1. If in the following diagram the substance is in solid form during stage 1, at what stage do the particles have no intermolecu
lar bonding?
(THERE IS A PIC AND IT'S THE FIRST ONE!!!)
1
3
5
Intermolecular bonding occurs during all stages.
2. Which of the following equations is not balanced?
N2 + 3H2 → 2NH3
MgI2 + Br2 → MgBr2 + I2
KClO3 → KCl + O2
HCl + NaOH → NaCl + H2O
3. In the alpha decay of Polonium-218, what daughter element is produced?
(PIC IS PIC 2)
4. Using the general equation above for a nuclear reaction, supply the following answer:
(PIC IS PIC 3)
There are (96,146,242) curium neutrons.
(THERE IS A PIC AND IT'S THE FIRST ONE!!!)
1
3
5
Intermolecular bonding occurs during all stages.
2. Which of the following equations is not balanced?
N2 + 3H2 → 2NH3
MgI2 + Br2 → MgBr2 + I2
KClO3 → KCl + O2
HCl + NaOH → NaCl + H2O
3. In the alpha decay of Polonium-218, what daughter element is produced?
(PIC IS PIC 2)
4. Using the general equation above for a nuclear reaction, supply the following answer:
(PIC IS PIC 3)
There are (96,146,242) curium neutrons.
2 answers:
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Answer:
Velocity, u = 14.7 m/s
Explanation:
It is given that, a driver can probably survive an acceleration of 50 g that lasts for less than 30 ms, but in a crash with a 50 g acceleration lasting longer than 30 ms, a driver is unlikely to survive.
Let v is the highest speed that the car could have had such that the driver survived. Using a = -50 g and t = 30 ms
Using first equation of kinematics as :
In case of crash the final speed of the driver is, v = 0
u = 14.7 m/s
So, the highest speed that the car could have had such that the driver survived is 14.7 m/s. Hence, this is the required solution.
Answer:
pH = 6.82
Explanation:
To solve this problem we can use the<em> Henderson-Hasselbach equation</em>:
- pH = pKa + log
We're given all the required data to <u>calculate the original pH of the buffer before 0.341 mol of HCl are added</u>:
- pKa = -log(Ka) = -log(2.9x10⁻⁸) = 7.54
- [HOCl] = [NaOCl] = 0.500 mol / 0.125 L = 4 M
- pH = 7.54 + log
- pH = 7.54
By adding HCl, w<em>e simultaneously </em><u><em>increase the number of HOCl</em></u><em> and </em><u><em>decrease NaOCl</em></u>:
- pH = 7.54 + log
- pH = 7.54 + log
- pH = 6.82