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Reptile [31]
2 years ago
12

A team of geologists is surveying on a mountain. They come across a newly formed stream. The stream is flowing at an average rat

e of 420 cubic meters per minute. At the average rate of flow the stream will produce 14,700cubic meters of water in
Chemistry
1 answer:
zhannawk [14.2K]2 years ago
6 0

Answer:

35 minutes

Explanation:

The flow rate of the stream is the rate of volume of water with respect to time. It expresses the volume of water flowing per unit time.

Flow rate = \frac{volume}{time}

From the given question,

average flow rate of the stream = 420 cubic meters per minute, and the required volume = 14700 cubic meters.

So that;

time = \frac{volume}{flow rate}

       = \frac{14700}{420}

       = 35

time = 35 minutes

Therefore, with the given average flow rate, the steam will produce 14,700 cubic meters of water in 35 minutes.

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5.972 × 10^24 kg

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As the Amplitude of the wave decreases what be does the frequency do​
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Answer: pic of the test

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4 0
2 years ago
Select correct examples of linear molecules for five electron groups. Select correct examples of linear molecules for five elect
Shalnov [3]

Explanation:

Formula to calculate hybridization is as follows.

                Hybridization = \frac{1}{2}[V+N-C+A]

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

So, hybridization of BeCl_{2} is as follows.

              Hybridization = \frac{1}{2}[V + N - C + A]

                                    = \frac{1}{2}[2 + 2]

                                    = 2

Hybridization of BeCl_{2} is sp. Therefore, BeCl_{2} is a linear molecule. There will be only two electron groups through which Be is attached.

Similarly, hybridization of XeF_{2} is calculated as follows.

         Hybridization = \frac{1}{2}[V + N - C + A]

                                    = \frac{1}{2}[8 + 2]

                                    = 5

Therefore, hybridization of XeF_{2} is sp^{3}d. Therefore, [tex]XeF_{2} is also a linear molecule. Though there are three lone pair of electrons present on a xenon atom and it is further attached with fluorine atoms through two electron pairs. Hence, there are in total five electron groups.

Thus, we can conclude that out of the given options XeF_{2} is the correct examples of linear molecules for five electron groups.

7 0
2 years ago
One mole of an ideal gas with a volume of 1.0 L and a pressure of 5.0 atm is allowed to expand isothermally into an evacuated bu
Deffense [45]

Answer:

w= - 1.7173 kJ, q= 1.7173 kJ, q(rev) = 1717.3 J = 1.7173 kJ.

Explanation:

Okay, from the question we are given the information below;

Number of moles, n= 1 mole; initial volume, v(1) = 1.0 litres (L); pressure (p) = 5atm, final volume(v2) = 2.0 Litres(L) ; the workdone, w= not given; the heat, q and q(rev)= not given and the gas was said to expand isothermally.

So, this question is a question from the part of chemistry known as thermodynamics. Therefore, grip yourself we are delving into thermodynamics 'waters' now.

For expansion isothermally; the workdone, w= -nRT ln v2/v1.

Where T= temperature= 25° C = 298 k and R= gas constant.

Therefore; workdone, w = - 1 × 8.314 × 298 × ln(2/1).

Workdone,w= - 1717.32204643. =

- 1717.3 Joules (J).

==> Workdone,w= - 1.7173 kJ.

Then, we are to find q. q can be solved by using the first law of thermodynamics, which by mathematical representation is:

∆U= q + w. Where ∆U= change in internal enegy. Since the question is dealing with isothermal expansion, there is this rule that says for an isothermal expansion ∆U = 0.

Hence, 0 =q + [- 1717.3 Joules (J)].

q=1717.3 J = 1.7173 kJ.

Finally, the q(rev) which is= nRT ln (v2/V1).

q(rev) = 1 × 8.314 × 298 ln (2/1).

q(rev) = 1717.3 J = 1.7173 kJ.

PS: please note the negative signs in the workdone and the positive sign in the q(rev).

7 0
3 years ago
calculate the concentration of silver in a 100.0 mL solution of silver bromide at 25oC (298 K) when the Ksp is 5.0 x 10-13
Sliva [168]

Answer:

Concentration AgBr at saturation = 7.07 x 10⁻⁷M

Explanation:

Given AgBr(s) => Ag⁺(aq) + Br⁻(aq) ; Ksp = 5 x 10⁻¹³ = [Ag⁺][Br⁻]

    I       ---                0              0

    C     ---               +x             +x

     E     ---                 x               x

[Ag⁺][Br⁻] = (x)(x) = x² = 5 x 10⁻¹³ => x = SqrRt(5 x 10⁻¹³) = 7.07 x 10⁻⁷M

4 0
2 years ago
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