All categories

3 years ago

Ughh please help me

Chemistry

2 answers:

Alexxx [7]3 years ago

6 0

Answer:

Substance 2

Explanation:

I know this because substances with a stronger molecular attraction are less susceptible to melting quicker than substances who have a weaker attraction with each particle. Therefore, substance 2, which melted after substance 1, has a stronger molecular attraction then substance 1.

(copy and paste above if you want)

TEA [102]3 years ago

5 0

Answer:

Substance 2

Explanation:

The stronger the intermolecular forces are, the more energy is required, so the higher the melting point is.

You might be interested in

Select the most likely product for this reaction:

Advocard [28]

Answer:

The answer is B- LiNO3(aq)+H2O(I)

Explanation:

I just did the assignment also UNUS ANNUS

3 0

3 years ago

Read 2 more answers

What did Rutherford’s gold foil experiment indicate?

Klio2033 [76]

The answer would be D.

4 0

3 years ago

Carbon is found in all known forms of life. Which statement best describes why carbon is uniquely suited to forming macromolecul

Helen [10]

C
Carbon and its compound's stability means it is useful in synthesizing organic molecules which are used in the bodies of organisms.

6 0

2 years ago

Read 2 more answers

Benzoic acid is a natural fungicide that naturally occurs in many fruits and berries. The sodium salt of benzoic acid, sodium be

AlexFokin [52]

Answer:

a. pH = 2.52

b. pH = 8.67

c. pH = 12.83

Explanation:

The equation of the titration between the benzoic acid and NaOH is:

C₆H₅CO₂H + OH⁻ ⇄ C₆H₅CO₂⁻ + H₂O (1)

a. To find the pH after the addition of 20.0 mL of NaOH we need to find the number of moles of C₆H₅CO₂H and NaOH:

$\eta_{NaOH} = C*V = 0.250 M*0.020 L = 5.00 \cdot 10^{-3} moles$

$\eta_{C_{6}H_{5}CO_{2}H}i = C*V = 0.300 M*0.050 L = 0.015 moles$

From the reaction between the benzoic acid and NaOH we have the following number of moles of benzoic acid remaining:

$\eta_{C_{6}H_{5}CO_{2}H} = \eta_{C_{6}H_{5}CO_{2}H}i - \eta_{NaOH} = 0.015 moles - 5.00 \cdot 10^{-3} moles = 0.01 moles$

The concentration of benzoic acid is:

$C = \frac{\eta}{V} = \frac{0.01 moles}{(0.020 + 0.050) L} = 0.14 M$

Now, from the dissociation equilibrium of benzoic acid we have:

C₆H₅CO₂H + H₂O ⇄ C₆H₅CO₂⁻ + H₃O⁺

0.14 - x x x

$Ka = \frac{[C_{6}H_{5}CO_{2}^{-}][H_{3}O^{+}]}{[C_{6}H_{5}CO_{2}H]}$

$Ka = \frac{x*x}{0.14 - x}$

$6.5 \cdot 10^{-5}*(0.14 - x) - x^{2} = 0$ (2)

By solving equation (2) for x we have:

x = 0.0030 = [C₆H₅CO₂⁻] = [H₃O⁺]

Finally, the pH is:

$pH = -log([H_{3}O^{+}]) = -log (0.0030) = 2.52$

b. At the equivalence point, the benzoic acid has been converted to its conjugate base for the reaction with NaOH so, the equilibrium equation is:

C₆H₅CO₂⁻ + H₂O ⇄ C₆H₅CO₂H + OH⁻ (3)

The number of moles of C₆H₅CO₂⁻ is:

$\eta_{C_{6}H_{5}CO_{2}^{-}} = \eta_{C_{6}H_{5}CO_{2}H}i = 0.015 moles$

The volume of NaOH added is:

$V = \frac{\eta}{C} = \frac{0.015 moles}{0.250 M} = 0.060 L$

The concentration of C₆H₅CO₂⁻ is:

$C = \frac{\eta}{V} = \frac{0.015 moles}{(0.060 L + 0.050 L)} = 0.14 M$

From the equilibrium of equation (3) we have:

C₆H₅CO₂⁻ + H₂O ⇄ C₆H₅CO₂H + OH⁻

0.14 - x x x

$Kb = \frac{[C_{6}H_{5}CO_{2}H][OH^{-}]}{[C_{6}H_{5}CO_{2}^{-}]}$

$(\frac{Kw}{Ka})*(0.14 - x) - x^{2} = 0$

$(\frac{1.00 \cdot 10^{-14}}{6.5 \cdot 10^{-5}})*(0.14 - x) - x^{2} = 0$

By solving the equation above for x, we have:

x = 4.64x10⁻⁶ = [C₆H₅CO₂H] = [OH⁻]

The pH is:

$pOH = -log[OH^{-}] = -log(4.64 \cdot 10^{-6}) = 5.33$

$pH = 14 - pOH = 14 - 5.33 = 8.67$

c. To find the pH after the addition of 100 mL of NaOH we need to find the number of moles of NaOH:

$\eta_{NaOH}i = C*V = 0.250 M*0.100 L = 0.025 moles$

From the reaction between the benzoic acid and NaOH we have the following number of moles remaining:

$\eta_{NaOH} = \eta_{NaOH}i - \eta_{C_{6}H_{5}CO_{2}H} = 0.025 moles - 0.015 moles = 0.010 moles$

The concentration of NaOH is:

$C = \frac{\eta}{V} = \frac{0.010 moles}{0.100 L + 0.050 L} = 0.067 M$

Therefore, the pH is given by this excess of NaOH:

$pOH = -log([OH^{-}]) = -log(0.067) = 1.17$

$pH = 14 - pOH = 12.83$

I hope it helps you!

4 0

3 years ago

How many molecules of CO are produced when 2.73 moles of HPO3 react? 4HPO3 + 12C ----> 2H2 + 12CO + P4

Zina [86]

According to the balanced chemical equation:
4 HPO₃ + 12 C → 2 H₂ + 12 CO + P₄
4 moles of HPO₃ ---gives---> 12 moles of CO
2.73 moles of HPO₃ ---gives---> ? moles of CO
so number of moles of CO = $\frac{(2.73 * 12)}{4}$ = 8.19 moles of CO
Number of molecules of CO = number of moles * Avogadro's number
= 8.19 * (6.022 * 10²³) = 4.93 * 10²⁴ molecules

3 0

3 years ago

Ughh please help me ​

Ughh please help me