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Novosadov [1.4K]
2 years ago
13

Based on the results of your titration, what volume of the sample would an adult need to consume to reach the recommended daily

amount? 60 mg is the recommened daily amount of vitamin C.
Ascorbic acid Volume of DCPIP in Total Volume of DCPIP

concentration (g/L) final syringe (mL) used (mL)

0.00 2.18 mL 0.18 mL

0.50 3.1 mL 0.6 mL

1.00 4.6 mL 2.2 mL

1.50 5.56 mL 3.39 mL

2.00 6.37 mL 4.37 mL

Unknown Trial 1 3.21 mL 1.21 mL

Unknown Trial 2 3.11 mL 1.11 mL

Unknown Trial 3 3.15 mL 1.06 mL

Chemistry
1 answer:
Mekhanik [1.2K]2 years ago
5 0

Answer:  85.7 mL

Explanation:

Given the information from the question as plotted in the graph i will be uploading along side this answer,

Average of total volume of DCPIP used is

= (1.21 + 1.11 + 1.06)mL / 3

= 1.12 mL

and corresponding ( ascorbic acid ) is 0.70 g/L

Two parameter given as volume of DCPIP in final syringe and total volume of DCPIP are quite ambiguous

700mg ⇒ 1 L

THEREFORE volume that contains 60mg =  (1000/700) × 60 = 85.7 mL

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Oxygen has greater electronegativity than hydrogen, because of that oxygen is partially negative and hydrogen is partially positive.
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2 years ago
What do you think happens to Difluoroethane at –24°C? Provide evidence to support your claim.
balandron [24]

Answer:

The following subsections explain the explanation according to the particular circumstance.

Explanation:

  • The boiling point seems to be the temperature beyond which the working fluid as well as the boiling phase would be at a predetermined pressure or voltage at equilibrium among one another and.  
  • The vapor or boiling temperature of 1,1 difluoroethane seems to be -25oC at 1 atm, although as a gas it can remain at a higher temperature around -24oC.
6 0
2 years ago
Most Bic lighters hold 5.0ml of liquified butane (density = 0.60 g/ml). Calculate the minimum size container you would need to "
Hatshy [7]

Answer:

Volume of container = 0.0012 m³ or 1.2 L or 1200 ml

Explanation:

Volume of butane = 5.0 ml

density = 0.60 g/ml

Room temperature (T) = 293.15 K

Normal pressure (P) = 1 atm = 101,325 pa

Ideal gas constant (R) = 8.3145 J/mole.K)

volume of container V = ?

Solution

To find out the volume of container we use ideal gas equation

PV = nRT

P = pressure

V = volume

n = number of moles

R = gas constant

T = temperature

First we find out number of moles

<em>As Mass = density × volume</em>

mass of butane = 0.60 g/ml ×5.0 ml

mass of butane = 3 g

now find out number of moles (n)

n = mass / molar mass

n = 3 g / 58.12 g/mol

n = 0.05 mol

Now put all values in ideal gas equation

<em>PV = nRt</em>

<em>V = nRT/P</em>

V = (0.05 mol × 8.3145 J/mol.K × 293.15 K) ÷ 101,325 pa

V = 121.87 ÷ 101,325 pa

V = 0.0012 m³ OR 1.2 L OR 1200 ml

8 0
2 years ago
a.) The diameter of a uranium atom is 3.50 Å . Express the radius of a uranium atom in both meters and nanometers. b.) How many
Svetradugi [14.3K]

Answer:

Th answer to your question is:

a)  3.5 x10⁻¹⁰ meters; 0.35 nm

b) 6857142.86 atoms

c) Volume = 2.06 x 10⁻²³ cm³

Explanation:

a) data

Uranium atoms = 3.5A°

meters

           1 A° ----------------  1 x 10 ⁻¹⁰ m

         3.5A° ---------------  x

 x = 3.5(1 x10⁻¹⁰)/ 1 = 3.5 x10⁻¹⁰ meters

          1 A° ------------------ 0.1 nm

        3.5 A° ---------------- 0.35 nm

b) 2.4 mm

Divide 2,40 mm / uranium diameter

But, first convert 3,5A° to mm   = 3.5 x 10⁻⁷ mm

# of uranium atoms = 2.4 / 3.5 x 10⁻⁷ = 6857142.86

c) volume in cubic cm

Convert 3.5A° to cm  = 3.5 x 10⁻⁸

Volume = 4/3 πr³ = (4/3) (3.14)(1.7 x10⁻⁸)³

Volume = 2.06 x 10⁻²³ cm³

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