All categories

2 years ago

What formula is used for solving problems involving Boyle’s law?

Chemistry

1 answer:

GaryK [48]2 years ago

5 0

Answer:

P1V1=P2V2

Explanation:

because volume is inversely proportional to pressure provided that temperature remains constant

You might be interested in

Which formula represents lead(II) hydroxide?

frez [133]

Pb(OH)2

This is the formula

5 0

3 years ago

What is the number of moles in 15.0 g AsH3?

pentagon [3]

We are given with the mass of Arsine ( $AsH_{3}$

The mass of arsine is 15g

there is a relation between moles, mass and molar mass of any compound which is

$moles=\frac{mass}{molarmass}$

The molar mass of Arsine = atomic mass of As + 3X atomic mass of H

the molar mass of Arsine = 74.92 + 3X 1 = 77.92 g/mol

Let us calculate the moles as

$moles=\frac{15}{77.92}=0.192mol$

7 0

3 years ago

If 16.0 grams of aluminum oxide were actually produced, what is the percent yield of the reaction below given that you start wit

Svetllana [295]

Answer: The percent of yield is 50.03%.

Explanation:

First, we need to balance the equation:

$4Al + 3O_{2}$ ⇒ $2Al_{2}O_{3}$

We need to remember that the chemical equations are written in moles, so we have to convert the amounts in grams to moles, using the molecular weight of every compound: Al2O3 (101.96 g/mol), Al (29.98 g/mol) and O2 (31.99 g/mol).

In consequence, the amounts in moles of every compound will be:

$16 gAl_{2}O_{3} * \frac{1 mol}{101.96 g} =0.157 mol Al_{2}O_{3}$

$10 g Al * \frac{1mol}{29.98 g}= 0.333 mol Al$

$19 g O_{2}*\frac{1 mol}{31.99 g} =0.594 mol O_{2}$

Now, we have to find out which is the limit reagent or in other words, which of the reagents will be consumed first, taking into account the stoichiometric ratio of the balanced equation:

$0.333 mol Al * \frac{2 mol Al_{2}O_{3}}{4 mol Al} =0.1665 mol Al_{2}O_{3}$

$0.594 mol O_{2} * \frac{2 mol Al_{2}O_{3}}{3 mol O_{2}} =0.396 mol Al_{2}O_{3}$

As you can see, the maximum amount (theoretically) of Al2O3 that can be produced is 0.1665 mol.

Finally, we have to use the yield formula to calculate the percent yield of the reaction:

$Percent of yield = \frac{actual yield}{theoretical yield} * 100 = \frac{0.157 mol Al_{2}O_{3}}{0.1665 mol Al_{2}O_{3}} * 100 = 94.25$

Therefore, the percent of yield is 50.03%.

4 0

3 years ago

I need help with these

svp [43]

Answer: 46. E 47. D 48. C 49. A

50. A 51. D 52. B 53. B 54. C

Explanation: solution attached.

5 0

3 years ago

WHY is there a difference between how an electrolytes and non electrolytes affect collegiative properties?

natulia [17]

<span>Colligative properties are dependent upon the number of molecules or ions present in solution. Therefore, 1 mole of Na2SO4 will produce 3 moles of ions and so it will have 3 times as much of an effect as 1 mole of sugar, which is not an electrolyte and can't dissociate to an appreciable extent.</span>

4 0

3 years ago