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meriva
3 years ago
13

Which situation is the best example of interia

Chemistry
1 answer:
AveGali [126]3 years ago
6 0
One's body movement to the side when a car makes a sharp turn. Tightening of seat belts in a car when it stops quickly. A ball rolling down a hill will continue to roll unless friction or another force stops it.
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Which is the most likely to be reduced
ollegr [7]

Answer:

B

Explanation:

4 0
3 years ago
Read 2 more answers
What is the simplest aromatic compound?<br> toluene<br> benzene<br> cyclohexane<br> cyclopropane
pav-90 [236]
It’s Benzene, I do believe :) hope this helped, good luck!
3 0
3 years ago
determine the ph of a buffer that is 0.55 M HNO2 and 0.75 M KNO2. tha value of Ka for HNO2 is 6.8*10^-4
Mariana [72]

Answer:

pH = 3.3

Explanation:

Buffer solutions minimize changes in pH when quantities of acid or base are added into the mix. The typical buffer composition is a weak electrolyte (wk acid or weak base) plus the salt of the weak electrolyte. On addition of acid or base to the buffer solution, the solution chemistry functions to remove the acid or base by reacting with the components of the buffer to shift the equilibrium of the weak electrolyte left or right to remove the excess hydronium ions or hydroxide ions is a way that results in very little change in pH of the system. One should note that buffer solutions do not prevent changes in pH but minimize changes in pH. If enough acid or base is added the buffer chemistry can be destroyed.

In this problem, the weak electrolyte is HNO₂(aq) and the salt is KNO₂(aq). In equation, the buffer solution is 0.55M HNO₂ ⇄ H⁺ + 0.75M KNO₂⁻ . The potassium ion is a spectator ion and does not enter into determination of the pH of the solution. The object is to determine the hydronium ion concentration (H⁺) and apply to the expression pH = -log[H⁺].

Solution using the I.C.E. table:

              HNO₂ ⇄    H⁺   +   KNO₂⁻

C(i)        0.55M       0M      0.75M

ΔC            -x            +x          +x

C(eq)  0.55M - x       x     0.75M + x    b/c [HNO₂] / Ka > 100, the x can be                                    

                                                             dropped giving ...

           ≅0.55M        x       ≅0.75M        

Ka = [H⁺][NO₂⁻]/[HNO₂] => [H⁺] = Ka · [HNO₂]/[NO₂⁻]

=> [H⁺] = 6.80x010⁻⁴(0.55) / (0.75) = 4.99 x 10⁻⁴M

pH = -log[H⁺] = -log(4.99 x 10⁻⁴) -(-3.3) = 3.3

Solution using the Henderson-Hasselbalch Equation:

pH = pKa + log[Base]/[Acid] = -log(Ka) + log[Base]/[Acid]

= -log(6.8 x 10⁻⁴) + log[(0.75M)/(0.55M)]

= -(-3.17) + 0.14 = 3.17 + 0.14 = 3.31 ≅ 3.3

3 0
3 years ago
A composition of reflections over parallel lines is the same as a __________.
8090 [49]

Answer : Option A) Translation


Explanation : A composition of reflections over parallel lines is the same as a <u>Translation.</u>


To identify if the composition of reflections over parallel lines are same as translation or not?


We can check using a picture of some shape in the plane. Place the picture on the right side of two vertical parallel. Now, we can see the reflected the shape over the nearest parallel line, then check the reflection over the other parallel line. We see that the shape winds up in the same orientation, like it was just shifted over to the right. Hence, it is translation.

6 0
3 years ago
Read 2 more answers
What is the electric field (in N/C) at a point 5.0 cm from the negative charge and along the line between the two charges?
Natali [406]

Answer: E = 2.455 x 10^5 N/C

Explanation:

q1 = 1.2x10^-7C

q2 = 6.2x10^-8C

Electric field, E = kQ/r²

where k = 9.0x10^9

since the location is (27 - 5)cm from q1

hence electric field, E1 = k*q1/r²

E1= (9x10^9 x 1.2x10^-7)/(0.22)² = 22314.05 N/C

for q2:

E1 = k*q2/r²

E2 at 5cm

E2 = (9x10^9 x 6.2x10^-8)/(0.05)² = 223200 N/C

Hence, the total electric field at 5cm position is

E = E1 + E2

E = 22314.05 + 223200 = 245514.05 N/C

E = 2.455 x 10^5 N/C

4 0
3 years ago
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