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FrozenT [24]
3 years ago
15

The hollow bones of birds,which keep birds lightweight for flying, is an example of an???

Chemistry
1 answer:
MatroZZZ [7]3 years ago
8 0
An "adaptation."  It's the natural way that birds of flight adapt to their needs or environment which makes them more efficient.  :)
You might be interested in
Determine % yield if a student obtains 45 g of product in an experiment and the theoretical amount is determined to be 50 g. *
ycow [4]

Answer:

90%

Explanation:

Percentage yield = ?

Theoretical yield = 50g

Actual yield = 45g

To calculate the percentage yield of a compound, we'll have to use the formula of percentage yield which is the ratio between the actual yield to theoretical multiplied by 100

Percentage yield = (actual yield / theoretical yield) × 100

Percentage yield = (45 / 50) × 100

Percentage yield = 0.9 × 100

Percentage yield = 90%

The percentage yield of the substance is 90%

6 0
3 years ago
Please help<br> Explain why a molecule that has polar bonds can be a polar molecule.
Zepler [3.9K]

Answer:

Explanation:

Well, obviously a molecule with polar bonds can be polar in itself. It's like saying I am an atheltic person who can just reach the basketball rim with my head and also I can dunk.

But if the question is how can a molecule that in non-polar have polar bonds, well, its because the polar bonds' dipole cancels each other out. It's like a tight rope. If a person pulls in one direction, it intuitively, the rope would go in that direction. However, if a person pulls in the other direction with the same amount of force, the rope stays still. This is the same case. Although molecules can have different electronegativities, the pull of electrons in one direction is cancelled out by a pull in the opposite direction, making the net dipole 0.

This is common for main VSERP shaped molecules like linear, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral.

8 0
3 years ago
Write the Iconic Bond to Metals
Nutka1998 [239]

<u>Answer:</u> The ionic compound formed is magnesium chloride having formula MgCl_2

<u>Explanation:</u>

Ionic compound is defined as the compound which is formed by complete transfer of electrons from one atom to another atom.

The atom which looses the electron is known as electropositive atom and the atom which gains the electron is known as electronegative atom. This bond is usually formed between a metal and a non-metal.

Magnesium is the 12th element of the periodic table having electronic configuration of 1s^22s^22p^63s^2

This element will loose 2 electrons to form Mg^{2+} ion

Chlorine is the 17th element of the periodic table having electronic configuration of 1s^22s^22p^63s^23p^5

This element will gain 1 electron to form Cl^{-} ion

So, for every 1 atom of magnesium, 2 atoms of chlorine are required. Thus, the chemical formula becomes MgCl_2

Hence, the ionic compound formed is magnesium chloride having formula MgCl_2

6 0
3 years ago
If 15.5 mL of 0.225 M aqueous magnesium chloride is added to 37.5 mL of 0.250 M aqueous lead(II) nitrate, then what mass of lead
Evgen [1.6K]

Answer:

B. 0.971 g

Explanation:

When MgCl₂(aq) reacts with Pb(NO₃)₂(aq), PbCl₂(s) and Mg(NO₃)₂(aq) are produced:

MgCl₂(aq) + Pb(NO₃)₂(aq) →, PbCl₂(s) + Mg(NO₃)₂(aq)

Thus, we need to find imiting reactant finding moles of each reactant:

<em>Moles MgCl₂:</em>

15.5mL = 0.0155L * (0.225 mol / L) = 3.49x10⁻³ moles

<em>Moles Pb(NO₃)₂:</em>

37.5mL = 0.0375L * (0.250mol / L) = 9.38x10⁻³ moles

As the ratio of the reactants is 1:1, the moles of PbCl₂ are 3.48x10⁻³ moles.

We need to convert thes moles to mass using molar mass of PbCl₂ (278.1g/mol), thus:

3.48x10⁻³ moles * (278.1g/mol) =

0.968g of PbCl₂ are precipitate

Thus, right answer is:

<h3>B. 0.971 g</h3>
4 0
3 years ago
If the rate of disappearance of N2O5 is equal to 1.40 mol/min at a particular moment, what is the rate of appearance of NO2 at t
kramer

Answer:

2.8 mol/min  is the rate of appearance of NO_2 at that moment.

Explanation:

2N_2O_5\rightarrow 4NO_2+O_2

Rate of the reaction = R

R=\frac{-1}{2}\frac{d[N_2O_5]}{dt}=\frac{1}{4}\frac{d[NO_2]}{dt}=\frac{1}{1}\frac{d[O_2]}{dt}

Rate of disappearance of N_2O_5 is =-\frac{d[N_2O_5]}{dt}= 1.40 mol/min

R=\frac{-1}{2}\frac{d[NO_2]}{dt}

R=\frac{1}{2}\times 1.40 mol/min=0.70 mol/min

Rate of appearance of NO_2 is =\frac{1}{4}\frac{d[NO_2]}{dt}:

R=\frac{1}{4}\frac{d[NO_2]}{dt}

\frac{d[NO_2]}{dt}=4\times 0.70 mol/min=2.8 mol/min

2.8 mol/min  is the rate of appearance of NO_2 at that moment.

7 0
3 years ago
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