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DedPeter [7]
3 years ago
14

Pipettes are used when measuring the volumes of liquids with high degree of precision. The following volumes were obtained durin

g the calibration of a 10.00 mL pipette: 10.15 mL, 9.95 mL, 9.99 mL, and 10.02 mL. Calculate the standard deviation for these measurements.
Chemistry
1 answer:
never [62]3 years ago
6 0

Answer:

0.075

Explanation:

First obtain the mean of the measurement;

Mean = 10.15 + 9.95 + 9.99 + 10.02/4 = 10.03

Then obtain d^2= (mean-score)^2 for each score;

(10.15-10.03)^2 = 0.0144

(9.95-10.03)^2 = 0.0064

(9.99-10.03)^2 = 0.0016

(10.02-10.03)^2= 0.0001

∑d^2= 0.0144 + 0.0064 + 0.0016 + 0.0001

∑d^2= 0.0225

Variance = ∑d^2/N = 0.0225/4 = 0.005625

Standard deviation= √0.005625

Standard deviation= 0.075

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What is the volume of 3.89 grams of helium gas?
lukranit [14]

Answer: There are 971.77 millimoles in 3.89 grams of Helium

Explanation:

i think its right im not really sure

7 0
3 years ago
If 5.0 g of copper metal reacts with a solution of silver nitrate, how many grams of silver metal are recovered?
sukhopar [10]

Answer:

16.9g

Explanation:Cu+2AgNO3→2Ag+Cu(NO3)2  

Cu will likely have a +2 oxidation state. It is higher in the activity series than Ag, so it is a stronger reducing agent and will reduce Ag in a displacement reaction. Then you need to balance the coefficients knowing than NO3 is -1 and Ag is +1.

Then to calculate the theoretical yield you need to compare moles of the reactants:

m(Cu)=5g

M(Cu)=63.55

n(Cu)=5/63.55=0.0787

By comparing coefficients you require twice as much silver: 0.157mol

n(Ag)=0.157

M(Ag)=107.86

m(Ag)=0.157x107.86=16.9g

Hence, the theoretical yield of this reaction would be 16.9g

3 0
3 years ago
Experiments were carried out in which a beam of cathode rays was first bent by a magnetic field and then bent back by an electro
Sergio039 [100]

a. the ratio of mass to charge of an electron

Explanation:

The experiment permitted the direct measurement of the ratio of mass to charge of an electron.

  • The charge to mass ratio of an electron was determine by accelerating a beam of cathode rays in magnetic and electric fields.
  • No matter the gas used in the tube or the nature of the material of the electrodes, the rays were found to have constant charge to mass ratio of 1.76 x 10¹¹coulombkg⁻¹.

learn more:

Subatomic particles brainly.com/question/2757829

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6 0
3 years ago
Nitrogen dioxide (NO2) cannot be obtained in a pure form in the gas phase because it exists as a mixture of NO2 and N2O4. At 16°
Pavel [41]

Answer:

PNO₂ = 0.49 atm

PN₂O₄ = 0.45 atm

Explanation:

Let's begin with the equation of ideal gas, and derivate from it an equation that  involves the density (ρ = m/V).

PV = nRT

n = m/M (m is the mass, and M the molar mass)

PV = \frac{m}{M}RT

PxM = \frac{m}{V}RT

PxM = ρRT

ρ = PxM/RT

With the density of the gas mixture, we can calculate the average of molar mass (Mavg), with the constant of the gases R = 0.082 atm.L/mol.K, and T = 16 + 273 = 289 K

2.7 = \frac{0.94xMavg}{0.082x289}

0.94Mavg = 63.9846

Mavg = 68.0687 g/mol

The molar mass of N is 14 g/mol and of O is 16 g/mol, than M_{NO2} = 46 g/mol and M_{N2O4} = 96 g/mol. Calling y the molar fraction:

Mavg = M_{NO2}y_{NO2} + M_{N2O4}y_{N2O4}

And,

y_{NO2} + y_{N2O4} = 1

y_{N2O4} = 1 - y_{NO2}

So,

68.0687 = 46y_{NO2} + 92x(1 - y_{NO2})

68.0687 - 92 = 46y_{NO2} - 92y_{NO2}

46y_{NO2} = 23.9313

y_{NO2} = 0.52

y_{N2O4} = 0.48

The partial pressure is the molar fraction multiplied by the total pressure so:

PNO₂ = 0.52x0.94 = 0.49 atm

PN₂O₄ = 0.48x0.94 = 0.45 atm

8 0
3 years ago
In the citric acid cycle (also known as the Krebs cycle), acetyl CoA is completely oxidized. From the following compounds involv
REY [17]

Answer:

We place ( O₂, CO₂, coenzyme A and acetyl CoA) into bin [not input or output];

We place (ADP, NAD⁺, Glucose) into [ Net Input] bin.;

And (ATP, NADH and Pyruvate) intothe [Net Output] bin.

Explanation:

Citric Acid Cycle is also known as Tricarboxylic Acid Cycle, Krebs Cycle and Szent-Györgyi-Krebs cycle.

It is a series of enzyme-catalyzed chemical reactions, which of central importance in all living cells that use oxygen as part of cellular respiration.

Here are few things to know about the CAC:

• The Krebs cycle uses the products of Glycolysis to produce 2 ATP molecules for each molecule of Glucose.

• Oxygen is required.

• The Krebs Cycle occurs in cellular power plant — Mitochondria of the cell.

• Each glucose molecule nets 2 pyruvic acids. The Krebs Cycle breaksdown 1 pyruvic acid one at å time.

• CAC begins and ends with Oxaloacetic Acid - a 4 carbon molecule.

• It is an 8 step cycle that begins when the Acetyl CoA that is a product of Glycolysis is 'picked up' by Oxalacetic Acid.

• Lipids, Proteins and Carbohydrates can all be metabolized in the Krebs Cycle.

• The NADH and FADH2 that are high energy co- enzymes (helper molecules) are by-products of the Krebs cycle and continue on to fuel the Electron Transport Chain.

The clues above helps us to know where to drag the compounds into;

We place these into bin of [Net Input]: ADP, NAD⁺, Glucose

We place these into the bin of [Net Output]: ATP, NADH and Pyruvate,

We place the following into [not input or output] bin: O₂, CO₂, coenzyme A and acetyl CoA

3 0
3 years ago
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