All categories

3 years ago

What happens when a crystal potassium nitrate is added to a saturated solution as it cooled

1 answer:

3 0

Crystals will begin to form. crystals form from certain molecules in the liquid gather together as they attempt to become stable. they do this in a uniform and repeating pattern that forms the crystal.in nature, crystals can form when liquid rock, called magma, cools.if it cools slowly then crystals mayform. hope it helps :)

You might be interested in

Pls answer this ASAP thank you

vaieri [72.5K]

Answer:

The anwer is not D the anwer is A

Explanation:

7 0

3 years ago

The pressure of nitrogen gas at 35°C is changed from 0.89 atm to 4.3 atm. What will be its final temperature in Kelvin?

Alja [10]

Answer: The final temperature in Kelvin is 1488

Explanation:

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=0.89atm\\T_1=35^0C=(35+273)K=308K\\P_2=4.3atm\\T_2=?$

Putting values in above equation, we get:

$\frac{0.89}{308}=\frac{4.3}{T_2}\\\\T_2=1488K$

Hence, the final temperature in Kelvin is 1488

8 0

3 years ago

Calculate the standard entropy change for the following reactions at 25°C.

Art [367]

Answer:

(a) ΔSº = 216.10 J/K

(b) ΔSº = - 56.4 J/K

Explanation:

We know the standard entropy change for a given reaction is given by the sum of the entropies of the products minus the entropies of reactants.

First we need to find in an appropiate reference table the standard molar entropies entropies, and then do the calculations.

(a) C2H5OH(l) + 3 O2(g) ⇒ 2 CO2(g) + 3 H2O(g)

Sº 159.9 205.2 213.8 188.8

(J/Kmol)

ΔSº = [ 2(213.8) + 3(188.8) ] - [ 159.9 + 3(205.) ] J/K

ΔSº = 216.10 J/K

(b) CS2(l) + 3 O2(g) ⇒ CO2(g) + 2 SO2(g)

Sº 151.0 205.2 213.8 248.2

(J/Kmol)

ΔSº = [ 213.8 + 2(248.2) ] - [ 151.0 + 3(205.2) ] J/K = - 56.4 J/K

Sº 173.3 205.2 213.8 188.8

(J/Kmol)

ΔSº = [ 12(213.8) + 6(188.8) ] - [ 2(173.3) + 15( 205.2) ] = 273.8 J/K

Whenever possible we should always verify if our answer makes sense. Note that the signs for the entropy change agree with the change in mol gas. For example in reaction (b) we are going from 4 total mol gas reactants to 3, so the entropy change will be negative.

Note we need to multiply the entropies of each substance by its coefficient in the balanced chemical equation.

5 0

3 years ago

Seriously what do i put in the x and y table im stuck and need to move to the next assignment

dem82 [27]

Remember the relationship

P is indirectly proportional to V

The pairs are

(2,4)
(1,8)
(4,2)
(12,2/3)
(2/3,12)

And so on

6 0

2 years ago

In an experiment, 8.50 g of methane, CH4, was reacted with 15.9 g of oxygen gas to produce carbon dioxide and water. Determine t

exis [7]

Answer:

89.3 %

Explanation:

M(CH4) = 12+ 4*1 = 16 g/mol

M(O2) = 2*16 = 32 g/mol

M(CO2) = 12 + 2*16 = 44 g/mol

8.50 g * 1 mol/16 g = 0.5313 mol CH4

15.9 g * 1 mol/32 g = 0.4969 mol O2

9.77 g * 1 mol/44 g = 0.2220 mol CO2

1) CH4 + 2O2 -----> CO2 + 2H2O

from reaction 1 mol 2 mol

given 0.5313 mol (0.4969 mol)

1 mol CH4 --- 2 mol O2

0.5313 mol CH4 --- x mol O2

x= 2*0.5313 = 1.0626 mol O2

We can see that for given amount of CH4 we do not have enough O2, so O2 is a limiting reactant.

2) CH4 + 2O2 -----> CO2 + 2H2O

from reaction 2 mol 1 mol

given 0.4969 mol x mol

x = 0.4969*1/2 = 0.2485 mol CO2 theoretical yield

Practical yield CO2 = 0.2220 mol

Theoretical yield CO2 = 0.2485 mol

% yield = (0.2220/0.2485)*100% = 89.3 %

7 0

3 years ago