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babunello [35]
3 years ago
13

A membrane Is a flexible outer layer found on:

Chemistry
1 answer:
ratelena [41]3 years ago
8 0

Answeridk:

Explanation:isk

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What is the median of the following set: 2, 4, 6, 8, and 10?
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Answer:9

Explanation:

(3+8+10+15)/4 = 36/4 = 9

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3 years ago
WILL MARK BRAINLIEST!!!!
Lyrx [107]

Answer:

5.702 mol K₂SO₄

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Compounds
  • Moles

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[Given] 993.6 g K₂SO₄

[Solve] moles K₂SO₄

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

[PT] Molar Mass of K: 39.10 g/mol

[PT] Molar Mass of S: 32.07 g/mol

[PT] Molar mass of O: 16.00 g/mol

Molar Mass of K₂SO₄: 2(39.10) + 32.07 + 4(16.00) = 174.27 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 993.6 \ g \ K_2SO_4(\frac{1 \ mol \ K_2SO_4}{174.27 \ g \ K_2SO_4})
  2. [DA] Divide [Cancel out units]:                                                                         \displaystyle 5.7015 \ mol \ K_2SO_4

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 4 sig figs.</em>

5.7015 mol K₂SO₄ ≈ 5.702 mol K₂SO₄

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3 years ago
According to Sam's soap recipe, she needs 855
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7 0
3 years ago
The normal boiling point of bromine is 58.8°C, and its enthalpy of vaporization is 30.91 kJ/mol. What is the approximate vapor p
saul85 [17]

Answer : The vapor pressure of bromine at 10.0^oC is 0.1448 atm.

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of bromine at 10.0^oC = ?

P_2 = vapor pressure of propane at normal boiling point = 1 atm

T_1 = temperature of propane = 10.0^oC=273+10.0=283.0K

T_2 = normal boiling point of bromine = 58.8^oC=273+58.8=331.8K

\Delta H_{vap} = heat of vaporization = 30.91 kJ/mole = 30910 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{1atm}{P_1})=\frac{30910J/mole}{8.314J/K.mole}\times (\frac{1}{283.0K}-\frac{1}{331.8K})

P_1=0.1448atm

Hence, the vapor pressure of bromine at 10.0^oC is 0.1448 atm.

4 0
3 years ago
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