All categories

2 years ago

What is thermodynamics?

Chemistry

1 answer:

Pavel [41]2 years ago

6 0

Answer:

The answer to your question is B

You might be interested in

The coolant in automobiles is often a 50/50 % by volume mixture of ethylene glycol, C2H6O2, and water. At 20°C, the density of e

Misha Larkins [42]

Explanation:

Let the volume of the solution be 100 ml.

As the volume of glycol = 50 = volume of water

Hence, the number of moles of glycol = $\frac{mass}{molar mass}$

= $\frac{density \times volume}{molar mass}$

= $\frac{1.1088 \times 50}{62 g/mol}$

= 0.894 mol

Hence, number of moles of water = $\frac{50 \times 0.998}{18}$

= 2.77

As glycol is dissolved in water.

So, the molality = $0.894 \times \frac{1000}{49.92}$

= 17.9

Therefore, the expected freezing point = $-1.86 \times 17.9$

= $-33.31^{o}C$

Thus, we can conclude that the expected freezing point is $-33.31^{o}C$ .

6 0

2 years ago

275 g of a compound with the molecular formula ccl2f2 contains how many molecules of the compound

madreJ [45]

brainly.com/question/8581600#respond

3 0

2 years ago

. A large quantity of very dilute aqueous HCl solution is neutralized by the addition of the stoichiometric amount of a 10-mol-%

ad-work [718]

Answer:

Make the question more clear for me

Explanation:

5 0

3 years ago

Write the mole ratios of (2ca+o2➡2cao)

tresset_1 [31]

N(Ca)/2 = n(O)/1 = n(CaO)/2

The calcium and the Calcium Oxide are divided by 2 because of their coefficients
there is no number in front of the oxygen so it is over one.
Hope this helped!!

6 0

2 years ago

Read 2 more answers

The enthalpy of solution (dissolving) of sodium hydroxide is given below. Determine the change in temperature of a coffee cup ca

geniusboy [140]

Answer:

The change in temperature of a coffee cup calorimeter is 8.87°C.

Explanation:

Volume of the water = V = 150 g

Density of the water , d =1.0 g/mL

Mass of the water = M

$M=d\times V=1.00 g/mL\times 150 ml =150.0 g$

Mass of solution = m = M = 150.0 g

$NaOH(s)\rightarrow NaOH(aq),\Delta H =-44.51 kJ/mol$

Moles of NaOH = $\frac{5.00 g}{40 g/mol}=0.125 mol$

Energy released when 0.125 moles of NaOH added in water = Q

$Q=0.125 moles\times (-44.51 kJ/mol)=-5.5638 kJ=-5,563.8 J$

1 kJ = 1000 J

Heat gained by water = Q' = -Q ( conservation of energy)

$Q'= 5,563.8 J$

Specific heat of solution = c = 4.184 J/g°C

Change in temperature of the solution = $\Delta T$

$Q'=mc\times \Delta T$

$5,563.8 J=150.0 g\times 4.184 J/g^oC\times \Delta T$

$\Delta T=\frac{5,563.8 J}{150.0 g\times 4.184 J/g^oC}=8.87^oC$

The change in temperature of a coffee cup calorimeter is 8.87°C.

7 0

3 years ago