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Vaselesa [24]
2 years ago
11

How does heat move between objects that are touching?

Chemistry
1 answer:
Tom [10]2 years ago
6 0

Answer:

<em>Conduction</em>

Explanation:

Conduction is heat transfer between two objects that are touching each other. When two objects touch and one has a higher temperature than the other; heat is transferred to the object with the lower temperature. ... In radiation, heat is transferred from one object to another by electromagnetic waves of energy. <em>Edit </em><em>:</em> <em>Please </em><em>mark </em><em>brainliest</em><em> </em><em>if </em><em>it </em><em>helped,</em><em> </em><em>thanks.</em>

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O, li, c which electrons experience the greatest effective nuclear charge? express your answer as a chemical formula.
Nataly_w [17]
<span>The effective nuclear charge of an atom = total electrons - inner electrons For O, ENC = 8 - 2 = 6 For Li, ENC = 3 - 2 = 1 For C, ENC = 6 - 2 = 4 The electrons in O experience the greatest effective nuclear charge and that is why O is smaller than C (which is smaller than Li).</span>
5 0
3 years ago
Use the Henderson-Hasselbalch equation and your knowledge of ionization to help you answer this question. Aspirin is a weak acid
Leya [2.2K]

Answer:

The protonated form is predominant when aspirin is absorbed more readily. The ratio of conjugate base to acid is 1 to 100.

Explanation:

Aspirin is more readily absorbed when it is protonated, that is when pH is lower than pKa (<em>more H⁺ available in the medium</em>). We can confirm this using Henderson-Hasselbalch equation for pH = 1.5:

pH = pKa +log(\frac{conjugate base}{weak acid} )\\1.5=3.5 +log(\frac{conjugate base}{weak acid} )\\-2.0=log(\frac{conjugate base}{weak acid} )\\10^{-2.0} =\frac{conjugate base}{weak acid}\\\frac{conjugate base}{weak acid}=0.01=\frac{1}{100}

When aspirin is absorbed more readily the ratio of conjugate base to acid is 1 to 100, being the acid the <em>predominant</em> form.

6 0
3 years ago
The Great Burdock plant’s seeds have spines on them that attach to the fur of animals that brush against it. The seed then trave
Sphinxa [80]
The answer is B. Commensalism.

<span>Commensalism is a relationship between two organisms in which only one of them has benefit, and the other one is not affected. In this example, the Great Burdock's plants spread their seeds using animals, so they benefit from this relationship. On the other hand, animals neither have benefits not are harmed from the relationship.</span>
3 0
3 years ago
Read 2 more answers
How many moles are in 7.2x10^15 atoms of Pb?​
fiasKO [112]
<h3>Answer:</h3>

1.2 × 10⁻⁸ mol Pb

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 7.2 × 10¹⁵ atoms Pb

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                     \displaystyle 7.2 \cdot 10^{15} \ atoms \ Pb(\frac{1 \ mol \ Pb}{6.022 \cdot 10^{23} \ atoms \ Pb})
  2. [DA] Multiply/Divide [Cancel out units]:                                                            \displaystyle 1.19562 \cdot 10^{-8} \ mol \ Pb

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

1.19562 × 10⁻⁸ mol Pb ≈ 1.2 × 10⁻⁸ mol Pb

6 0
2 years ago
Naphthalene, C10H8, melts at 80.2°C. If the vapour pressure of the liquid is 1.3 kPa at 85.8°C and 5.3 kPa at 119.3°C, use th
sweet-ann [11.9K]

(a) One form of the Clausius-Clapeyron equation is

ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂); where in this case:

  • P₁ = 1.3 kPa
  • P₂ = 5.3 kPa
  • T₁ = 85.8°C = 358.96 K
  • T₂ = 119.3°C = 392.46 K

Solving for ΔHv:

  • ΔHv = R * ln(P₂/P₁) / (1/T₁ - 1/T₂)
  • ΔHv = 8.31 J/molK * ln(5.3/1.3) / (1/358.96 - 1/392.46)
  • ΔHv = 49111.12 J/molK

(b) <em>Normal boiling point means</em> that P = 1 atm = 101.325 kPa. We use the same formula, using the same values for P₁ and T₁, and replacing P₂ with atmosferic pressure, <u>solving for T₂</u>:

  • ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂)
  • 1/T₂ = 1/T₁ - [ ln(P₂/P₁) / (ΔHv/R) ]
  • 1/T₂ = 1/358.96 K - [ ln(101.325/1.3) / (49111.12/8.31) ]
  • 1/T₂ = 2.049 * 10⁻³ K⁻¹
  • T₂ = 488.1 K = 214.94 °C

(c)<em> The enthalpy of vaporization</em> was calculated in part (a), and it does not vary depending on temperature, meaning <u>that at the boiling point the enthalpy of vaporization ΔHv is still 49111.12 J/molK</u>.

3 0
3 years ago
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