Answer:
P = 83.16 Watt
Explanation:
Given data:
Mass of box = 2 Kg
velocity of box = 29.7 m/s
Time = 21 sec
Power = ?
Solution:
Formula:
Power = work done/time
Now we will calculate the work done.
W = force × distance ...... (1)
d = speed /time
d = 29.7 m/s× 21 s
d = 623.7 m
Force:
acceleration = 29.7 m/s / 21 s
a = 1.4 m/s²
Force = m×a
Force = 2 kg×1.4 m/s²
Force = 2.8 N
Now we will put the values in equation 1.
W = F × d
W = 2.8 N ×623.7 m
W = 1746.36 Nm
Now for power:
P = W / time
P = 1746 Nm / 21 s
P = 83.16 Watt
In alpha and beta decay the parent element and the daughter element will be different isotope and not the same.
This is because as alpha particles comes out; the new element will have two less in atomic number; that is the atomic number will reduce by two and the mass number will reduce by four. While in a beta particle decay, the new element will have one higher in atomic number; that is the atomic number will increase by one.
Answer: 20 + 1 + 3.8 + 2.4 +27 = 54.2
Explanation: 20 gallons plus 1 gallon plus 3.8 plus 2.4 plus 27 equals to a total number of 54.2 gallons for the propane.
Answer: “Formaldehyde” is emitted from most manufactured building materials and furniture.
Answer:
10.88 g
Explanation:
We have:
[CH₃COOH] = 0.10 M
pH = 5.25
Ka = 1.80x10⁻⁵
V = 250.0 mL = 0.250 L
The pH of the buffer solution is:
![pH = pKa + log(\frac{[CH_{3}COONa*3H_{2}O]}{[CH_{3}COOH]})](https://tex.z-dn.net/?f=%20pH%20%3D%20pKa%20%2B%20log%28%5Cfrac%7B%5BCH_%7B3%7DCOONa%2A3H_%7B2%7DO%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D%29%20)
(1)
By solving equation (1) for [CH₃COONa*3H₂O] we have:
![log [CH_{3}COONa*3H_{2}O] = pH - pKa + log [CH_{3}COOH]](https://tex.z-dn.net/?f=%20log%20%5BCH_%7B3%7DCOONa%2A3H_%7B2%7DO%5D%20%3D%20pH%20-%20pKa%20%2B%20log%20%5BCH_%7B3%7DCOOH%5D%20)
![log [CH_{3}COONa*3H_{2}O] = 5.25 - (-log(1.80 \cdot 10^{-5})) + log (0.10) = -0.495](https://tex.z-dn.net/?f=%20log%20%5BCH_%7B3%7DCOONa%2A3H_%7B2%7DO%5D%20%3D%205.25%20-%20%28-log%281.80%20%5Ccdot%2010%5E%7B-5%7D%29%29%20%2B%20log%20%280.10%29%20%3D%20-0.495%20)
![[CH_{3}COONa*3H_{2}O] = 10^{-0.495} = 0.32 M](https://tex.z-dn.net/?f=%5BCH_%7B3%7DCOONa%2A3H_%7B2%7DO%5D%20%3D%2010%5E%7B-0.495%7D%20%3D%200.32%20M)
Hence, the mass of the sodium acetate tri-hydrate is:
![m = moles*M = [CH_{3}COONa*3H_{2}O]*V*M = 0.32 mol/L*0.250 L*136 g/mol = 10.88 g](https://tex.z-dn.net/?f=m%20%3D%20moles%2AM%20%3D%20%5BCH_%7B3%7DCOONa%2A3H_%7B2%7DO%5D%2AV%2AM%20%3D%200.32%20mol%2FL%2A0.250%20L%2A136%20g%2Fmol%20%3D%2010.88%20g)
Therefore, the number of grams of CH₃COONa*3H₂O needed to make an acetic acid/sodium acetate tri-hydrate buffer solution is 10.88 g.
I hope it helps you!
