Answer: The concentration of the OH-, CB = 0.473 M.
Explanation:
The balanced equation of reaction is:
2HCl + Ca(OH)2 ===> CaCl2 + 2H2O
Using titration equation of formula
CAVA/CBVB = NA/NB
Where NA is the number of mole of acid = 2 (from the balanced equation of reaction)
NB is the number of mole of base = 1 (from the balanced equation of reaction)
CA is the concentration of acid = 1M
CB is the concentration of base = to be calculated
VA is the volume of acid = 23.65 ml
VB is the volume of base = 25mL
Substituting
1×23.65/CB×25 = 2/1
Therefore CB =1×23.65×1/25×2
CB = 0.473 M.
3090000000nm
since there's 1m = 1000000000nm
Let the mass of the solute be x
So, the equation would be
x/2.5+x ×100 = 23.22
x/2.5+x = 23.22/100
100x = 58.05 + 23.22x
100x - 23.22x = 58.05
76.78x = 58.05
x =0.756 ≈ 0.76 litres
Answer:
Answer is given below:
Explanation:
<em>Given Data:</em>
mass = 80kg
acceleration = 4 ms
force = 800N
<em>Find out:</em>
friction = ?
<em>Formula</em><em>:</em>
F-friction = weight - f-net
<em>Solution:</em>
weight = (80)(10)
= 800 N
F-net = ma =(80)(4) = 320N
F-friction = weight - F-net
=800 N - 320N
=480N
<em>Answer</em> :
Friction = 480 N
Answer: The distance is slightly less than 3.5 m
Explanation: assuming wall and target are the same thing, and the bullet has constant velocity, the bullet will travel 7 m in half a second, so half that distance is 3.5 m.
In reality, the bullet is decelerating (at an unknown rate) so the distance is slightly less than 3.5 m.
There is also a vertical velocity component, which means it hits the target/wall at an angle. The trajectory is such that it hits the wall above the shooter because the ricochet hits at ~the level at which it left the firearm.
If the wall was absent, the bullet would have described a parabola which brough it back to the initial level after 7m. This could be calculated, but it means that the actual distance between the shooter and the wall is slightly less than 3.5 m
In addition, the collision with the wall is not 100% elastic, so the velocity aftercthe ricochetvis further reduced.
A calculation would be complex because these confounding factors are not completely independent of each other, but all reduce the average velocity and therefore the distance.
Therefore it is only possible to say that the distance was somewhat less than 3.5 m
