. Given the mass of oxygen gas, we get the moles of ethane needed by converting this mass to mole (dividing my 32 g/mol), then multiply by 1/3.5 (stioch ratio) and the molar mass of ethane (30 g/mol). The answer is 16.69 grams ethane. 
6 0
3 years ago
What mass of aluminum sulfate is required to precipitate all of the Ba?? out of 45.0 mL of 0.548 M barium nitrate solution? 3 Ba
Verizon [17]

Answer:

2.82 g

Explanation:

Step 1: Write the balanced precipitation reaction

3 Ba(NO₃)₂ (aq) + Al₂(SO₄)₃ (aq) ⇒ 3 BaSO₄(s) + 2 Al(NO₃)₃(aq)

Step 2: Calculate the reacting moles of Ba(NO₃)₂

45.0 mL (0.0450 L) of 0.548 M Ba(NO₃)₂ react.

0.0450 L × 0.548 mol/L = 0.0247 mol

Step 3: Calculate the moles of Al₂(SO₄)₃ that react with 0.0247 moles of Ba(NO₃)₂

The molar ratio of Ba(NO₃)₂ to Al₂(SO₄)₃ is 3:1. The reacting moles of Al₂(SO₄)₃ are 1/3 × 0.0247 mol = 8.23 × 10⁻³ mol

Step 4: Calculate the mass corresponding to 8.23 × 10⁻³ moles of Al₂(SO₄)₃

The molar mass of Al₂(SO₄)₃ is 342.2 g/mol.

8.23 × 10⁻³ mol × 342.2 g/mol = 2.82 g

5 0
3 years ago
what is the concentration of an NaOH solution that requires 50 mL of a 1.25 M H2SO4 solution to neutralize 78.0 ml of NaOH​
Dimas [21]

Answer:

  • The answer is the concentration of an NaOH = 1.6 M

Explanation:

The most common way to solve this kind of problem is to use the formula  

  • C₁ * V₁ = C₂ * V₂

In your problem,

For NaOH

C₁ =??     v₁= 78.0 mL = 0.078 L

For H₂SO₄

C₁ =1.25 M     v₁= 50.0 mL = 0.05 L

but you must note that for the reaction of NaOH with H₂SO₄

2 mol of NaOH raect with 1 mol H₂SO₄

So, by applying in above formula

  • C₁ * V₁ = 2 * C₂ * V₂
  • (C₁ * 0.078 L) = (2*  1.25 M * 0.05 L)
  • C₁ = (2*  1.25 M * 0.05 L) / (0.078 L) = 1.6 M  

<u>So, the answer is the concentration of an NaOH = 1.6 M</u>

3 0
3 years ago
A 29.4 ml sample of 0.347 m triethylamine, (c2h5)3n, is titrated with 0.375 m hydrobromic acid. at the equivalence point, the ph
Debora [2.8K]

A 29.4 ml sample of 0.347 m triethylamine, (c2h5)3n, is titrated with 0.375 m hydrobromic acid. at the equivalence point, the pH is 5.81.

Given,

Volume of Triethylamine (C₂H₅)₃N = 29.4 ml = 0.0294 L

Molarity of (C₂H₅)₃N = 0.275 M

Molarity of HBr = 0.375M

Solution:

(C₂H₅)₃N + HBr ⇔ (C₂H₅)₃ NH⁺  +  Br⁻

⇒ Volume of HBr = mole of (C₂H₅)₃N / Molarity of HBr

⇒ Volume of HBr = 0.00572 mol / 0.375 mol/l

⇒ Volume of HBr = 0.0152 L

∴ Total Volume = 0.0294 + 0.0152

⇒ Total Volume = 0.0446 L

Concentration of (C₂H₅)₃NH⁺ = \frac{0.00572}{0.0446} ⇒ 0.128 M,

(C₂H₅)₃NH⁺ + H₃O⁺ ⇄ (C₂H₅)₃N + H₃O⁺

let's assume, (C₂H₅)₃N = x,

⇒ H₃O⁺ = x, and

⇒ (C₂H₅)₃NH⁺ = 0.128 - x

∴ Now, k = kw / kb

⇒ k = 1.9 × 10⁻¹¹

and, k = [(C₂H₅)₃N][H₃O⁺] / (C₂H₅)₃NH⁺

⇒ 1.9 × 10⁻¹¹ = x^{2} / (0.128 - x)

Thus,  x = 1.55 × 10⁻⁶ M,

Hence, pH = - log[x]

⇒ - log [1.55 × 10⁻⁶ ]

⇒ - log (1.55) - log (10⁻⁶)

⇒ - log (1.55) + 6log10

⇒ - 0.19 + 6

⇒ 5.81 = pH

Therefore, pH = 5.81.

To learn more about equivalence point here

brainly.com/question/14782315

#SPJ4

7 0
2 years ago
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