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3 years ago

45.0 mL of a Naoh solution is titrated with 0.12 M HCl. If 29.6 mL is required for neutralization,

Chemistry

1 answer:

OLga [1]3 years ago

5 0

Answer:

M(NaOH) = 0.0789 M

Explanation:

NaOH + HCl ---> NaCl + H2O

M(NaOH)*V(NaOH) =M(HCl)*V(HCl)

M(NaOH) *45.0 mL (NaOH) = 0.12 M(HCl) * 29.6 mL(HCl)

M(NaOH) = 0.12*29.6/45.0 = 0.0789 M

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Put the elements in order from the highest electronegativity at the top of your list, to the lowest at the bottom. Germanium (Ge

Illusion [34]

Br
Se
As
Ge
Ga
Rb
Elements in the top right corner of the periodic table have the highest electronegativity. Elements on the right side have a higher electronegativity than those on the left, same with the ones on the top in comparison to those on the bottom.

8 0

3 years ago

It is known that the kinetics of recrystallization for some alloy obey the Avrami equation andthat the value of n in the exponen

Paul [167]

Answer:

rate of recrystallization = 4.99 × 10⁻³ min⁻¹

Explanation:

For Avrami equation:

$y = 1-e ^{(-kt^n)} \\ \\ e^{(-kt^n)} = 1-y\\ \\ -kt^n = In(1-y) \\ \\ k = \dfrac{-In(1-y)}{t^n}$

To calculate the value of k which is a dependent variable for the above equation ; we have:

$k = \dfrac{-In(1-0.40)}{200^{2.5}}$

$k = 9.030 \times 10 ^{-7}$

The time needed for 50% transformation can be determined as follows:

$y = 1-e ^{(-kt^n)} \\ \\ e^{(-kt^n)} = 1-y\\ \\ -kt^n = In(1-y) \\ \\ t =[ \dfrac{-In(1-y)}{k}]^{^{1/n}}$

$t_{0.5} =[ \dfrac{-In(1-0.4)}{9.030 \times 10^{-7}}]^{^{1/2.5}}$

= 200.00183 min

The rate of reaction for Avrami equation is:

$rate = \dfrac{1}{t_{0.5}}$

$rate = \dfrac{1}{200.00183}$

rate = 0.00499 / min

rate of recrystallization = 4.99 × 10⁻³ min⁻¹

8 0

3 years ago

Mg(OH)2 + HCl --> MgCl2 + H2O

Rashid [163]

I am guessing you want us to balance this equation so.
To balance, we add another molecule of HCl to the left side of the equation and another molecule of water (H20) to the right side of the equation to give:

<span>Mg(OH)2 + 2HCl = MgCl2 + 2H20 </span>

6 0

3 years ago

The balanced equation for combustion in an acetylene torch is shown below:

vichka [17]

Answer:

70mol

Explanation:

The equation of the reaction is given as:

2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O

Given parameters:

Number of moles of acetylene = 35.0mol

Number of moles of oxygen in the tank = 84.0mol

Unknown:

Number of moles of CO₂ produced = 35.0mol

Solution:

From the information given about the reaction, we know that the reactant that limits this combustion process is acetylene. Oxygen is given in excess and we don't know the number of moles of this gas that was used up. We know for sure that all the moles of acetylene provided was used to furnish the burning procedure.

To determine the number of moles of CO₂ produced, we use the stoichiometric relationship between the known acetylene and the CO₂ produced from the balanced chemical equation:

From the equation:

2 moles of acetylene produced 4 moles of CO₂

∴ 35.0 mol of acetylene would produced:

$\frac{35 x 4}{2}$ = 70mol