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Ierofanga [76]
3 years ago
15

19. When sugar is dissolved in water

Chemistry
1 answer:
Mars2501 [29]3 years ago
6 0
C.
The solute is dissolved in the solvent
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Una disolución contiene 50 gramos de KOH en 0.25 L de Determine la Molaridad de la disolución
snow_tiger [21]

Answer:

1. 3.56 M.

2. 0.99 M.

Explanation:

¡Hola!

1. En este caso, dado que la molaridad de una solution es calculada por medio de la siguiente ecuación:

M=\frac{n}{V}

Es posible calcular la molaridad de 50 gramos de hidróxido de potasio primero calculando las moles en dicha masa por medio de la masa molar:

n=50.0gKOH*\frac{1molKOH}{56.11gKOH}=0.891molKOH

Luego, dado el volumen de la solución, podemos calcular la molaridad:

M=\frac{0.891mol}{0.25L}=3.56M

2. En este segundo ejercicio, procedemos de la misma manera, pues primero calculamos las moles de nitrato de potasio:

n=75gKNO_3\frac{1molKNO_3}{101.1gKNO_3}=0.742mol

Luego, calculamos la molaridad justo como se hizo anteriormente:

M=\frac{0.742mol}{0.35L} \\\\M=0.99M

Best regards!

3 0
3 years ago
c) What is the pH of the buffer system in part a when 0.030 moles of strong acid are added (without a change in volume)
pishuonlain [190]

Answer:

remain the same

Explanation:

The pH of the buffer system remain the same when 0.030 moles of strong acid are added because buffer system has the property to resist any change in the pH  when acid or base is added to the solution. In buffer system, one molecule is responsible for neutralizing the pH of the solution by giving H+ or OH-.This molecule is known as buffer agent. If more base is added, the molecule provide H+ and when more acid is added to the solution, then the molecule add OH- to the solution.

7 0
4 years ago
The radiator in a car is filled with a solution of 60% antifreeze and 40% water. The manufacturer of the antifreeze suggests tha
lakkis [162]

Answer:

0.6 Liters of coolant should be drained and 0.6 Liter of water should be added.

Explanation:

Volume of the radiator = 3.6 L

Percentage of antifreeze = 60%

Total volume of anti freeze in radiator = 60% of 3.6 L :

\frac{60}{100}\times 3.6 = 2.16 L

Percentage of water= 40%

Given,optimal cooling of the engine is obtained with only 50% antifreeze.

So, now we want to reduce the percentage of antifreeze from 60% to 50 %

Volume of coolant removed = x

Volume of water added = x

Volume of anti freeze removed = 60% of(x) = 0.6x

Volume of antifreeze left in radiator  =50% of 3.6 L = \frac{50}{100}\times 3.6=1.8 L

1.8 Liter is the desired volume of the antifreeze.

Total volume - Removed volume = desired  volume

2.16 L - 60% of( x) = 1.8 L

2.16 L-0.6x=1.8 L

x=\frac{2.16 l- 1.8 L}{0.6}=0.6 L

0.6 Liters of coolant should be drained and 0.6 Liter of water should be added.

4 0
4 years ago
A hydrocarbon with general formaul cxhy is burned completely in air yielding 0.18 g of water and 0.44 g of carbon dioxide. Which
Savatey [412]

Answer:

CH2

Explanation:

When a hydrocarbon is burnt in air or oxygen, there are two products only, these are water and carbon iv oxide.

Fuel + Oxygen—-> Water + Carbon iv oxide

We can get the formula through calculations as follows.

From the mass of carbon iv oxide produced, we can get the number of moles of carbon produced. We first divide the mass by the molar mass of carbon iv oxide. The molar mass of carbon iv oxide is 44g/mol

The number of moles of carbon iv oxide is 0.44/44 = 0.01

Since there is only one carbon atom in CO2, the number of moles of carbon is same as above I.e 0.01 moles

From the number of moles of water, we can get the number of moles of hydrogen. To get the number of moles of water, we need to divide the mass of water by its molar mass. Its molar mass is 18g/mol. The number of moles here is thus 0.18/18 = 0.01moles

But there are 2 atoms of hydrogen in 1 mole of water and thus, the number of moles of hydrogen is 2 * 0.01= 0.02moles

The empirical formula can be obtained by dividing the number of moles of each by the smallest which is that of 0.01

H = 0.02/0.01 = 2

C = 0.01/0.01 = 1

From the calculations, x = 1 and y = 2

The empirical formula is thus CH2

8 0
4 years ago
Read 2 more answers
If an object has a volume of 5 mL and a mass of 50 g, what is the density of the object?
mr Goodwill [35]
Using the formula density = mass ÷ volume, I got my answer as 10g/cm^3 or 10g/mL
6 0
3 years ago
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