How many grams of naoh are present in 350.0 ml of a 5.69 m naoh(aq) solution?

Gekata [30.6K]

<u>Answer:</u> The value of <em>i</em> is 1.4 and 40% dissociation of 100 particles of zinc sulfate will yield 60 undissociated particles.

<u>Explanation:</u>

The equation used to calculate the Vant' Hoff factor in dissociation follows:

$\alpha =\frac{i-1}{n-1}$

where,

$\alpha$ = degree of dissociation = 40% = 0.40

i = Vant' Hoff factor

n = number of ions dissociated = 2

Putting values in above equation, we get:

$0.40=\frac{i-1}{2-1}\\\\0.40=i-1\\\\i=1.4$

The equation used to calculate the degee of dissociation follows:

$\alpha =\frac{\text{Number of particles dissociated}}{\text{Total number of particles taken}}$

Total number of particles taken = 100

Degree of dissociation = 40% = 0.40

Putting values in above equation, we get:

$0.40=\frac{\text{Number of particles dissociated}}{100}\\\\\text{Number of particles dissociated}=(0.40\times 100)=40$

This means that 40 particles are dissociated and 60 particles remain undissociated in the solution.

Hence, 40% dissociation of 100 particles of zinc sulfate will yield 60 undissociated particles.

7 0

2 years ago

The smallest unit of an element that maintains properties of that element is called a(n)

Mazyrski [523]

Answer:

try c atom i hope this helps!! : )

Explanation:

7 0

3 years ago

What volume of water vapor would be produced from the combustion of 815.74 grams of propane (C3H8) with 1,006.29 grams of oxygen

d1i1m1o1n [39]

3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane ( $C_3H_8$ ) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Stoichiometric calculations:

$C_3H_8(g) + 5 O_2(g)$ → $3 CO_2(g) + 4 H_2O(g)$

From the equation of the reaction, the mole ratio of propane to oxygen is 1:5.

Mole of 815.74 grams of propane = $\frac{ 815.74}{44.1 }$

Mole of 815.74 grams of propane = 18.49750567 moles

Mole of 1,006.29 grams of oxygen = $\frac{ 1,006.29}{32 }$

Mole of 1,006.29 grams of oxygen = 31.4465625 moles

Going by the mole ratio, it appears propane is limiting while oxygen is in excess.

From the equation, 1 mole of propane produces 4 moles of water vapour. Thus, the equivalent mole of water vapour will be:

18.49750567 moles x 4 = 73.99 moles.

Using the ideal gas equation:

PV = nRT

v = (73.99 x 0.08206 x 623) ÷ 0.96

v = 3940.2

Hence, 3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane ( $C_3H_8$ ) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

Learn more about the ideal gas here:

brainly.com/question/27691721

#SPJ1

7 0

2 years ago

Measuring Volume

Dvinal [7]

Seems right, but if it’s just asking for one i would pick B

7 0

1 year ago

How do the lithosphere and the asthenosphere work together?

miskamm [114]