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3 years ago

What might be a good metaphor for this statement? All the birds get fed when the corn gets crushed

Chemistry

1 answer:

Oksana_A [137]3 years ago

5 0

Poor people get money when we eat the rich or something along those lines

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A cyclist rode at an average speed of 10 mph for 15 miles. How long was the ride?

xxMikexx [17]

If the cyclist rode at an average speed of 10mph for 15 miles...

we can solve by dividing the distance by the speed to get time using the equation...

Δspeed = Δdistance / Δtime

Δtime = Δdistance / Δspeed

Δtime = 15 miles / 10 mph = 1.5 hours

7 0

3 years ago

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Would you weigh more or less on the moon than you do on Earth, and why?

VladimirAG [237]

Answer:You would weigh less on the moon because there is less gravity on the moon.

Explanation:

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3 years ago

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atmospheric pressure at elevations of 8000 feet averages about 0.72 atmospheres. Would a cabin pressurized at 500 mm hg meet fed

NeTakaya

Answer:

Yes, but it must be kept at that value and do not let it to decrease more.

Explanation:

Hello.

In this case, in order to substantiate whether the cabin meet the federal standards, we need to convert the 500 mmHg to atm and compare the result with 0.72 atm by knowing that 1 atm equals 760 mmHg:

$500mmHg*\frac{1atm}{760mmHg} \\\\=0.66atm$

Thus, since 0.66 atm is 0.06 atm away from the federal standard we can infer that it may meet the federal standard, however, it would not be recommended to let the pressure decrease more than that.

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3 years ago

true or false: the use of fossil fuels for energy production beneficial to the environment and not cause environmental problems

MissTica

Answer:

False
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3 years ago

If PbI2(s) is dissolved in 1.0MNaI(aq) , is the maximum possible concentration of Pb2+(aq) in the solution greater than, less th

fredd [130]

Answer:

$\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

Less than the concentration of Pb2+(aq) in the solution in part ( a )

Explanation:

From the question:

We assume that s to be the solubility of PbI₂.

The equation of the reaction is given as :

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹

[Pb²⁺] = s

Then [I⁻] = 2s

$K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} = 4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{The mathematical expressionthat can be used to determine the value of }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

The Concentration of Pb²⁺ in water is calculated as :

$\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

$\mathbf{s =\sqrt [3]{\dfrac{7*10^{-9}}{4}}}$

$\mathbf{s} =\sqrt[3]{1.75*10^{-9}}$

$\mathbf{s} =\mathbf{1.21*10^{-3} \ mol/L }$

The Concentration of Pb²⁺ in 1.0 mol·L⁻¹ NaI

$\mathbf{PbCl{_2}} \leftrightharpoons \ \ \ \ \ \ \ \mathbf{Pb^{2+}} \ \ \ \ \ + \ \ \ \ \ \ \ \mathbf{2 I^-}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf0} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+2x}$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{+x} \ \ \ \ \ \ \ \ \ \ \ \ \mathbf{1.0+2x}$

The equilibrium constant:

$K_{sp} =[Pb^{2+}}][I^-]^2 \\ \\ K_{sp} = s*(1.0*2s)^2 =7*1.0^{-9} \\ \\ s = 7*10^{-9} \ \ m/L$

It is now clear that maximum possible concentration of Pb²⁺ in the solution is less than that in the solution in part (A). This happens due to the common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.

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4 years ago