Please help! This is for the 9.06 Acid Neutralization Lab! Due today!

A motorcycle travel at 6.0 m/s. After 3.0 second, the motorcycle travel at 15.0 m/s. Which of the following was the average acce

Lerok [7]

Hello!

$\large\boxed{a = 3m/s^{2}}$

Use the following equation to solve for the average acceleration of the motorcycle:

$a = \frac{v_{f}-v_{i}}{t}$

Plug in the given final, initial velocities, and the time:

$a = \frac{15-6}{3}\\\\a = \frac{9}{3}\\\\a = 3m/s^{2}$

7 0

3 years ago

For the reaction o(g) + o2(g) → o3(g) δh o = −107.2 kj/mol given that the bond enthalpy in o2(g) is 498.7 kj/mol, calculate the

Aneli [31]

The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.

8 0

3 years ago

A 10.0 mL sample of 0.25 M NaOH(aq) is titrated with 0.10 M HCl(aq) (adding HCl to NaOH). Determine which region on the titratio

Anna11 [10]

Answer:

1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.

2) The pH of the solution after adding HCl is 12.6

Explanation:

10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.

$nNaOH=\frac{0.25mol}{L} .10.0 \times 10^{-3} L=2.5 \times 10^{-3}mol$

$nHCl=\frac{0.10mol}{L} \times 15.0 \times 10^{-3} L=1.5 \times 10^{-3}mol$

There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.

NaOH + HCl ⇒ NaCl + H₂O

Initial 2.5 × 10⁻³ 1.5 × 10⁻³ 0 0

Reaction -1.5 × 10⁻³ -1.5 × 10⁻³ 1.5 × 10⁻³ 1.5 × 10⁻³

Final 1.0 × 10⁻³ 0 1.5 × 10⁻³ 1.5 × 10⁻³

The concentration of NaOH is:

$[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M$

NaOH is a strong base so [OH⁻] = [NaOH].

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log 0.040 = 1.4

pH = 14 - pOH = 14 - 1.4 = 12.6

5 0

3 years ago

An element has three naturally occurring isotopes. ten percent (.10) occurs as 55^x, fifteen percent (.15) occurs as isotope 56^

Leviafan [203]

The average atomic weight is, from the name itself, the average weight of all its naturally occurring isotopes. All you have to do is multiple the abundance of each isotope with its individual mass, then add them altogether.

Mass = (0.10*55)+(0.15*56)+(.75*57)
<em>Mass = 56.65 amu</em>

5 0

3 years ago

PLEASE HELP WHAT IS THE CORRECT ANSWER (if possible let me know why)

ANTONII [103]

Answer:I’m not 100% sure but I think it’s the first answer.

Explanation:

The pressure would increase as the molecules are pushing out word more and more, leading to more space being taken and more pressure. But then again, I’m not sure. I’m just a freshman with below average grades.

8 0

3 years ago