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3 years ago

Which type of seismic waves produces the most severe ground movement?

1 answer:

8 0

The surface waves are the type of seismic waves that produce the most severe ground movement. This wave is slow in nature and so produces a rolling effect similar to a surface wave in a pond. This kind of wave is far more devastating than the P waves and the S waves. The surface waves have the capacity to shake a building from side to side until it collapses. This kind of wave moves in a pattern similar to a circle. It actually originates at a point and then start moving outwards in a circle.

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Do you believe the research subjects in the original study were appropriate for this experiment

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3 years ago

Read 2 more answers

True or False: A theory is something that can never be proven correct or incorrect?

cestrela7 [59]

Answer:true

true

Explanation:

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2 years ago

In methane combustion, the following reaction pair is important: At 1500 K, the equilibrium constant Kp has a value of 0.003691

andre [41]

Explanation:

Let us assume that the value of $K_{r}$ = $2.82 \times 10^{5} \times e^({\frac{-9835}{T}}) m^{6}/mol^{2}s$

Also at 1500 K, $K_{r}$ = $2.82 \times 10^{5} \times e^({\frac{-9835}{1500}}) m^{6}/mol^{2}s$

$K_{r} = 400.613 m^{6}/mol^{2}s$

Relation between $K_{p}$ and $K_{c}$ is as follows.

$K_{p} = K_{c}RT$

Putting the given values into the above formula as follows.

$K_{p} = K_{c}RT$

$0.003691 = K_{c} \times 8.314 \times 1500$

$K_{c} = 2.9 \times 10^{-7}$

Also, $K_{c} = \frac{K_{f}}{K_{r}}$

or, $K_{f} = K_{c} \times K_{r}$

= $2.9 \times 10^{-7} \times 400.613$

= $1.1617 \times 10^{-4} m^{6}/mol^{2}s$

Thus, we can conclude that the value of $K_{f}$ is $1.1617 \times 10^{-4} m^{6}/mol^{2}s$ .

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3 years ago

In substitution reactions, (CH3)3C-I reacts at the same rate with Br- and Cl- even though Br- is a more reactive nucleophile tha

Kamila [148]

Answer:

A. (CH3)3C-I reacts by SN1 mechanism whose rate is independent of nucleophile reactivity.

Explanation:

We must recall that (CH3)3C-I is a tertiary alkyl halide. Tertiary alkyl halides preferentially undergo substitution reaction via SN1 mechanism.

In SN1 mechanism, the rate of reaction depends solely on the concentration of the alkyl halide (unimolecular mechanism) and is independent of the concentration of the nucleophile. As a result of this, both Br^- and Cl^- react at the same rate.

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3 years ago

Substance A has the following properties.

givi [52]

A curve of temperature vs. time for the entire heating process.

The sample is heated up to 100.°C, therefore, the heat and time required to heat the sample to its boiling point, the heat and time required to boil the sample, and the heat and time required to heat the sample from its boiling point to 100.°C are needs to be calculated.

i ) Calculating the heat and time required to heat the sample to its boiling point:

Boiling point = 85°C

C(liquid) = 2.5 J/g °C

The heat required up to melting the sample is calculated in the previous parts. Therefore, the heat required to heat the sample from -20°C to 85°C can be calculated as,

Therefore, T f = 85°C and T i = - 20°C

Plug in the values in the specific heat formula to calculate the heat energy required to heat the sample to its melting point,

q3 = 25 g × 2.5 J/g °C × [85 - (-20)]°C

= 25 J/°C ×[85+20]°C

= 6562.5 J

The total heat energy required for heating the sample from initial temperature to boiling point is:-

q1 + q2 + q3 = 500 J + 4500 J + 6562.5 J

= 11562.5 J

The Rate of heating = 450 J/min

450. J = 1 min

11562.5 J = ? min

11562.5 J × 1min/450 J = 25.69 min

ii) Calculating the heat and time required to boil the sample:

∆H Vap = 500 J/g

The boiling is the phase change from liquid to gas at 85°C, therefore, the heat required to boil the sample can be determined

q4= m × ∆Hvap

= 25 g × 500 J/g

= 12500 J

Thus, total heat required to this phase change is q1 + q2 + q3 + q4 = 500 J + 4500 J +6562.5 J + 12500 J = 24062.5 J

The Rate of heating = 450 J / min

450 J = 1 min

24062.5 J = ? min

24062.5J × 1min / 450 J = 53.47 min

iii) Calculating the heat and time required to heat the sample from its boiling point to 100°C

C gas = 0.5 J / g °C

The heat required to boil the sample is calculated in the previous parts. Therefore, the heat required to heat the sample from 85°C to 100°C can be calculated as,

Therefore, T f = 100.°C and T i = 85°C

q5 = 25 g × 0.5 J / g °C × [100 - 85] °C

= 25 J / °C ×15 °C

= 187.5 J

The total heat energy required for heating the sample from initial temperature to 100°C is

q1 + q2 + q3 + q4 + q5 = 500 J + 4500 J + 2625J + 12500 J + 187.5 J

=24250 J

The Rate of heating = 450 J / min

450. J = 1 min

24250 J=? min

Thus, heating the sample to 100.°C takes a total of 53.89 min.

iv) Draw a curve of temperature vs. time for the entire heating process:-

Temperature °C Temperature K Heat energy (J) Time (min)

-40 °C 233 0 0

-20 °C 253 500 1.11

Melting -20 °C 253 5000 11.11

85 °C 358 11562.5 25.69

Boiling 85 °C 358 24062.5 53.475

100 °C 373 24250 53.89

Hence, the graph for the result is in the image.

Learn more about temperature here:-brainly.com/question/24746268

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4 0

1 year ago