HURRY PLEASE

Calcular la cantidad de NaOH necesaria para preparar medio litro de disolución 2,5 N. (Dato: peso molecular del NaOH = 40 g/mol)

kolbaska11 [484]

Answer:

The amount of NaOH required to prepare a solution of 2.5N NaOH.

The molecular mass of NaOH is 40.0g/mol.

Explanation:

Since,

NaOH has only one replaceable -OH group.

So, its acidity is one.

Hence,

The molecular mass of NaOH =its equivalent mass

Normality formula can be written as: $Normality=\frac{mass of solute NaOH}{its equivalent mass} * \frac{1}{volume of solution in L} \\$

Substitute the given values in this formula to get the mass of NaOH required.

$2.5N=\frac{mass of NaOH}{40g/mol} *\frac{1}{1L} \\mass of NaOH=2.5N*40gmol\\ = 100.0g$

Hence, the mass of NaOH required to prepare 2.5N and 1L. solution is 100g

8 0

2 years ago

What is the final temperature of the solution formed when 1.52 g of NaOH is added to 35.5 g of water at 20.1 °C in a calorimeter

Inessa [10]

Answer : The final temperature of the solution in the calorimeter is, $31.0^oC$

Explanation :

First we have to calculate the heat produced.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = -44.5 kJ/mol

q = heat released = ?

m = mass of $NaOH$ = 1.52 g

Molar mass of $NaOH$ = 40 g/mol

$\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{1.52g}{40g/mole}=0.038mole$

Now put all the given values in the above formula, we get:

$44.5kJ/mol=\frac{q}{0.038mol}$

$q=1.691kJ$

Now we have to calculate the final temperature of solution in the calorimeter.

$q=m\times c\times (T_2-T_1)$

where,

q = heat produced = 1.691 kJ = 1691 J

m = mass of solution = 1.52 + 35.5 = 37.02 g

c = specific heat capacity of water = $4.18J/g^oC$

$T_1$ = initial temperature = $20.1^oC$

$T_2$ = final temperature = ?

Now put all the given values in the above formula, we get:

$1691J=37.02g\times 4.18J/g^oC\times (T_2-20.1)$

$T_2=31.0^oC$

Thus, the final temperature of the solution in the calorimeter is, $31.0^oC$

4 0

3 years ago

If the pressure on a gas at -73°C is doubled but its volume is held constant, what will its final temperature be in degrees Cels

Gre4nikov [31]

Answer:

127°C

Explanation:

This excersise can be solved, with the Charles Gay Lussac law, where the pressure of the gas is modified according to absolute T°.

We convert our value to K → -73°C + 273 = 200 K

The moles are the same, and the volume is also the same:

P₁ / T₁ = P₂ / T₂

But the pressure is doubled so: P₁ / T₁ = 2P₁ / T₂

P₁ / 200K = 2P₁ / T₂

1 /2OOK = (2P₁ / T₂) / P₁

See how's P₁ term is cancelled.

200K⁻¹ = 2/ T₂

T₂ = 2 / 200K⁻¹ → 400K

We convert the T° to C → 400 K - 273 = 127°C

4 0

2 years ago

How many atoms are in 17.0 moles of Pb?

Oksanka [162]

No. of atoms=mols*avagadros no.
N=n*No
N=17 * 6.022 *10^23
No. Of atoms=(17) (6.022*10^23)

8 0

3 years ago

When salt is dissolved in water, water is the

stich3 [128]

Water is solvent, salt is solute, reactants are H2O and NaCl, and the whole thing is solution

5 0

2 years ago