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2 years ago

You have three elements, A, B, and C, with the following electronegativity values:

Chemistry

1 answer:

DerKrebs [107]2 years ago

6 0

#AB

Electronegativity difference=3.3-2.9=0.4.

It's a covalent bond.
Gaseous or solid substance.

#AC

Electronegativity difference=3.3-0.7=2.6

Its an ionic bond.
Solid substance.

#BC

Electronegativity difference=2.9-0.7=2.3

It's an ionic bond
Solid substance

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In an acid-base titration experiment, 50 mL of a 0.05 M solution of acetic acid (Ka= 1.75 x 10-5 ) was titrated with a 0.05M sol

Viktor [21]

Answer:

(3) 5.36

Explanation:

Since this is a titration of a weak acid before reaching equivalence point, we will have effectively a buffer solution. Then we can use the Henderson-Hasselbalch equation to answer this question.

The reaction is:

HAc + NaOH ⇒ NaAc + H₂O

V NaOH = 40 mL x 1 L/1000 mL = 0.040 L

mol NaOH reacted with HAc = 0.040 L x 0.05 mol/L = 0.002 mol

mol HAC originally present = 0.050 L x 0.05 mol/L = 0.0025 mol

mol HAc left after reaction = 0.0025 - 0.002 = 0.0005

Now that we have calculated the quantities of the weak acid and its conjugate base in the buffer, we just plug the values into the equation

pH = pKa + log ((Ac⁻)/(HAc))

(Notice we do not have to calculate the molarities of Ac⁻ and HAc because the volumes cancel in the quotient)

pH = -log (1.75 x 10⁻⁵) + log (0.002/0.0005) = 5.36

THe answer is 5.36

5 0

3 years ago

Read 2 more answers

Calculate the energy (in J/atom) for vacancy formation in silver, given that the equilibrium number of vacancies at 800 C is 3.6

MAXImum [283]

Answer:

the energy vacancies for formation in silver is $\mathbf{Q_v = 3.069*10^{-4} \ J/atom}$

Explanation:

Given that:

the equilibrium number of vacancies at 800 °C

i.e T = 800°C is 3.6 x 10¹⁷ cm3

Atomic weight of sliver = 107.9 g/mol

Density of silver = 9.5 g/cm³

Let's first determine the number of atoms in silver

Let silver be represented by N

SO;

$N = \dfrac{N_A* \rho _{Ag}}{A_{Ag}}$

where ;

$N_A =$ avogadro's number = $6.023*10^{23} \ atoms/mol$

$\rho _{Ag}$ = Density of silver = 9.5 g/cm³

$A_{Ag}$ = Atomic weight of sliver = 107.9 g/mol

$N = \dfrac{(6.023*10^{23} \ atoms/mol)*( 9.5 \ g/cm^3)}{(107.9 \ g/mol)}$

N = 5.30 × 10²⁸ atoms/m³

However;

The equation for equilibrium number of vacancies can be represented by the equation:

$N_v = N \ e^{^{-\dfrac{Q_v}{KT}}$

From above; Considering the natural logarithm on both sides; we have:

$In \ N_v =In N - \dfrac{Q_v}{KT}$

Making $Q_v$ the subject of the formula; we have:

${Q_v = - {KT} In( \dfrac{ \ N_v }{ N})$

where;

K = Boltzmann constant = 8.62 × 10⁻⁵ eV/atom .K

Temperature T = 800 °C = (800+ 273) K = 1073 K

$Q _v =-( 8.62*10^{-5} \ eV/atom.K * 1073 \ K) \ In( \dfrac{3.6*10^{17}}{5.3 0*10^{28}})$

$\mathbf{Q_v = 2.38 \ eV/atom}$

Where;

1 eV = 1.602176565 × 10⁻¹⁹ J

Then

$Q_v = (2.38 \ * 1.602176565 * 10^{-19} ) J/atom }$

$\mathbf{Q_v = 3.069*10^{-4} \ J/atom}$

Thus, the energy vacancies for formation in silver is $\mathbf{Q_v = 3.069*10^{-4} \ J/atom}$

8 0

3 years ago

Mature elephants consume an average of 600 pounds of food per day. If an elephant

MAXImum [283]

Answer:

$\large \boxed{\text{4 600 000 kg or 4600 Mg}}$

Explanation:

1. Calculate the adult life

If the elephant is an adult from its 10th birthday until the day before its 56th birthday, its

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2. Convert years to days

$\text{Adult life} = \text{46 yr} \times \dfrac{\text{365.25 da}}{\text{1 yr}} = \text{16 800 da}$

3. Convert days to pounds of feed

$\text{Feed} = \text{16 800 da} \times \dfrac{\text{600 lb}}{\text{1 da}} = 1.01 \times 10^{7} \text{ lb}$

4. Convert pounds to kilograms and megagrams

$\text{Feed} =1.01 \times 10^{7} \text{ lb} \times \dfrac{\text{1 kg}}{\text{2.20 lb}} =\textbf{4 600 000 kg} = \textbf{4600 Mg}\\\\\text{The elephant will eat $\large \boxed{\textbf{4 600 000 kg or 4600 Mg}}$ of food.}$

Note: The answer can have only two significant figures because that is all you gave for age of the elephant.

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