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2 years ago

How many liters of water would it require to 36 grams of the substance?

Chemistry

2 answers:

WARRIOR [948]2 years ago

6 0

It would required 3 liters of substance

Bas_tet [7]2 years ago

3 0

3 liters

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The vapor pressure of diethyl ether (ether) is 463.57 mm Hg at 25 °C. A nonvolatile, nonelectrolyte that dissolves in diethyl et

Alexxx [7]

<u>Answer:</u> The vapor pressure of solution is 459.17 mmHg

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For testosterone:</u>

Given mass of testosterone = 7.752 g

Molar mass of testosterone = 288.4 g/mol

Putting values in equation 1, we get:

$\text{Moles of testosterone}=\frac{7.752g}{288.4g/mol}=0.027mol$

<u>For diethyl ether:</u>

Given mass of diethyl ether = 208.0 g

Molar mass of diethyl ether = 74.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of diethyl ether}=\frac{208.0g}{74.12g/mol}=2.81mol$

Mole fraction of a substance is calculated by using the equation:

$\chi_A=\frac{n_A}{n_A+n_B}$

$\chi_{\text{testosterone}}=\frac{n_{\text{testosterone}}}{n_{\text{testosterone}}+n_{\text{diethyl ether}}}$

$\chi_{\text{testosterone}}=\frac{0.027}{0.027+2.81}\\\\\chi_{\text{testosterone}}=0.0095$

The formula for relative lowering of vapor pressure will be:

$\frac{p^o-p_s}{p^o}=i\times \chi_{\text{solute}}$

where,

$p^o$ = vapor pressure of solvent (diethyl ether) = 463.57 mmHg

$p^s$ = vapor pressure of the solution = ?

i = Van't Hoff factor = 1 (for non electrolytes)

$\chi_{\text{solute}}$ = mole fraction of solute (testosterone) = 0.0095

Putting values in above equation, we get:

$\frac{463.57-p^s}{463.57}=1\times 0.0095\\\\p^s=459.17mmHg$

Hence, the vapor pressure of solution is 459.17 mmHg

7 0

3 years ago

If a neutral atom has 10 proton how many electrons would it have?

olchik [2.2K]

It should have 10 electrons

7 0

3 years ago

Read 2 more answers

What is the name for NH

rjkz [21]

Answer:

Due to the lack of context for your question I’m not completely sure which you are talking about but NH can be Imidogen or it could be ammonium which is NH4^1+

4 0

3 years ago

1 Na2CO3(aq) + 1 CaCl2(aq) → 1 CaCO3(s) + 2 NaCl(aq) 4. Use the balanced chemical equation from the last question to solve this

LenKa [72]

<h3>Answer:</h3>

0.6 g NaCl

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] Na₂CO₃ (aq) + CaCl₂ (aq) → CaCO₃ (s) + 2NaCl (aq)

[Given] 0.5 g Na₂CO₃ reacted with excess CaCl₂

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

[RxN] Na₂CO₃ → 2NaCl

Molar Mass of Na - 22.99 g/mol

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of Na₂CO₃ - 2(22.99) + 12.01 + 3(16.00) = 105.99 g/mol

Molar Mass of NaCl - 22.99 + 35.45 = 58.44 g/mol

<u>Step 3: Stoichiometry</u>

Set up: $\displaystyle 0.5 \ g \ Na_2CO_3(\frac{1 \ mol \ Na_2CO_3}{105.99 \ g \ Na_2CO_3})(\frac{2 \ mol \ NaCl}{1 \ mol \ Na_2CO_3})(\frac{58.44 \ g \ NaCl}{1 \ mol \ NaCl})$
Multiply/Divide: $\displaystyle 0.551373 \ g \ NaCl$