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1 year ago

How many moles of solute would be dissolved in .500 kg of solvent to make a 2.50 molal NaOH solution?

1 answer:

3 0

Molality is a measure of concentration that relates the moles of solute to the kilograms of solvent, it is described by the following equation:

$Molality=\frac{molSolute}{kgSolvent}$

We are given the molality(2.50m) and kilograms of solvent(0.500kg), so we solve for moles of solute from the equation:

$molSolute=Molality\times kgSolvent$ $\begin{gathered} molSolute=2.50m\times0.500kg \\ molSolute=1.25molSolute \end{gathered}$

To make a 2.50molal NaOH solution would be needed 1.25moles of solute

Answer: 1.25 moles

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Check all items common to bases. minerals metal gas H+ OH- nitrogen

Darya [45]

OH- is common to bases.

Explanation:

The base is a is an ionic compounds which when placed in aqueous solution dissociates in to a cation and an anion OH-.

The presence of OH- in the solution shows that the solution is basic or alkaline.

From Bronsted and Lowry concept base is a molecule that accepts a proton for example in NaOH, Na is a proton donor and OH is the proton acceptor.

A base accepts hydrogen ion and the concentration of OH is always higher in base.

There is a presence of conjugate acid and conjugate base in the Bronsted and Lowry acid and base.

Conjugate acid is one which is formed when a base gained a proton.

Conjugate base is one which is formed when an acid looses a proton.

And from the Arrhenius base Theory, the base is one that dissociates in to water as OH-.

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3 years ago

. In a titration, a 25.0 mL sample of 0.150 M HCl is neutralized with 44.45 mL of Ba(OH)2. a. Write the balanced molecular equat

choli [55]

Answer:

Equation of reaction:

a) 2HCl + Ba(OH)2 ==> CaCl2 + 2H2O

b) Molarity of base = 0.042 M.

Explanation:

Using titration equation

CAVA/CBVB = NA/NB

Where NA is the number of mole of acid = 2

NB is the number of mole of base = 1

CA is the molarity of acid =0.15M

CB is the molarity of base = to be calculated

VA is the volume of acid = 25 ml

VB is the volume of base = 44.45mL

Substituting

0.15×25/CB×44.45 = 2/1

Therefore CB =0.15×25×1/44.45×2

CB = 0.042 M.

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3 years ago

What is △n for the following equation in relating Kc to Kp?

Nimfa-mama [501]

Answer:

-1

Explanation:

The relation between Kp and Kc is given below:

$K_p= K_c\times (RT)^{\Delta n}$

Where,

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant , 0.082057 L atm.mol⁻¹K⁻¹

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

$2Na_{(s)}+2H_2O_{(l)}\rightleftharpoons 2NaOH_{(aq)}+2H_2_{(g)}$

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants) = (2+1)-(2+2) = -1

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