Answer:
Equal
Explanation:
When the bowling ball is first dropped, it has a maximum potential energy but minimum kinetic energy. The height is max, so the potential energy will be greatest. Velocity is 0, so kinetic energy will be 0.
The answer to this question is 6.25ml
To answer this question, you need to calculate the azithromycin drug doses for this patient. The calculation would be: 25kg * 10mg/kg/d= 250mg/d
Then multiply the doses with the available drug. It would be:
250 mg/d / (200mg/5ml)= 6.25ml/d
Answer:
(A) A layer of gases that surrounds the planet
Explanation:
An atmosphere is the layer of gases that surrounds a planet. In the case of Earth, the atmosphere contains the air we breath, as well as other gases that protect us from the Sun's harmful rays.
Please mark as brainliest ;)
Answer:
pH = 4.164
Explanation:
The first process is to find the initial moles for the base (B) & the acid (HA)
i.e.
![= \dfrac{27 mL \times 0.0758 \ moles \ \ of \ B}{1000 \ mL}](https://tex.z-dn.net/?f=%3D%20%5Cdfrac%7B27%20mL%20%5Ctimes%200.0758%20%5C%20moles%20%5C%20%5C%20of%20%5C%20B%7D%7B1000%20%5C%20mL%7D)
![=0.0020466](https://tex.z-dn.net/?f=%3D0.0020466)
![\simeq 2.047\times 10^{-3} \ moles \ of \ B](https://tex.z-dn.net/?f=%5Csimeq%202.047%5Ctimes%2010%5E%7B-3%7D%20%5C%20moles%20%5C%20of%20%5C%20B)
![= \dfrac{27 mL \times 0.0553 \ moles \ \ of \ HA}{1000 \ mL}](https://tex.z-dn.net/?f=%3D%20%5Cdfrac%7B27%20mL%20%5Ctimes%200.0553%20%5C%20moles%20%5C%20%5C%20of%20%5C%20HA%7D%7B1000%20%5C%20mL%7D)
![=0.0014931](https://tex.z-dn.net/?f=%3D0.0014931)
![\simeq 1.493\times 10^{-3} \ moles \ of \ HA](https://tex.z-dn.net/?f=%5Csimeq%201.493%5Ctimes%2010%5E%7B-3%7D%20%5C%20moles%20%5C%20of%20%5C%20HA)
The acid with base reaction is expressed as;
HA + B → A⁻ + HB⁺
to 1.493 × 10⁻³ 2.047 × 10⁻³ - -
- 1.493 × 10⁻³ 1.493 × 10⁻³ 1.493 × 10⁻³ 1.493 × 10⁻³
0 5.54 × 10⁻⁴ 1.493 × 10⁻³ 1.493 × 10⁻³
From observation; both the acid & base weak
Given that:
The pKa for base = 4.594
The pKa for acid = 3.235
Recall that;
pKa = -log Ka
So; Ka = ![\mathbf{10^{-Ka}}](https://tex.z-dn.net/?f=%5Cmathbf%7B10%5E%7B-Ka%7D%7D)
By applying this:
For Base; Ka =
= 2.5468 × 10⁻⁵
For Acid: Ka =
= 5.821 × 10⁻⁴
After the reaction; we have the base with its conjugate acid & conjugate base of acid; Thus, since the conjugate acid of the base possesses a higher value of K, it is likely it would be the one to define the pH of the solution.
By analyzing the system, we have:
HB⁺ + H₂O ↔ B + H₃O⁺
![\dfrac{5.54\times 10^{-4}}{0.1 \ L}](https://tex.z-dn.net/?f=%5Cdfrac%7B5.54%5Ctimes%2010%5E%7B-4%7D%7D%7B0.1%20%5C%20L%7D)
to 0.01493 M 0.00554 M
- x x x
0.01493 - x 0.00554 - x x
Thus;
![2.5468 \times 10^{-5} = \dfrac{(0.00554 -x)\times x}{(0.01493-x)}](https://tex.z-dn.net/?f=2.5468%20%5Ctimes%2010%5E%7B-5%7D%20%3D%20%5Cdfrac%7B%280.00554%20-x%29%5Ctimes%20x%7D%7B%280.01493-x%29%7D)
Using the common ion effect;
0.00554 - x
0.00554 &
0.01493 - x
0.01493
∴
![x = \dfrac{2.5468 \times 10^{-5} \times 0.01493}{0.00554}](https://tex.z-dn.net/?f=x%20%3D%20%5Cdfrac%7B2.5468%20%5Ctimes%2010%5E%7B-5%7D%20%5Ctimes%200.01493%7D%7B0.00554%7D)
x = [H₃O⁺] = 6.8635 × 10⁻⁵
∴
pH = -log(6.8635 × 10⁻⁵)
pH = 4.164