Help me please ( I'll give brainlist to whoever answers right)

A certain substance melts at a temperature of . But if a sample of is prepared with of urea dissolved in it, the sample is found

pshichka [43]

Answer:

2.2 °C/m

Explanation:

It seems the question is incomplete. However, this problem has been found in a web search, with values as follow:

" A certain substance X melts at a temperature of -9.9 °C. But if a 350 g sample of X is prepared with 31.8 g of urea (CH₄N₂O) dissolved in it, the sample is found to have a melting point of -13.2°C instead. Calculate the molal freezing point depression constant of X. Round your answer to 2 significant digits. "

So we use the formula for freezing point depression:

ΔTf = Kf * m

In this case, ΔTf = 13.2 - 9.9 = 3.3°C

m is the molality (moles solute/kg solvent)

350 g X ⇒ 350/1000 = 0.35 kg X

31.8 g Urea ÷ 60 g/mol = 0.53 mol Urea

Molality = 0.53 / 0.35 = 1.51 m

So now we have all the required data to solve for Kf:

ΔTf = Kf * m

3.3 °C = Kf * 1.51 m

Kf = 2.2 °C/m

5 0

3 years ago

How many atoms are in 12 moles of silver?

lord [1]

Answer:

NA=72.264x1023

Explanation:

I got you

7 0

3 years ago

How many molecules are present in 4.5 moles of H2O?

Gekata [30.6K]

Answer:

Mole of the H2O = 4.5

Number of molecules =4.5 multipled by avogadro's number.

6 0

3 years ago

What is the value of quantum numbers (n, l, ml, ms) of 2p^3?

eduard

Answer:

Explanation:

First digit of the 2p^3 gives you value of n, in this case its = 2, So, n= 2

Second alphabet gives you the value of l,

l=0 =s

l=1 =p

l=3=d

l=4=f

since "p" is the alphabet in 2p^3, so in your case lt shoudlbe = 1 right?

ml= -l to +l , that is -1, 0, +1

Ms= +1/2 or -1/2 alaways remains same foe evrything.

6 0

2 years ago

A 20.0 g piece of aluminum at 5.00 C is dropped into 20.2 g of water at 90.00 C. The final temperature is 75.00 C. Use the First

bekas [8.4K]

Answer:

The specific heat of aluminium is 0.906 J/g°C

Explanation:

Step 1: data given

Mass of aluminium = 20.0 grams

Temperature = 5.00 °C

Mass of water = 20.2 grams

Temperature of water = 90.00 °C

The final temperature = 75.00 °C

Specific heat of water = 4.184 J/g°C

Step 2: calculate the specific heat of aluminium

heat won = heat lost

Qaluminium = -Qwater

Q = m*c* ΔT

m(aluminium * c(aluminium) *ΔT(aluminium = -m(water) * c(water) *ΔT(water)

⇒with m(aluminium) = mass of aluminium = 20.0 grams

⇒with c(aluminium) = the specific heat of aluminium = TO BE DETERMINED

⇒with ΔT(aluminium) = the change of temperature = T2 - T1 = 75.00 °C - 5.00 °C = 70.00 °C

⇒with m(water) = the mass of water = 20.2 grams

⇒with c(water) = the specific heat of water = 4.184 J/g°C

⇒with ΔT(water) = T2 - T1 = 75.00°C - 90.00 °C = -15.00 °C

20.0 * c(aluminium) * 70.00 = -20.2 * 4.184 * -15.00

c(aluminium) = 0.906 J/g°C

The specific heat of aluminium is 0.906 J/g°C

7 0

3 years ago