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3 years ago

HELPP ASAPP

Chemistry

1 answer:

Trava [24]3 years ago

3 0

Answer:

The answer for the calculation is as follows

17.48373

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The [OH-] of a solution is 7.89 10^-12 M. What is the pH of the solution? it acidic or basic? *

kompoz [17]

Answer: pH = 2,897 , basic $[H+][OH-] = 10^{-14} ==> [H+] = \frac{10^{-14}}{7,89*10^{-12} } =\frac{1}{789} \\pH= -lg([H+]) = 2,897 \\pH basic$

Explanation:

6 0

3 years ago

Use one of your experimentally determined values of k, the activation energy you determined, and the Arrhenius equation to calcu

stealth61 [152]

Answer:

Explanation:

Use one of your experimentally determined values of k, the activation energy you determined, and the Arrhenius equation to calculate the value of the rate constant at 25 °C. Alternatively, you can simply extrapolate the straight line plot of ln(k) vs. 1/T in your notebook to 1/298 , read off the value of ln(k), and determine the value of k. Please put your answer in scientific notation. slope=-12070, Ea=100kJ/mol, k= 0.000717(45C), 0.00284(55C), 0.00492(65C), 0.0165(75C), 0.0396(85C)

Explanation;

According to Arrhenius equation:

i.e. ln(k2/k1) = -Ea/R (1/T2 - 1/T1)

Where, k1 = 0.000717, T1 = 45 oC = (45+273) K = 318 K

T2 = 25 oC = (25 + 273) K = 298 K

i.e. ln(k2/0.000717) = -12070 (1/298 - 1/318)

i.e. ln(k2/0.000717) = -2.54738

i.e. k2/0.000717 = $e^{-2.54738}$

= 0.078286

Therefore, the required constant (k2) = 0.078286 * 0.000717 = $5.61*10^-^5$

6 0

3 years ago

What is the velocity of a 0.5-kg ball<br> that has a momentum of 3 kg.m/s?

maria [59]

Answer:

6m/s

Explanation:

Data obtained from the question include:

Mass = 0.5kg

Momentum = 3 kg.m/s

Velocity =.?

Momentum is simply the product of mass and velocity as shown by the equation below:

Momentum = Mass x Velocity

Velocity = Momentum /mass

Velocity = 3kg.m/s / 0.5kg

Velocity = 6m/s

Therefore, the velocity of the ball is 6m/s

3 0

3 years ago

What must happen in a chemical reaction?

klasskru [66]

The anwser is atoms are destroyed

6 0

2 years ago

The total volume required to reach the endpoint of a titration required more than the 50 mL total volume of the buret. An initia

Margaret [11]

Answer:

The endpoint volume is 50.52 ± 0.14 mL

Explanation:

In a titration always is necessary to subtract the blank volume to the titrant volume to obtain the real volume of the titrant. Thus in this case, the total endpoint volume is the sum of the initial volume delivered and the second volume delivered, minus the blank volume:

V = (49.16±0.06 mL) + (1.69±0.04 mL) - (0.33±0.04 mL)

V = (49.16 + 1.69 - 0.33) ± (0.06+0.04+0.04) mL

V = 50.52 ± 0.14 mL

It is necessary to consider the sum of the errors too.

7 0

3 years ago