I NEED HELP ASAP!! I DOMT KNOW HOW TO DO THIS

The boiling point of water is 100.00 °C at 1 atmosphere.

Oksana_A [137]

Answer:

The solution's boiling point is 100,2°C and its molality is 0,13 m

Explanation:

This is the colligative propertie about elevation of boiling point

ΔT = Kb . m . i where

ΔT is the difference between T° at boiling point of the solution - T° at boiling point of the solvent pure

Kb means ebulloscopic constant (0,52 °C.kg/m .- a known value for water)

m means molality (moles of solute in 1kg of solvent)

i means theVan 't Hoff factor ( degree of dissociation for a compound)

IT HAS NO UNITS

NiI2 ---> Ni2+ + 2I- (we have 1 Ni2+ and 2 I-), the i for this, is 3

The 11,11 g of the salt are in 272,2g of water but I need to know how many mass of the salt is in 1000 g of water (1000 g is 1 kg) so the rule of three is:

272,2g ____ 11,11g

1000g _____ (1000 g . 11,11g) / 272,2g = 40,81g

As the molar mas of NiI2 is 312.5 g/mol, the moles of salt are, mass/molar mass, 40,81g /312.5 g/mol = 0,130 moles

T° of b p sl - 100°C = 0,52 °C.kg/m . 0,130 m/kg . 3

T° of boiling point solution = (0,52 °C.kg/m . 0,130 m/kg . 3) + 100°C

T° of boiling point solution = 100,2°C

4 0

3 years ago

In the Haber process for ammonia synthesis, K " 0.036 for N 2 (g) ! 3 H 2 (g) ∆ 2 NH 3 (g) at 500. K. If a 2.0-L reactor is char

lisabon 2012 [21]

Answer : The partial pressure of $N_2,H_2\text{ and }NH_3$ at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution : Given,

Initial pressure of $N_2$ = 1.42 bar

Initial pressure of $H_2$ = 2.87 bar

$K_p$ = 0.036

The given equilibrium reaction is,

$N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)$

Initially 1.42 2.87 0

At equilibrium (1.42-x) (2.87-3x) 2x

The expression of $K_p$ will be,

$K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}$

Now put all the values of partial pressure, we get

$0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}$

By solving the term x, we get

$x=0.287\text{ and }3.889$

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of $NH_3$ at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of $N_2$ at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of $H_2$ at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of $N_2$ + Partial pressure of $H_2$ + Partial pressure of $NH_3$

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

6 0

3 years ago

What property of water allows it to stick to a dry surface

son4ous [18]

You are looking for ADHESION

4 0

3 years ago

Which phrases accurately describe this rock. check all that apply

Nina [5.8K]

What are the phrases I don’t see any pictures??

7 0

2 years ago

Do electrons have more or less mass than neutrons

nata0808 [166]

Electrons have less mass than a neutron.

3 0

2 years ago