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qwelly [4]
4 years ago
8

What is thermal boundary layers

Chemistry
1 answer:
bagirrra123 [75]4 years ago
8 0

Answer:

the layer of a liquid or gaseous heat-transfer agent between the free stream and a heat-exchange surface. In this layer the temperature of the heat-transfer agent changes from that of the wall to that of the free stream

Explanation:

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1. Replace ? with a whole number to make the statements true.
Ira Lisetskai [31]
<span>1.) a. 20 ÷ 4 ? means ? × 4 = 20     </span>→ replace ? with 5.<span>
b. 2,725 ÷ 5 ? means ? × 5 = 2,725  </span>→ replace ? with 545<span>
c. ? ÷ 5 = 0 </span>→ replace ? with 0.

2) 1,475 / 25 = 59
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5 0
4 years ago
What were the units of light energy emitted by blackbody radiation originally called?
Dafna11 [192]
The units of light energy emitted by blackbody radiation is originally called as quanta. Black body radiation is a type of radiation within a boy in thermodynamic equilibrium. This is a theoretical body. In 1900, Planck proposed that light and other electromagnetic waves were emitted in discrete packets of energy which were called "quanta". This was the original unit used.
5 0
4 years ago
A sample initially contains 8.0 moles of a radioactive isotope. How much of the sample remains after four half-lives? Express yo
Tju [1.3M]

Answer:

0.50 mol

Explanation:

The half-life is <em>the time required for the amount of a radioactive isotope to decay to half that amount</em>.

Initially, there are 8.0 moles.

  • After 1 half-life, there remain 1/2 × 8.0 mol = 4.0 mol.
  • After 2 half-lives, there remain 1/2 × 4.0 mol = 2.0 mol.
  • After 3 half-lives, there remain 1/2 × 2.0 mol = 1.0 mol.
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6 0
3 years ago
1. How many valence electrons are in an atom of phosphorus? (atomic number 15)
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1. D. 5
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Hope this helps you!!
Have a wonderful day!
7 0
4 years ago
Combustion analysis of 63.8 mg of a c, h and o containing compound produced 145.0 mg of co2 and 59.38 mg of h2o. what is the emp
Makovka662 [10]
<span>: The empirical formula for the compound is C3H60 (see below) CO2 is the only product containing C, C produced = 145.0 mg CO2 x (1 g / 1000 mg) x (1 mole CO2 / 44.0 g CO2) x (1 mole C / 1 mole CO2) = 0.00330 moles C. H2O is the only product containing H, H produced = 59.38 mg H2O x (1 g / 1000 mg) x (1 mole H2O / 18.0 g H2O) x (2 moles H / 1 mole H2O) = 0.00660 moles H. Oxygen is in both and the unknown reacts with oxygen(in the air) 0.00330 moles C x (12.0 g C / 1 mole C) = 0.0396 g C = 39.6 mg C 0.00660 moles H x (1.01 g H / 1 mole H) = 0.00667 g H = 6.7 mg H Because the unknown weighed 63.8 mg and consists off justC, H, and O, then mass O = g unknown - g C - g H = 63.8 mg - 39.6 mg - 6.7 mg = 17.5 mg = 0.0175 g 0.0175 g O x (1 mole O / 16.0 g O) = 0.00109 moles O The mole ratio of C:H:O is: C = 0.00330 H = 0.00660 O = 0.00109 Divide by the smallest you get: C = 0.00330 / 0.00109 = 3.03 H = 0.00660 / 0.00109 = 6.06 O = 0.00109 / 0.00109 = 1.00</span>
6 0
4 years ago
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