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3 years ago

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In 2011, a computer named Watson competed against two humans on a popular television quiz show and won. Watson listened to each

question and responded.
1. What similarities and differences are there between a computer like Watson and the human nervous system?

1 answer:

Molodets [167]3 years ago

6 0

Answer:

A computer is made up of wires which is like nerves in our body they send signals throughout our bodies and tell us what to do about a situation. The hard drive of a computer is just like our brain which is the main component of the nervous system it controls involuntary, voluntary, movement and coordination. Also, just like the keys and mouse of a computer we have sensory neurons which pick up what is around us by touch.

Explanation:

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Answer: D: a volcano forming a new island in a chain

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The map shows the distribution of three species of organisms observed during a population survey. Identify the pattern of distri

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ZILLDIFFEQMODAP11 3.1.021. My Notes Ask Your Teacher A tank contains 150 liters of fluid in which 10 grams of salt is dissolved.

SSSSS [86.1K]

Answer:

Therefore, the number of grams of salt in the tank at time t is $A(t) = 150-140 e^{-\frac{t}{30} }$

Explanation:

Given:

Tank A contain $V_{1} = 150$ lit

Rate $\alpha = 5 \frac{L}{min}$

Dissolved salt $A = 10$ gm

Salt pumped in one minute is $4 \frac{L}{min}$

Salt pumped out is $\frac{5L}{150L} = \frac{1}{30}$ of initial amount added salt.

To find $A(t)$

$\frac{dA}{dt} = Rate _{in} - Rate _{out}$

$A' = 5 - \frac{A}{30}$

$A' + \frac{A}{30} = 5$

Solving above equation,

$I .F = e^{\int\limits {p} \, dt }$

$y = e^{\int\limits {\frac{1}{30} } \, dt }$

$y = e^{\frac{t}{30} }$

$(Ae^{\frac{t}{30} } )' = 5 e^{\frac{t}{30} } + c$

Integrating on both side,

$Ae^{\frac{t}{30} } = 5 \times 30 e^{\frac{t}{30} } +c$

Add $e^{-\frac{t}{30} }$ on above equation,

$A = 150 + ce^{-\frac{t}{30} }$

Here given in question, $A(t=0) = 10$

$10 =150 +c$

$c = -140$

Put value of constant in above equation, and find the number of grams of salt in the tank at time t.

$A(t) = 150-140 e^{-\frac{t}{30} }$

Therefore, the number of grams of salt in the tank at time t is $A(t) = 150-140 e^{-\frac{t}{30} }$

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3 years ago

A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (

djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

$Kb = \frac{[ HSO3-][0H-]}{SO3-2}$

$Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7$

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation $Kb=\frac{[HSO3-][OH-]}{[SO3-2]}$

We noe have:

$1.485*10^-^7=\frac{x*x}{(0.056-x)}$

$x = [OH-] = 9.11*10^-^5$

$pOH = -log(OH) = -log(9.11*10^-^5)$

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

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What is Le chatelier's principle?

ser-zykov [4K]

Answer:

Le Chatelier's principle, also called Chatelier's principle or "The Equilibrium Law", is a principle of chemistry used to predict the effect of a change in conditions on chemical equilibria.

hope it helps ,pls mark me as brainliest

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