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4 years ago

What is the oxidation number for s in s2o4?

1 answer:

3 0

The oxidation number of O is -2 and there's 4 of them so the total is -8. Since it's a molecule the sum of the oxidation numbers of the atoms must equal 0. If 4 oxygen atoms have an oxidation number of -8 then the two S atoms must is +8 to equal 0 so the oxidation number of the S is 8/2 = +4

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s e vapor pressure of the waterretnylene glycol solution at 32 "C? 5. What is the Boiling Point of the solution resulted from th

lyudmila [28]

Answer:

What is the Boiling Point of the solution resulted from the dissolving of 32.5 g of NaCl in 250.0 g of water? 102,31°C

Explanation:

This question involves the Elevation of boiling point

ΔT = Kb . m . i

(T°solution - T°solvent pure) = Ebulloscopic constant . molality . Van 't Hoff factor

Van 't Hoff factor is 2 for NaCl because you have two ions on disociation.

Kb for water is, 0,52 °C . kg/mol - It is a known value.

You know that water pure boils at 100°C so let's build the formula

(T°solution - 100°C) = 0,52 °C.kg/mol . molality . 2

We have to find out the molality (moles of solute/1kg solvent)

In 250 g H2O we have 32,5 g NaCl so the rule of three will be

250 g H2O ______32,5 g NaCl

1000g __________ (1000g . 32,5g) / 250g = 130 g NaCl

Molar mass NaCl : 58,45 g/m

Moles NaCl : mass/ molar mass --> 130g /58,45 g/m = 2,22 m

(T°solution - 100°C) = 0,52 °C.kg/mol . 2,22 mol/gk . 2

T°solution - 100°C = 2,31 °C

T° at boiling point in the solution = 2,31°C + 100°C = 102,31°C

3 0

3 years ago

What is Chlorine in its excited state?

weeeeeb [17]

Chlorine in its excited state is 2-8-6-1

3 0

4 years ago

2NH4Cl(aq) + Ba(OH)2(aq) -> BaCl2(aq) + 2NH3(aq) + 2H2O(1)

Akimi4 [234]

How do I answer that.

7 0

3 years ago

If 1.00 mol of argon is placed in a 0.500-L container at 29.0 ∘C , what is the difference between the ideal pressure (as predict

Mashcka [7]

Answer:

Ideal ,P=49.52 atm

Real ,P=47.62 atm

Explanation:

Given that

n= 1 mol

V= 0.5 L

T= 29 ∘C = 29 +273 K

T= 302 K

For ideal gas

P V = n R T

P x 0.5 = 1 x 0.0821 x 302

P=49.52 atm

For real gas

$\left ( P+\dfrac{an^2}{v^2} \right )\left ( v-nb \right )=nRT$

Now by putting the values

$\left ( P+\dfrac{1.345\times 1^2}{0.5^2} \right )\left ( 0.5-1\times 0.03219 \right )=1\times 0.0821\times 302$

$\left ( P+\dfrac{1.345}{0.5^2} \right )=53$

P=47.62 atm

4 0

3 years ago

If you combine 27.1 g of a solute that has a molar mass of 27.1 g/mol with 100.0 g of a solvent, what is the molality of the res

baherus [9]

Answer: The molality of the solution is 0.1 m.

Explanation:

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute = 27.1 g

$M_{solute}$ = Molar mass of solute = 27.1 g/mol

$W_{solvent}$ = Mass of solvent = 100 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{27.1\times 1000}{27.1\times 100}\\\\\text{Molality of solution}=0.1m$

Hence, the molality of the solution is 0.1 m.

8 0

3 years ago

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