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Elanso [62]
3 years ago
11

Which color changes represent a positive reaction for the presence of starch using the iodine test?

Chemistry
1 answer:
-BARSIC- [3]3 years ago
3 0
Looking at a ph level color chart, it should be moving to more acidic if it’s positive
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What layers are composed of metals? Choose all that apply.<br> Inner core<br> Outer core<br> Mantle
svlad2 [7]

Answer:

to be honest ask your teacher

3 0
2 years ago
What is the product if an atom of Po-209<br> undergoes alpha decay?
g100num [7]
<h3>Answer:</h3>

Lead-205 (Pb-205)

<h3>Explanation:</h3>

<u>We are given;</u>

  • An atom of Po-209

We are supposed to identify its product after an alpha decay;

  • Polonium-209 has a mass number of 209 and an atomic number of 84.
  • When an element undergoes an alpha decay, the mass number decreases by 4 while the atomic number decreases by 2.
  • Therefore, when Po-209 undergoes alpha decay it results to the formation of a product with a mass number of 205 and atomic number of 82.
  • The product from this decay is Pb-205, because Pb-205 has a mass number of 205 and atomic number 82.
  • The equation for the decay is;

²⁰⁹₈₄Po → ²⁰⁵₈₂Pb + ⁴₂He

  • Note; An alpha particle is represented by a helium nucleus, ⁴₂He.

7 0
3 years ago
If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcoho
ankoles [38]

If you place 1.0 L of ethanol (C2H5OH) in a small laboratory that is 3.0 m long, 2.0 m wide, and 2.0 m high, will all the alcohol evaporate? If some liquid remains, how much will there be? The vapor pressure of ethyl alcohol at 25 °C is 59 mm Hg, and the density of the liquid at this temperature is 0.785g/cm^3 .

will all the alcohol evaporate? or none at all?

Answer:

Yes, all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore  be zero.

Explanation:

Given that:

The volume of alcohol which is placed in a small laboratory = 1.0 L

Vapor pressure of ethyl alcohol  at 25 ° C = 59 mmHg

Converting 59 mmHg to atm ; since 1 atm = 760 mmHg;

Then, we have:

= \frac{59}{760}atm

= 0.078 atm

Temperature = 25 ° C

= ( 25 + 273 K)

= 298 K.

Density of the ethanol = 0.785 g/cm³

The volume of laboratory = l × b × h

= 3.0 m × 2.0 m × 2.5 m

= 15 m³

Converting the volume of laboratory to liter;

since 1 m³ = 100 L; Then, we  have:

15 × 1000 = 15,000 L

Using ideal gas equation to determine the moles of ethanol in vapor phase; we have:

PV = nRT

Making n the subject of the formula; we have:

n = \frac{PV}{RT}

n = \frac{0.078 * 15000}{0.082*290}

n = 47. 88 mol of ethanol

Moles of ethanol in 1.0 L bottle can be calculated as follows:

Since  numbers of moles = \frac{mass}{molar mass}

and mass = density × vollume

Then; we can say ;

number of moles = \frac{density*volume }{molar mass of ethanol}

number of moles =\frac{0.785g/cm^3*1000cm^3}{46.07g/mol}

number of moles = \frac{&85}{46.07}

number of moles = 17.039 mol

Thus , all the ethanol present in the laboratory will evaporate since the mole of ethanol present in vapor is greater. The volume of ethanol left will therefore be zero.

5 0
2 years ago
What is the number of molecules of c2h5oh in a 3m solution that contains 4.00kg h2o?
Veronika [31]
 The number  of C2H5OH  in a 3 m solution that contain 4.00kg H2O is calculate as below

M = moles of the solute/Kg  of water

that is 3M = moles of solute/ 4 Kg
multiply  both side by  4

moles of the  solute  is therefore = 12  moles

by use of Avogadro law constant

1 mole =6.02 x10^23  molecules

what  about 12 moles

=12 moles/1 moles  x 6.02 x10^23 = 7.224 x10^24 molecules
3 0
3 years ago
On a cool morning (12"C), a balloon is filled with 1.5 L of helium. By mid afternoon, the temperature has soared to 32°C. What i
ddd [48]

Answer:

1.6 L

Explanation:

Using Charle's law  

\frac {V_1}{T_1}=\frac {V_2}{T_2}

Given ,  

V₁ = 1.5 L

V₂ = ?

T₁ = 12 °C

T₂ = 32 °C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (12 + 273.15) K = 285.15 K  

T₂ = (32 + 273.15) K = 305.15 K  

Using above equation as:

\frac{1.5}{285.15}=\frac{V_2}{305.15}

V_2=\frac{1.5\cdot \:305.15}{285.15}

New volume = 1.6 L

8 0
3 years ago
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