In the reaction between 1 molecule of bromine and 2 molecules of potassium chloride, there are six atoms in the products.
Let's consider the balanced equation for the reaction between 1 molecule of bromine and 2 molecules of potassium chloride. This is a single replacement reaction.
Br₂ + 2 KCl ⇒ 2 KBr + Cl₂
We obtain as products, 2 molecules of potassium bromide and 1 molecule of chlorine.
- 1 molecule of KBr has 2 atoms, so 2 molecules contribute with 4 atoms.
- 1 molecule of Cl₂ has 2 atoms.
- The 4 atoms from KBr and the 2 atoms from Cl₂ make a total of 6 atoms.
In the reaction between 1 molecule of bromine and 2 molecules of potassium chloride, there are six atoms in the products.
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Answer:
Distance = 13.9 meters
Explanation:
Given the following data;
Maximum speed = 150 km/hr to meters per seconds = 150 * 1000/3600 = 41.67 m/s
Decelerating speed = 3m/s
To find the distance travelled with this speed;
Distance = maximum speed/decelerating speed
Distance = 41.67/3
Distance = 13.9 meters
Therefore, the bus would travel a distance of 13.9 meters before stopping.
A. Average speed is weighted mean (1 × 2 + 2 × 3 + 3 × 5 + 4 × 7 + 3 × 9 + 2 × 12.5)/15 = (2 + 6 + 15 + 28 + 27 + 25)/15 = 103/15 = 6.867 b. RMS is square root of 1/15 times sum of squares of speeds Sum of squares is 4 + 9 + 9 + 25 + 25 + 25 + 49 + 49 + 49 + 49 + 81 + 81 + 81 +156.25 + 156.25 = 848.5
c. RMS speed = √(848.5/15) = 7.521
Most likely the speed is the peak in the speed distribution, which is 7.
Answer:
please give me brainlist and follow
Explanation:
At the bottom of the hill, the baby carriage will likely have less momentum Therefore, option D is correct. Solution: ... Therefore, at the bottom of the hill, the heavy truck will have more momentum and baby carriage will have less momentum.
Answer:
F = ⅔ F₀
Explanation:
For this exercise we use Coulomb's law
F = k q₁q₂ / r²
let's use the subscript "o" for the initial conditions
F₀ = k q² / r²
now the charge changes q₁ = q₂ = 2q and the new distance is r = 3 r
we substitute
F = k 4q² / 9 r²
F = k q² r² 4/9
F = ⅔ F₀
