All categories

2 years ago

Which solution has the highest boiling point at standard pressure?

1 answer:

6 0

The correct answer is 3: 0.10 M K3PO4(aq). The higher the concentration (or molarity), the higher the boiling point. Thus, the solution with the most moles will have the highest boiling point. SO4, PO4, and NO3 are all polyatomic ions, so by definition, they only have one mole. In K3PO4, K3 has 3 moles and PO4 has 1 mole, meaning all together it has 4 moles (more than any of the other options).

You might be interested in

13 moles of water to molecules

FromTheMoon [43]

I’m confused what are you asking exactly?

6 0

3 years ago

A fixed number of moles of an ideal gas are kept in a container of volume V and an absolute temperature T. If T and V are both d

liubo4ka [24]

Answer:

If the temperature and volume ot a gas increases, the r.m.s. velocity of the molecules in the gas will be 2 times the original r.m.s. molecular velocity.

If T doubles while V is held constant, the new net internal energy of the gas will be 2 times the original internal energy of the gas.

Explanation:

Temperature and root mean square velocity are directly proportional to one anoth. If the temperature increases, root mean square velocity also increases and vice versa, while temperature is also directly proportional to the internal energy of the gas molecules, higher the temperature, higher will be the internal energy and lower the temperature so internal energy will be decreased.

8 0

3 years ago

Balance each of the following examples of heterogeneous equilibria and write each Kc expression. Then calculate the value of Kc

marysya [2.9K]

Explanation:

1) $2 Al(s) + 2 NaOH(aq) + 6 H_2O(l) \longleftrightarrow 2 Na[Al(OH)_4](aq) + 7 H_2(g)$

$Kc=\frac{[Na[Al(OH)_4]]^2*[H_2]^7}{[NaOH]^2}$

The Kc for the reverse reaction is the inverse of the Kc of the reaction:

$Kc_{reverse}=\frac{1}{Kc}=\frac{1}{11}=0.091$

2) $H_2O(l) + SO_3(g) \longleftrightarrow H_2SO_4 (aq)$

$Kc=\frac{[H_2SO_4]}{[SO_3]^2}$

The Kc for the reverse reaction is the inverse of the Kc of the reaction:

$Kc_{reverse}=\frac{1}{Kc}=\frac{1}{0.0123}=81.3$

3) $P_4(s) + 3 O_2(g) \longleftrightarrow P_4O_6(s)$

$Kc=\frac{1}{[O_2]^3}$

The Kc for the reverse reaction is the inverse of the Kc of the reaction:

$Kc_{reverse}=\frac{1}{Kc}=\frac{1}{1.56}=0.641$

5 0

3 years ago

In the background information, it was stated that CaF2 has solubility, at room temperature, of 0.00160 g per 100 g of water. How

____ [38]

Answer:

2.05*10⁻⁵ moles of CF₂ can dissolve in 100 g of water.

12.82 moles of CaF₂ will dissolve in exactly 1.00 L of solution

Explanation:

First, by definition of solubility, in 100 g of water there are 0.0016 g of CaF₂. So, to know how many moles are 0.0016 g, you must know the molar mass of the compound. For that you know:

Ca: 40 g/mole
F: 19 g/mole

So the molar mass of CaF₂ is:

CaF₂= 40 g/mole + 2*19 g/mole= 78 g/mole

Now you can apply the following rule of three: if there are 78 grams of CaF₂ in 1 mole, in 0.0016 grams of the compound how many moles are there?

$moles=\frac{0.0016 grams*1 mole}{78 grams}$

moles=2.05*10⁻⁵

2.05*10⁻⁵ moles of CF₂ can dissolve in 100 g of water.

Now, to answer the following question, you can apply the following rule of three: if by definition of density in 1 mL there is 1 g of CaF₂, in 1000 mL (where 1L = 1000mL) how much mass of the compound is there?

$mass of CaF_{2}=\frac{1000 mL*1g}{1mL}$

mass of CaF₂= 1000 g

Now you can apply the following rule of three: if there are 78 grams of CaF₂ in 1 mole, in 1000 grams of the compound how many moles are there?

$moles=\frac{1000 grams*1 mole}{78 grams}$

moles=12.82

12.82 moles of CaF₂ will dissolve in exactly 1.00 L of solution

5 0

2 years ago

If 45.0 mL of ethanol (density =0.789g/mol) initially at 6.0°C mix with 45.0 mL of water (density =1.0 g/mol) initially at 28.0°

Likurg_2 [28]

The final temperature of the mixture : 21.1° C

<h3>Further explanation </h3>

The law of conservation of energy can be applied to heat changes, i.e. the heat received / absorbed is the same as the heat released

Q in(gained) = Q out(lost)

Heat can be calculated using the formula:

Q = mc∆T

Q = heat, J

m = mass, g

c = specific heat, joules / g ° C

∆T = temperature difference, ° C / K

Q ethanol=Q water

mass ethanol=

$\tt mass=\rho\times V\\\\mass=0.789\times 45=35.505~g$

mass water =

$\tt mass=1~g/ml\times 45~ml=45~g$

then the heat transfer :

$\tt 35.505\times 2.42~J/g^oC\times (t-6)=45\times 4.18~J/g^oC\times (28-t)\\\\85.922t-515.533=5266.8-188.1t\\\\274.022t=5782.33\rightarrow t=21.1^oC$

5 0

2 years ago