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strojnjashka [21]
3 years ago
7

A solution is made from ethanol, C2H5OH and water is 0.75 m. How many grams of ethanol are contained per 250. g of water?

Chemistry
1 answer:
kumpel [21]3 years ago
8 0

Answer:

m_{solute}=8.6g

Explanation:

Hello,

In this case, the unit of concentration is molality which is defined by:

m=\frac{n_{solute}}{m_{solvent}}

Whereas the mass of the solvent is measured in kilograms. In such a way, with the given data, we first compute the kilograms of water:

m_{solvent}=250g*\frac{1kg}{1000g} =0.25kg

Then, we solve for the moles of the solute:

n_{solute}=0.75\frac{mol}{kg}*0.25kg =0.19mol

Finally since the molar mass of ethanol is 46 g/mol, we compute the grams for the given solution:

m_{solute}=0.19*\frac{46g}{1mol} \\\\m_{solute}=8.6g

Best regards.

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I HAVE TIME LIMIT I NEED CORRECT ANSWERS!
Basile [38]

Answer:

5) oxygen

7)0.5480 M

9)5s

10) 2 σ and 2 π bonds

12) All of the carbon-oxygen bonds in a carbonate ion are weaker than the carbon-oxygen bonds in a carbon dioxide molecule

19) 0.400 g

Explanation:

For question 5

C2H6 + 7/2 O2 -----> 2CO2 + 3H2O

If 1 mole of ethane gives 2 moles of carbon dioxide

13 moles of ethane will give 13×2 = 26 moles of carbon dioxide

If 3.5 moles of oxygen gives 2 moles of carbon dioxide

42 moles of oxygen will give 42×2/3.5 = 24 moles of carbon dioxide

The reactant that gives the lower number of moles of product is the limiting reactant. This means that oxygen is the limiting reactant here.

7) concentration of acid CA =2.456 M

Concentration of base CB= ???

Volume of acid VA= 15.41 ml

Volume of base VB= 34.53 ml

Number of moles of acid NA= 2

Number of moles of base NB= 1

Ca(OH)2(aq) + 2HCl(aq) ----> CaCl2(aq) + 2H2O (l)

CB= CA VA NB/VB NA

CB= 2.456 × 15.41 ×1/34.53×2

CB= 0.548 M

9) For question 9, we have to look at the electronic configuration of Rb. We have [Kr]5s1. The outermost 5s1 level will have the least effective nuclear charge and is easily lost.

10) There are two sigma bonds and two pi bonds in CO2

12) The C-O bond length in the carbonate ion is 136 pm while the C-O bond length in CO2 is 116 pm. The longer the bond, the weaker the bond hence the C-O bonds in the carbonate ion are weaker than those in the carbon dioxide molecule.

19) From

m= mass of solute

M= molar mass of solute

C= concentration of solute

V= volume of solution

m/M = CV

m= MCV

m= 40.0gmol-1 × 50.0/1000 × 0.200

m= 0.4 g

6 0
3 years ago
Can y’all Please help !!!!
ycow [4]

Answer:

yes

Explanation:

<em>pp popo</em>

4 0
3 years ago
If 700 g of water at 90 °C loses 27 kJ of heat, what is its final temperature?​
Phoenix [80]

Answer:

If 700 g of water at 90 °C loses 27 kJ of heat, its final temperature is 106.125 °C

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

In this way, between heat and temperature there is a direct proportional relationship (Two magnitudes are directly proportional when there is a constant so that when one of the magnitudes increases, the other also increases; and the same happens when either of the two decreases .). The constant of proportionality depends on the substance that constitutes the body and its mass, and is the product of the specific heat and the mass of the body. So, the equation that allows to calculate heat exchanges is:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature, ΔT= Tfinal - Tinitial

In this case:

  • Q= 27 kJ= 27,000 J (being 1 kJ=1,000 J)
  • c=4.186 \frac{J}{g* C}
  • m=700 g
  • ΔT= Tfinal - Tinitial= Tfinal - 90 °C

Replacing:

27,000 J=4.186 \frac{J}{g* C}*400 g* (Tfinal - 90C)\\

Solving:

27,000 J=1,674.4 \frac{J}{C}* (Tfinal - 90C)

\frac{27,000 J}{1,674.4 \frac{J}{C}} =(Tfinal - 90C)

16.125 °C= Tfinal - 90 °C

Tfinal= 16.125 °C + 90 °C

Tfinal= 106.125 °C

<u><em>If 700 g of water at 90 °C loses 27 kJ of heat, its final temperature is 106.125 °C</em></u>

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