After sugar dissolves in water, the sugar is no

PLS HURRY: 217 mL of Helium gas at 551oC is brought to standard temperature (0oC) while the pressure is held constant. What is t

krok68 [10]

71.9 mL is the answer

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3 years ago

What mass of carbon dioxide is produced from the complete combustion of 5.30x10-3 g of methane

Alenkinab [10]

0.0159 g of carbon dioxide

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3 years ago

Read 2 more answers

a) Calculatethe molality, m, of an aqueous solution of 1.22 M sucrose, C12H22O11. The density of the solution is 1.12 g/mL.b) Wh

Contact [7]

Answer:

a) 1,74 molal

b) 37,2 %

c) 0,03

Explanation:

We are going to define sucrose as solute, water as solvent and the mix of both, the solution.

Let´s start with the data:

$Molarity = M = \frac{1,22 mol solute}{lts solution}$

We can assume as a calculus base, 1 liter of solution. So, in 1 liter of solution we have 1,22 moles of solute:

$1 lts solution * \frac{1,22 moles solute}{lts solution}=1,22 moles solute$

Knowing that the molality (m) is defined as mol of solute/kgs solvent, we have to calculate the mass of solvent on the solution. Remember our calculus base (1 lts of solution). In 1 lts of solution we have 1120 grams of solution.

$1 lts solution * \frac{1,12 grs solution}{mL solution}*\frac{1000 mL solution}{1 lts solution} = 1120 grs of solution$

With the molecular weight of solute (Sum of: for carbon = 12*12=144; for hydrogen = 1*22=22 and for oxygen = 16*11=176. Final result = 342 grs per mol), we can obtain the mass of solute:

$1,22 mol solute*\frac{342 grs solute}{1 mol solute} = 417,24 grs solute$

Now, the mass of solvent is: mass solvent = mass of solution - mass of solute. So, we have: 1120 - 417,24 = 702,76 grs of solvent = 0,70276 Kgs of solvent

$molality = m = \frac{1,22 mol solute}{0,70276 kgs solvent}= 1,74 molal$

For b) question we have that the mass percent of solute is hte ratio between the mass of solute and the mass of solution. So,

$%(w/w) = \frac{417,24 grs solute}{1120 grs solution} = 37,2%$

For c) question we have that the mole fraction of solute is the ratio between moles of solute and moles of solution. Let's calculate the moles of solution as follows: Moles solution = moles solute + moles solvent. First we have that the moles of solvent are (remember that the molecular weight of water for this calculus is 18 grs per mol):

$702,76 grs solvent*\frac{1 mol solvent}{18 grs solvent} = 39,04 moles solvent$

So, we have the moles of solution: 1,22 moles of solute + 39,04 moles of solvent = 40,26 moles of solution

Finally, we have:

$Mol frac solute = \frac{1,22 mol solute}{40,26 mol solution}= 0,03$

6 0

3 years ago

A 37.2 g sample of copper at 99.8 °C is carefully placed into an insulated container containing 188 g of water at 18.5 °C. Calcu

klasskru [66]

Answer:

T₂ = 19.95°C

Explanation:

From the law of conservation of energy:

$Heat\ Lost\ by\ Copper = Heat\ Gained\ by\ Water\\m_cC_c\Delta T_c = m_wC_w\Delta T_w$

where,

mc = mass of copper = 37.2 g

Cc = specific heat of copper = 0.385 J/g.°C

mw = mass of water = 188 g

Cw = specific heat of water = 4.184 J/g.°C

ΔTc = Change in temperature of copper = 99.8°C - T₂

ΔTw = Change in temperature of water = T₂ - 18.5°C

T₂ = Final Temperature at Equilibrium = ?

Therefore,

$(37.2\ g)(0.385\ J/g.^oC)(99.8\ ^oC-T_2)=(188\ g)(4.184\ J/g.^oC)(T_2-18.5\ ^oC)\\99.8\ ^oC-T_2 = \frac{(188\ g)(4.184\ J/g.^oC)}{(37.2\ g)(0.385\ J/g.^oC)}(T_2-18.5\ ^oC)\\\\99.8\ ^oC-T_2 = (54.92) (T_2-18.5\ ^oC)\\54.92T_2+T_2 = 99.8\ ^oC + 1016.02\ ^oC\\\\T_2 = \frac{1115.82\ ^oC}{55.92}$

T₂ = 19.95°C

6 0

3 years ago

Hey guys, I was wondering if you could give me an explanation on aqueous electrolysis.

Crazy boy [7]

Aqueous compounds are a mixtures of two electrolytes – the compound and water.

Does this help a little?

6 0

3 years ago