Helpppppp!!!!! It’s due today <br> URGENT!!!!!

An ethylene glycol solution contains 16.2 g of ethylene glycol (c2h6o2) in 85.4 ml of water. calculate the boiling point of the

Andre45 [30]

<span>Answer: (16.2 g C2H6O2) / (62.0678 g C2H6O2/mol) / (0.0982 kg) = 3.9704 mol/kg = 3.9704 m a.) (3.9704 m) x (1.86 Â°C/m) = 7.38 Â°C change 0.00Â°C - 7.38 Â°C = - 7.38 Â°C b.) (3.9704 m) x (0.512 Â°C/m) = 2.03 Â°C change 100.00Â°C + 2.03 Â°C = 102.03 Â°C</span>

7 0

3 years ago

Will give brainliest!!!!

Alekssandra [29.7K]

Answer:

Its B :)

Hope this helps

7 0

3 years ago

The teacher tells your group to make a stock solution of sodium chloride, and then diluting it to

sergejj [24]

The idea here is that you need to figure out how many moles of magnesium chloride,

MgCl

2

, you need to have in the target solution, then use this value to determine what volume of the stock solution would contain this many moles.

As you know, molarity is defined as the number of moles of solute, which in your case is magnesium chloride, divided by liters of solution.

c

=

n

V

So, how many moles of magnesium chloride must be present in the target solution?

c

=

n

V

⇒

n

=

c

⋅

V

n

=

0.158 M

⋅

250.0

⋅

10

−

3

L

=

0.0395 moles MgCl

2

Now determine what volume of the target solution would contain this many moles of magnesium chloride

c

=

n

V

⇒

V

=

n

c

V

=

0.0395

moles

3.15

moles

L

=

0.01254 L

Rounded to three sig figs and expressed in mililiters, the volume will be

V

=

12.5 mL

So, to prepare your target solution, use a

12.5-mL

sample of the stock solution and add enough water to make the volume of the total solution equal to

250.0 mL

.

This is equivalent to diluting the

12.5-mL

sample of the stock solution by a dilution factor of

20

.

3 0

2 years ago

When a 3.00 grams sample of a compound containing only c, h, and o was completely burned, 1.17 grams of h2o and 2.87 grams of co

SVEN [57.7K]

Answer:- $CH_2O_2$

Solution:- From given masses of carbon dioxide and water we could calculate the moles that helps to calculate the moles of C and H.

Molar mass of carbon dioxide = 44 gram per mol

molar mass of water = 18.02 gram per mol

From given info, combustion of compound gives 1.17 grams of water and 2.87 grams of carbon dioxide. Let's calculate the moles of these:

$1.17gH_2O(\frac{1mol}{18.02g})$

= $0.0649molH_2O$

Similarly, $2.87gCO_2(\frac{1mol}{44g})$

= $0.0652molCO_2$

One mol of water has two moles of H. So, the moles of H would be two times the moles of water as calculated above.

So, moles of H = 2* 0.0649 = 0.1298 mol

One mol of carbon dioxide contains one mol of C. So, the moles of C would be equal to the moles of carbon dioxide calculated above.

moles of C = 0.0652 mol

Let's convert the moles of H and C to grams so that we could calculate the amount of oxygen present in the sample as:

grams of H in sample = 1.008 x 0.1298 = 0.1308 g

grams of C in sample = 12*0.0652 = 0.7824 g

If we subtract the sum of the masses of C and H from sample mass then it would give as the mass of oxygen since the sample has only C, H and O.

mass of O in sample = 3.00g - (0.1308 g + 0.7824 g)

= 3.00 g - 0.9132 g

= 2.0868 g

Let's convert these grams of oxygen to moles on dividing by it's atomic mass as:

$2.0868gO(\frac{1mol}{15.999g})$

= 0.130 mol O

Now, we have the moles of all the three atoms and we know that an empirical formula is the simplest whole number ratio of the moles of atoms. So, let's calculate the ratio. For this, we divide the moles of each by the least one of them.Looking at the moles, the least value is for carbon. So, let's divide the moles of each by the moles of C as:

C = $\frac{0.0652}{0.0652}$ = 1