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2 years ago

Calculate the molarity and molality of 49.0 wt% HF, assuming a density of 1.16 g/mL.

1 answer:

8 0

Answer:
Molarity 28.42
Molality = 24.5

To answer this question, it would be easier if you assume that the liquid volume is 1L. You also need to know that HF weight is 20 (fluor=19, helium=1). Then you need to calculate how many moles of HF inside the solution. The calculation would be:

1L x (1000ml/L) x 1.16g/ml x (49g/100g) / (20g/mole)= 28.42 mole

Then the solution molarity: 28.42mole / 1L= 28.42
Then the solution molality: 28.42mole / (1L x 1000ml/L x 1.16g/ml) = 24.5

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Ammonia gas occupies a volume of 2,725ml at a pressure of 701 kPa. What volume would it occupy at 101 kPa?

allsm [11]

B ........................

3 0

2 years ago

a gas that exerts a pressure of 215 torr in a container with a volume of 51.0 mL will exert a pressure of ? torr when transferre

zhannawk [14.2K]

To calculate the new pressure, we can use Boyle’s law to relate these two scenarios (Boyle’s law is used because the temperature is assumed to remain constant). Boyle’s law is:

P1V1 = P2V2,

Where “P” is pressure and “V” is volume. The pressure and volume of the first scenario is 215 torr and 51 mL, respectively, and the second scenario has a volume of 18.5 L (18,500 mL) and the unknown pressure - let’s call that “x”. Plugging these into the equation:

(215 torr)(51 mL) =(“x” torr)(18,500 mL)
x = 0.593 torr

The final pressure exerted by the gas would be 0.593 torr.

Hope this helps!

3 0

2 years ago

How much heat is lost when changing 65 g of water vapor (H2O) at 421 K to ice at 139 K?

poizon [28]

The heat change will be

Moles of water = mass / Molar mass = 65/ 18 = 3.61 mol

specific heat of ice =2.09J /g C

specific heat of water = 4.184 J/g C

Specific heat of vapour= 2.01 /g C

Heat of fusion = 3.33X10⁵ J /kg = 333 J /g

Heat of vaporization = 2.26 X10⁶J/kg = 2260J/g

Q1 = heat change when vapours get cooled to 373.15 K

Q2 = heat change when vapours get converted to liquid water

Q3 = heat change when liquid water cools to 273.15 K

Q4= heat change when liquid water freezes to ice

Q5= heat change when ice cools from 273.15K to 139 K

Q1= mass of water X specific heat of vapours X change in temperature

Q1 = 65 X 2.01 /g C X (421-373.15) = 6251.60 J = 6.252 kJ

Q2 = heat of vaporization X mass = 2260 X 65 = 146900 = 146.9 kJ

Q3 = mass X specific heat of water X change in temperature =

Q3 = 65 X 4.184 X (373.15-273.15) = 65 X 4.184 X 100 = 27196 J = 27.196kJ

Q4 = heat of fusion X mass =333X65 = 21645 J = 21.645 kJ

Q5 = mass X specific heat of ice X change in temperature

Q5 = 65 X 2.09 X (273.15-139) = 18224.3 J = 18.224 kJ

Total energy = 6.252 +146.9+27.196+ 21.645+ 18.224 = 220.217

As this is energy released so it will be expressed in negative

-220.217

from the given options the correct answer will be -219.4 kJ

The answer is little different as the reference values of specific heats or enthalpy may vary.

3 0

2 years ago

According to the following reaction, what volume of 0.244 M KCl solution is required to react exactly with 50.0 mL of 0.210 M Pb

svetoff [14.1K]

Answer:

86 mL

Explanation:

First find the moles of Pb (NO3)2

n=cv

where

c ( concentration)= 0.210 M

v ( volume in L) =0.05

n= 0.210 × 0.05

n= 0.0105

Using the mole ratio, we can find the moles of KCl by multiplying by 2

n (KCl) =0.0105 ×2

=0.021

v (KCl)= n/ c

= 0.021/ 0.244

=0.08606557377

=0.086 L

= 86 mL

8 0

2 years ago

Read 2 more answers

What are the relative numbers of h3o+ and oh- ions in an acidic and alkaline and a neutral solution?

lana [24]

Acid more H3O+ than OH-
Base less H3O+ than OH-

3 0

2 years ago

Read 2 more answers