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Bess [88]
3 years ago
14

Why is it necessary to centrifuge out any precipitate formed in the unknown solution and continue testing the remaining unknown

solution?
Physics
1 answer:
jonny [76]3 years ago
5 0

Answer:

Precipitation is the formation of a solid from a solution. It is necessary to centrifuge the precipitate to exert sufficient forces of gravity to bring the solid particles in the solution to come together and settle

Explanation:

When you centrifuge precipitate it enables the nucleation to form.

Centrifuging the precipitate helps in determining whether a certain element is present in a solution or not.

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I WILL GIVE YOU BRAINLIEST IF YOU ANSWER THOS QUESTION IN THE NEXT 5 MINUTES!!
schepotkina [342]

Answer:

the blue shopping cart.

Explanation:

The blue shopping cart doesnt have to worry about running someone over in the front. The red one does, so it slows down more.

7 0
2 years ago
By what factors does the speed of the elctron exceed that of the proton?<br>​
____ [38]

Answer:

» An electron is lighter than a proton.

<u>explanation</u><u>:</u>

{  =  \: \sf{an \: electron \: has \: formula \:  \: }}{ \bf{ {}^{0}_{  - 1}e }}

hence it's mass number is zero

{  =  \: \sf{an \: electron \: is \: helium \: particle \:  \: }}{ \bf{ {}^{4} _{2}He  }}

hence it's mass number is 4

<u>Therefore</u><u>,</u><u> </u><u>proton</u><u> </u><u>is</u><u> </u><u>heavier</u><u> </u><u>than</u><u> </u><u>electron</u>

» An electron has a small charge magnitude than a proton.

<u>Explanation</u><u>:</u>

An electron has charge of -1 while proton has charge of +2, therefore electron is less deflected by any energetic fields than a proton

8 0
3 years ago
a skier is gliding along 3.0 m/s on horizontal, frictionless snow. he suddenly starts down a 10 degrees incline. his speed at th
Elza [17]
... find length
(way 1) determine acceleration using force
only force act on skier is mg vertically. spilt vector we get force along the incline = mgsin(10) and f= ma so
ma = mgsin(10) or a = gsin(10)
a (along the incline)= gsin(10) = 10sin(10) = 1.74
v^2 = u^2 + 2as
15^2 = 3^2 + 2(1.74)s
s = 62.06 m

(way 2) using conservation of energy
energy (KE+PE) on top = energy at bottom
0.5m3^2 + mgh = 0.5m15^2 +0
h (height of incline) = (112.5 - 4.5)/10 = 19.8 m
length of incline = h/sin(10) = 62.2 m ; trigonometry

... find time
s = (u+v)t/2
t = 2s/(u+v) = 2(62.2)/(3+15) = 6.91 s
4 0
3 years ago
______ images are obtained when the plane mirror are kept parallel to the each other​
uranmaximum [27]

.infinite.

&

also

.virtual.

4 0
3 years ago
Very lost please help.
V125BC [204]
A) The acceleration is due to gravity at any given point if you look at it vertically, so -10 m/s^2.

b) sin(25) = V_y/V, so V_y = V*sin(25). We use V = V_0 + a t and then the final speed must be 0 because it stops at the highest point. So 0 = V_y - 10t. Solve for t and you get t = 32sin(25)/10 = 16sin(25)/5

c) Y = Y_0 + V_0t + (1/2)at^2, and then we plug the values: Y_m_a_x = 32sin(25)*t - (1/2)*10*t^2 and we already have the time from "b)", so Y_m_a_x = [(32sin(25))*(32sin(25)/10)] - 5(32sin(25)/10)^2; then we just rearrange it Y_m_a_x = 10[(32sin(25))^2/100] - 5 [(32sin(25))^2/100] and finally Y_m_a_x = 5[(32sin(25))^2/100] = (32sin(25))^2/20
6 0
3 years ago
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