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3 years ago

Please help!! any inappropriate answers will be reported

1 answer:

6 0

When the total surface area of the solute particles is increased, the solute dissolves more rapidly. Breaking a solute into smaller pieces increases its surface area and increases its rate of solution. So, any answer with “as surface area increases, solid dissolves faster” would be correct. :)

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The number used to measure distances outside of our solar system is a(n)

nevsk [136]

Answer:

Usually light years, or parsecs.

Explanation:

3 0

3 years ago

One of the compounds used to increase the octane rating of gasoline is toluene (pictured). Suppose 43.3 mL of toluene (d = 0.867

Thepotemich [5.8K]

<u>Answer:</u>

(A)

Density = Mass / Volume

Mass = Density × Volume

$= 0.867 g/mL \times 43.3mL = 37.5411 g Toluene$

$1C_6 H_5 CH_3 + 9 O_2 > 7 CO_2 + 4 H_2 O$

Mole ratio of toluene : Oxygen is 1 : 9

$37.5411 g \text { Toluene } \times \frac{1 \text {mol} \text {toluene}}{92 g \text { toluene}} \times \frac{9 {mol} O_{2}}{1 \text {mol} \text { toluene }} \times \frac{32 g O_{2}}{1 {mol} O_{2}}=117 g O_{2}(\text {Answer})$

(B)

1 mole of Toluene produces 7 moles of $CO_2$ gas and 4 moles of $H_2 O$ Vapour

So the mole ratio is 1 : 11

$37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene }} \times \frac{11 \mathrm{mol} \text { gas }}{1 \text { mol toluene }} $$\\\\=4.49 \text { mol gaseous products (Answer) } $$

(C)

1mole contains $6.022\times10^{23}$ molecules

$37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene}} \times \frac{4 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \text { toluene }} \times \frac{6.022 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}} $\\\\$=9.82 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O} \text { (Answer) } $$

6 0

3 years ago

A chemical test has determined the concentration of a solution of an unknown substance to be 2.41 M. a 100.0 mL volume of the so

Oksana_A [137]

Answer : The molar mass of unknown substance is, 39.7 g/mol

Explanation : Given,

Mass of unknown substance = 9.56 g

Volume of solution = 100.0 mL

Molarity = 2.41 M

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

$\text{Molarity}=\frac{\text{Mass of unknown substance}\times 1000}{\text{Molar mass of unknown substance}\times \text{Volume of solution (in mL)}}$

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

$2.41M=\frac{9.56g\times 1000}{\text{Molar mass of unknown substance}\times 100.0mL}$

$\text{Molar mass of unknown substance}=39.7g/mol$

Therefore, the molar mass of unknown substance is, 39.7 g/mol

5 0

3 years ago

What is climate? How does it differ from weather

ozzi

Weather refers to short term atmospheric conditions while climate is the weather of a specific region averaged over a long period of time. Climate change refers to long-term changes.

6 0

3 years ago

Read 2 more answers

Which formula is an empirical formula? (1) CH4 (2) C2 H6 (3) C3 H6 (4) C4 H10

rosijanka [135]

CH4 is an emprirical formula as it shows the simplest ratio of the numbers of different atoms present in the molecule. The empirical formula for CH4 is also the same as the molecular formula.
The other compunds can be simplified so they are not the empirical formula of compounds.
Hope this helps :).

3 0

3 years ago