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3 years ago

Although I am solid, I am so light that I can float in any liquid listed on the chart. What am I?-----

Chemistry

1 answer:

evablogger [386]3 years ago

3 0

Answer:

I know that Aerogel is the lightest metal in existence, but I don't think it would help much with your answer. I mean you can give it a try?

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A 3.25 L solution is prepared by dissolving 285 g of BaBr2 in water. Determine the molarity.

Effectus [21]

Answer:

0.295 mol/L

Explanation:

Given data:

Volume of solution = 3.25 L

Mass of BaBr₂ = 285 g

Molarity of solution = ?

Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

Number of moles of solute:

Number of moles = mass/ molar mass

Molar mass of BaBr₂ = 297.1 g/mol

Number of moles = 285 g/ 297.1 g/mol

Number of moles= 0.959 mol

Molarity:

M = 0.959 mol / 3.25 L

M = 0.295 mol/L

5 0

3 years ago

The answer is: variable

sergeinik [125]

Answer:

variable: not consistent or having a fixed pattern; liable to change.

Explanation:

6 0

3 years ago

222 g of MTBE (CO(CH3)4) are added to gasoline, resulting in a total volume of 2 L of reformulated gas (RFG). Assume that the de

Eva8 [605]

Answer:

Explanation:

From the information given:

(a)

$Concentration \ in \ mg/L = \dfrac{Mass \ of \ MTBE \ in \ mg}{Total \ volume (in \ L)}$

$Concentration \ in \ mg/L = \dfrac{222 \times 10^3 \ mg}{22}$

$Concentration \ in \ mg/L = 111 \times 10^3 \ mg/L$

(b)

$number \ of \ mole s= \dfrac{mass}{molar \ mass } \\ \\ number \ of \ mole s=\dfrac{222 \ g}{88.15 \ g/mol} \\ \\ \mathbf{= 2.518 mol}$

(c)

$w/w \ percentage = \dfrac{mass \ of \ MTBE }{mass \ of \ solution (RFG)}\times 100\%$

$where; \\ \\ mass \ of \ (RFG) = 2L \times 0.70 g/mL \\ \\ mass \ of \ (RFG) = 2000 ml \times 0.70 g/mL \\ \\ mass \ of \ (RFG) = 1400 g$

∴

$w/w \ percentage = \dfrac{222 \ g}{1400 \ g}\times 100\% = \mathbf{15.8\%}$

(d)

$Volume of MTBE =\dfrac{mass \ of \ MTBE}{density \ of \ MTBE}$

$Volume \ of \ MTBE = 300 \ mL\\$

∴

$v/v\% = \dfrac{volume \ of \ MTBE}{volume \ of \ RFG} \\ \\ v/v\% =\dfrac{300 \ mL}{2000 \ mL}\times 100\% \\ \\ \mathbf{v/v\% = 15.00\%}$

(e)

$From \the \ given \ information; \\ \\ 2.5184 \ moles\ of \ MTBE contain \ 2.5184 \ mole of oxygen$

∴

$mass of oxygen MTBE = 2.5284 mol \times 16\ g/mol \\ \\ mass of oxygen MTBE = 40.3 9 \ g\\ \\ mass\ of \ RFG = 1400 g$

∴

$\% w/w = \dfrac{mass \ of \ oxygen}{mass \ of RFG }=\dfrac{40.22 \ g}{1400 \ g} \times 100\%$

$\% w/w == 2.88\%$

3 0

3 years ago

Read 2 more answers

A sample of CO2 weighing 86.34g contains how many molecules?

irakobra [83]

Answer:

1.181 × 10²⁴ molecules CO₂

General Formulas and Concepts:

Chemistry - Atomic Structure

Reading a Periodic Table
Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

Step 1: Define

86.34 g CO₂

Step 2: Identify Conversion

Avogadro's Number

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol

Step 3: Convert

$86.34 \ g \ CO_2(\frac{1 \ mol \ CO_2}{44.01 \ g \ CO_2} )(\frac{6.022 \cdot 10^{23} \ molecules \ CO_2}{1 \ mol \ CO_2} )$ = 1.18141 × 10²⁴ molecules CO₂

Step 4: Check

We are given 4 sig figs. Follow sig fig rules and round.

1.18141 × 10²⁴ molecules CO₂ ≈ 1.181 × 10²⁴ molecules CO₂

4 0

3 years ago

An experiment is performed in which a crystalline substance is added to a beaker filled with room temperature water. The followi

Dafna1 [17]

A i think....don't take my word for it ompletely

3 0

3 years ago