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kirill115 [55]
3 years ago
6

What is the potential energy of a rollercoaster if it weighs 1565 kg and sits on a 115 m tall ride?

Chemistry
1 answer:
AnnyKZ [126]3 years ago
7 0

Answer:

Explanation:

Potential energy builds as the coaster is going up the loop-the-loop and converts to kinetic energy on the way down and out of the loop-the-loop. The potential energy of the coaster is at its lowest and the kinetic energy is at its highest as the coaster moves in the lowest point of the loop-the- loop.

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State the number of protons contained in the atomic number 8<br>​
Nat2105 [25]

Answer:

There are 8 protons

Explanation:

{ \bf{atomic \: number = proton \: number}}

6 0
3 years ago
Solution A is 0.44 M and reacts with 0.11 M of solution B. Assume that the value of x is 0, the value of y is 1, and r is 1.07 ×
xz_007 [3.2K]

Answer:

K, the rate constant = 9.73 × 10^(-1)/s

Explanation:

r = K × [A]^x × [B]^y

r = Rate = 1.07 × 10^(-1)/s

K = Rate constant

A and B = Concentration in mol/dm^-3

A = 0.44M

B = 0.11M

x = Order of reaction with respect to A = 0

y = Order of reaction with respect to B = 1

Solving, we get

r/([A]^x × [B]^y) = K

K = 1.07 × 10^(-1)/s/(0.44^0 × 0.11^1)= 0.9727

K = 0.9727

7 0
3 years ago
What would happen to life if earth were in a different location within the solar system
chubhunter [2.5K]
Closer=Burn
Farther=Freeze
We are the perfect distance away from the sun for it to sustain life.
3 0
4 years ago
Carbon dioxide, water vapor, and oxygen pass through ---------------
Lelechka [254]
The answer is B) stomata.
5 0
3 years ago
Read 2 more answers
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
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