How long does it take for fruit flies to develop from embryos to adults?

What animal wins march mammal madness in 2021?

Semmy [17]

Answer:

turtles i like turtles

Explanation:

6 0

3 years ago

Predict, using Boyle’s Law, what will happen to a balloon that an ocean diver takes to a pressure of 202 kPa (normal atmospheric

kaheart [24]

Explanation:

Balloon that an ocean diver takes to a pressure of 202 k Pa will get reduced in size that is the volume of the balloon will get reduced. This is because pressure and volume of the gas are inversely related to each other.

According to Boyle's law: The pressure of the gas is inversely proportional to the volume occupied by the gas at constant temperature(in Kelvins).

$pressure\propto \frac{1}{volume}$ (At constant temperature)

The pressure beneath the sea is 202 kPa and the atmospheric pressure is 101.3 kPa . This increase in pressure will result in decrease in volume occupied by the gas inside the balloon with decrease in size of a balloon. Hence, the size of the balloon will get reduced at 202 kPa (under sea).

7 0

3 years ago

State the reason that soft drinks is a mixture and water is a compound?

Yuri [45]

Carbonated drinks are a mixture of many compounds such as glucose(sugar) and others, while water is simply its own molecule.

7 0

2 years ago

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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:

A)

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

B)

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⁺.

3 0

3 years ago

If two identical atoms are bonded together, what kind of molecule is formed?

AlekseyPX

Answer:

C. A linear, nonpolar molecule

Explanation:

Molecules which are alike usually have the same degree of pull which results in them sharing electrons. This sharing of electrons is known as the molecules exhibiting Covalent bonding between them.

The equal pull also results in the cancelling out of electrons and favoring non polar bonds due to the absence of free electrons which would have been able to interact with H2O in a polar binding system.

6 0

3 years ago

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