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irakobra [83]
3 years ago
6

15

Chemistry
1 answer:
Fynjy0 [20]3 years ago
4 0

Answer:

larva hatching, larva with legs, young newt, adult newt, in that order

You might be interested in
How to balance the chemical equation H2 +N2
brilliants [131]

Answer:

N2 + 3H2 --> 2NH3

Explanation:

N =2                             N = 1*2 =2

H = 2*3 =6                   H = 3*2 = 6

7 0
3 years ago
You drop a rock weighing 23.2 g into a graduated cylinder that contains 55 mL. The level
snow_lady [41]

Answer:

<h2>3.31 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass = 23.2 g

volume = final volume of water - initial volume of water

volume = 62 - 55 = 7 mL

We have

density =  \frac{23.2}{7}  \\  = 3.314285

We have the final answer as

<h3>3.31 g/mL</h3>

Hope this helps you

8 0
3 years ago
The Mendeleev and Moseley periodic charts had
alisha [4.7K]

Answer: The Moseley chart was based on using the atomic mnumber of the element, not the chemical properties. Because of the similarities between elements of the same period, this often created problems in establishing an order; Moseley's work enabled the change of the atomic number from an arbitrary selection to a definable property, measurable through experimentation.

Explanation:

8 0
4 years ago
Hat are the main properties of solids (in contrast to liquids and gases)?
atroni [7]
Molecules huddle close together.
cannot form to any shape.
5 0
3 years ago
The solubility of lead(II) iodide is 0.064 g/100 mL at 20ºC. What is the solubility product for lead(II) iodide?
quester [9]

Answer:

Ksp=1.07x10^{-8}

Explanation:

Hello,

In this case, the dissociation reaction is:

PbI_2(s)\rightleftharpoons Pb^{2+}(aq)+2I^-(aq)

For which the equilibrium expression is:

Ksp=[Pb^{2+}][I^-]^2

Thus, since the saturated solution is 0.064g/100 mL at 20 °C we need to compute the molar solubility by using its molar mass (461.2 g/mol)

Molar solubility=\frac{0.064g}{100mL}*\frac{1000mL}{1L}*\frac{1mol}{461.2g}=1.39x10^{-3}M

In such a way, since the mole ratio between lead (II) iodide to lead (II) and iodide ions is 1:1 and 1:2 respectively, the concentration of each ion turns out:

[Pb^{2+}]=1.39x10^{-3}M

[I^-]=1.39x10^{-3}M*2=2.78x10^{-3}M

Thereby, the solubility product results:

Ksp=(1.39x10^{-3}M)(2.78x10^{-3}M)^2\\\\Ksp=1.07x10^{-8}

Regards.

6 0
4 years ago
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