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amid [387]
2 years ago
14

A 257-ml sample of a sugar solution containing 1.10 g of the sugar has an osmotic pressure of 31.5 mm hg at 39°c. what is the mo

lar mass of the sugar
Chemistry
1 answer:
dem82 [27]2 years ago
5 0
<span>D. sugar changes from white to a light amber color</span>
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The engine ran for 10 second
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A system receives 575 ) of heat and delivers 425 ) of work. Calculate the change in the internal energy. AE, of the system.​
vfiekz [6]

Answer:

ΔE = 150 J

Explanation:

From first law of thermodynamics, we know that;

ΔE = q + w

Where;

ΔE is change in internal energy

q is total amount of heat energy going in or coming out

w is total amount of work expended or received

From the question, the system receives 575 J of heat. Thus, q = +575 J

Also, we are told that the system delivered 425 J of work. Thus, w = -425 J since work was expended.

Thus;

ΔE = 575 + (-425)

ΔE = 575 - 425

ΔE = 150 J

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3 years ago
One of the products of photosynthesis is<br> carbon dioxide.<br> fire.<br> oxygen gas.<br> mass.
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One of the products of photosynthesis is carbon dioxide
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3 years ago
If the sample contained 2.0 moles of KClO3 at a temperature of 214.0 °C, determine the mass of the oxygen gas produced in grams
Westkost [7]

Answer : The mass of the oxygen gas produced in grams and the pressure exerted by the gas against the container walls is, 96 grams and 1.78 atm respectively.

Explanation : Given,

Moles of KCl_3 = 2.0 moles

Molar mass of O_2 = 32 g/mole

Now we have to calculate the moles of MgO

The balanced chemical reaction is,

2KClO_3\rightarrow 2KCl+3O_2

From the balanced reaction we conclude that

As, 2 mole of KClO_3 react to give 3 mole of O_2

So, 2.0 moles of KClO_3 react to give \frac{2.0}{2}\times 3=3.0 moles of O_2

Now we have to calculate the mass of O_2

\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2

\text{ Mass of }O_2=(3.0moles)\times (32g/mole)=96g

Therefore, the mass of oxygen gas produced is, 96 grams.

Now we have to determine the pressure exerted by the gas against the container walls.

Using ideal gas equation:

PV=nRT\\\\PV=\frac{w}{M}RT\\\\P=\frac{w}{V}\times \frac{RT}{M}\\\\P=\rho\times \frac{RT}{M}

where,

P = pressure of oxygen gas = ?

V = volume of oxygen gas

T = temperature of oxygen gas = 214.0^oC=273+214.0=487K

R = gas constant = 0.0821 L.atm/mole.K

w = mass of oxygen gas

\rho = density of oxygen gas = 1.429 g/L

M = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the ideal gas equation, we get:

P=1.429g/L\times \frac{(0.0821L.atm/mole.K)\times (487K)}{32g/mol}

P=1.78atm

Thus, the pressure exerted by the gas against the container walls is, 1.78 atm.

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Los fenomenos fisicos son cambios permamentes en la materia?
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Answer:

Explanation:

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