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valentinak56 [21]
3 years ago
5

in the event of a submarine disaster the___ can be used to rescue people from the depth up to 3500 feet.

Chemistry
2 answers:
gulaghasi [49]3 years ago
6 0
DSVR because it only makes sense. Try it out and if it works give me a Brainly


Alex_Xolod [135]3 years ago
3 0
<h2>Answer:</h2>

<u>A </u><u>deep-submergence rescue vehicle (DSRV)</u><u> can be used to rescue people from the depth up to 3500 feet.</u>

<h2>Explanation:</h2>

A deep-submergence rescue vehicle (DSRV) is a type of deep-submergence vehicle used for rescue of downed submarines. While DSRV is the term most often used by the United States Navy, other nations have different designations for their vehicles too. During the trials, the DSRV also dived successfully up to 666 metres.

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Hypochlorous acid decays in the presence of ultraviolet radiation. Assume that degradation occurs accord- ing to first-order kin
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Answer:

35.75 days

Explanation:

From the given information:

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current concentration C = 0.05 mg/L

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\mathsf{In( \dfrac{0.05}{3.65})= -(0.12) t}

㏑(0.01369863014) = -(0.12) t

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3 years ago
Tim has just jumped out of an airplane.The force of gravity pulls him downward,while air resistance pushes him upward
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3 years ago
Which of the following is not a valid set of quantum numbers
ASHA 777 [7]
I believe the answer is C, n = 3, l = 3, m = 3. The magnetic quantum number, or 

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For option C), the angular momentum quantum number of equal to ++2<span>, which means that <span>ml</span> can have a maximum value of </span>+2<span>. Since it is given as having a value of </span>+3**, this set of quantum numbers is not a valid one.

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3 0
3 years ago
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Suppose a 1.0 L solution contains 0.020 M in Cu(NO3)2, then 0.40 moles of NH3 are added. Assuming no change in volume, what is t
Rasek [7]

Answer:

4.6*10^{-14} M

Explanation:

Concentration of Cu^{2+} = [Cu(NO_3)_2] = 0.020 M

Constructing an ICE table;we have:

                                 Cu^{2+}+4NH_3_{aq} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)}

Initial (M)             0.020          0.40                        0

Change (M)         -  x                - 4 x                       x

Equilibrium (M)  0.020 -x        0.40 - 4 x              x

Given that: K_f =1.7*10^{13}

K_f } = \frac{[Cu(NH_3)_4]^{2+}}{[Cu^{2+}][NH_3]^4}

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Since x is so small; 0.40 -4x = 0.40

Then:

1.7*10^{13} = \frac{x}{(0.020-x)(0.0256)}

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1.7*10^{13}(5.12*10^{-4} - 0.0256x) = x

8.704*10^9-4.352*10^{11}x =x

8.704*10^9 = 4.352*10^{11}x

x = \frac{8.704*10^9}{4.352*10^{11}}

x = 0.0199999999999540

Cu^{2+}= 0.020 - 0.019999999999954

Cu^{2+} = 4.6*10^{-14} M

8 0
3 years ago
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