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3 years ago

How many mL of 0.25 M NaOH would have 0.18 moles of NaOH?

Chemistry

1 answer:

8_murik_8 [283]3 years ago

5 0

Answer:

720 mL

Explanation:

To answer this problem we'll use the definition of molarity:

Molarity = moles / liters

With that in mind we can calculate the volume in liters:

0.25 M = 0.18 moles / liters

liters = 0.72 L

Finally we convert liters to mL:

0.72 L * 1000 = 720 mL

Thus 720 mL of 0.25 M NaOH would have 0.18 moles of NaOH.

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If you use a horizontal force of 32.0 N to slide a 12.5 kg wooden crate across a floor at a constant velocity, what is the coeff

sergejj [24]

Answer:

Value of coefficient of kinetic friction is 0.26 .

Explanation:

Given:

Mass of wooden crate, m = 12.5 kg.

Horizontal force to keep the block moving with constant velocity, F = 32.0 N.

Since, the block is moving with constant velocity.

So, net force experience by it is zero.

Therefore, fore of friction is equal to applied force.

Now, force of friction , $F=\mu_kN$ ( here $\mu_k$ is coefficient of kinetic friction and N is normal force)

Therefore, $\mu_kN=\mu_k\times mg=\mu_k\times 12.5 \times 9.8=122.5\times \mu_k$

Now, both forces are equal.

$122.5\times \mu_k=32\\\mu_k=\dfrac{32}{122.5}=0.26$

The value of coefficient of kinetic friction is 0.26 .

Hence, it is the required solution.

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4 years ago

2A(g) + B(l) ⇌ 3C(aq) + D(s)

makvit [3.9K]

Answer:

[(0.75)^3(0.25)]÷[(0.50)^2(0.75)]

Explanation:

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3 years ago

What is the most important reason to publish results as part of the scientific process?

svp [43]

It allows other scientists to check findings is the most important reason to publish results as part of the scientific process.

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3 years ago

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Given LiCl, NaCl, KCl, MgCl2, and SrCl2, arrange the compounds in order of increasing lattice energy.

Oksanka [162]

Answer:- $KCl$

Explanations:- Lattice energy depends on two factors, charge and size.

High charge and small size gives higher lattice energy where as low charge and bigger size gives lower lattice energy.

in LiCl, NaCl and KCl, the anion is same and also the charges for Li, Na and K are also same. The deciding factor here is the size of cations. Since the size increases as we move down a group, the order of size of these three atoms is Li<Na<K.

The order of lattice energy is exactly opposite as it's increases as the size decreases.

Now, if we look at magnesium chloride and strontium chloride then again the anion is common but the metals have higher charge as compared to the alkali metals(Li, Na and K). So, lattice energy values must be higher for these two compounds. If we compare Mg and Sr then size of Mg is smaller and so the lattice energy would be greater for this.

Hence, the increasing order of lattice energy is $KCl$ .

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3 years ago

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A chemist titrates 130.0mL of a 0.4248 M lidocaine (C14H21NONH) solution with 0.4429 M HBr solution at 25 degree C . Calculate t

jeka57 [31]

Answer:

pH = 3.36

Explanation:

Lidocaine is a weak base to be titrated with the strong acid HBr, therefore at equivalence point we wil have the protonated lidocaine weak conjugate acid of lidocaine which will drive the pH.

Thus to solve the question we will need to calculate the concentration of this weak acid at equivalence point.

Molarity = mol /V ∴ mol = V x M

mol lidocaine = (130 mL/1000 mL/L) x 0.4248 mol/L = 0.0552 mol

The volume of 0.4429 M HBr required to neutralize this 0.0552 mol is

0.0552 mol x (1L / 0.4429mol) = 0.125 L

Total volume at equivalence is initial volume lidocaine + volume HBr added

0 .130 L +0.125 L = 0.255L

and the concentration of protonated lidocaine at the end of the titration will be

0.0552 mol / 0.255 L = 0.22M

Now to calculate the pH we setup our customary ICE table for weak acids for the equilibria:

protonated lidocaine + H₂O ⇆ lidocaine + H₃O⁺

protonated lidocaine lidocaine H₃O⁺

Initial(M) 0.22 0 0

Change -x +x +x

Equilibrium 0.22 - x x x

We know for this equilibrium

Ka = [Lidocaine] [H₃O⁺] / [protonaded Lidocaine] = x² / ( 0.22 - x )

The Ka can be calculated from the given pKb for lidocaine

Kb = antilog( - 7.94 ) = 1.15 x 10⁻⁸

Ka = Kw / Kb = 10⁻¹⁴ / 1.15 x 10⁻⁸ = 8.71 x 10⁻⁷

Since Ka is very small we can make the approximation 0.22 - x ≈ 0.22

and solve for x. The pH will then be the negative log of this value.

8.71 x 10⁻⁷ = x² / 0.22 ⇒ x = √(/ 8.71 X 10⁻⁷ x 0.22) = 4.38 x 10⁻⁴

( Indeed our approximation checks since 4.38 x 10⁻⁴ is just 0.2 % of 0.22 )

pH = - log ( 4.4x 10⁻⁴) = 3.36

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3 years ago