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3 years ago

Does a longer paper airplane fly farther than a wide one

Chemistry

1 answer:

storchak [24]3 years ago

6 0

Yes, Yes it does in fact

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Stannum has a body centered tetragonal with lattice constant, a = b = 5.83A and c = 3.18A. If the atomic radius is 0.145 nm, det

Fed [463]

Answer:

the atomic packing factor of Sn is 0.24

Explanation:

a = b = 5.83A and c = 3.18A.

Volume of unit cell = a²c

= (5.83)² * 3.18 * 10⁻²⁴ cm³

= 1.08 * 10⁻²²cm³

Volume of atoms =

$2 \times \frac{4}{3} \pi r^3$

(∴ BCC, effective number of atom is 2)

Volume of atoms =

$2 * \frac{4}{3} *3.14*(0.145*10^-^7cm)^3$

= 2.55*10⁻²³cm³

$\text {Atomic packing factor}=\frac{\text {volume occupied by atom}}{\text {volume of unit cell }}$

$=\frac{2.55*10^-^2^3}{1.08*10^-^2^2} \\\\=0.24$

<h3>therefore, the atomic packing factor of Sn is 0.24</h3>

4 0

4 years ago

What would you do to change a saturated solid/liquid solution to an unsaturated solution

Semenov [28]

Explanation:

Saturated solid/liquid solution is changed to an unsaturated solution by the dilution process.

When the saturated solution is diluted by addition large quantity of solvent it changed to an unsaturated solution.

Saturated solid/liquid solution can also be changed to an unsaturated solution by increasing the temperature or heating the solution.

A saturated solution is a solution where further solute will not dissolve in the solvent.

An unsaturated solution is a solution in which has the ability to dissolve more solute.

4 0

3 years ago

you are running a transformation reaction, and in your microfuge tube currently is your plasmid and your insert fragments (both

ElenaW [278]

A foreign DNA molecule can be incorporated into a bacterial plasmid during a transformation reaction.

<h3>How to explain the reaction?</h3>

With the aid of two enzymes, ligase and restriction enzymes, a foreign DNA molecule can be incorporated into a bacterial plasmid during a transformation reaction. Each enzyme detects a target DNA sequence and cuts it nearby, while ligase aids in connecting the DNA. When two bits of DNA have complimentary bases, it facilitates their joining.

Plasmid and the insert fragment are both present in the microfuge tube, and they both have compatible sticky ends. However, the ligase has been denatured and is no longer active because the prior student left it outside rather than freezing it; despite this, we had already put the ligase into the tube. Ligase aids in binding the plasmid and insert fragments together, but because it is denatured in this instance, it will no longer be able to do so. As a result, no transformation process will take place. And since ligase links DNA fragments together by catalyzing the development of connections between the nearby nucleotides, the two fragments will not be able to unite.

Learn more about reactions on:

brainly.com/question/11231920

#SPJ1

4 0

2 years ago

Si2C19<br> What is this compound

lutik1710 [3]

Answer:

BROMOTRIPHENYLSILANE

Explanation:

<h2>MARK ME AS BRAINLIST</h2>

PLZ FOLLOW ME

7 0

3 years ago

25.0cm3 of s saturated potassium hydroxide is neutralized by 35.0cm3 of hydrogen chloride acid of concentration 0.75 mol/dm3. Ca

Kamila [148]

Answer:

Concentration of the original $\rm KOH$ solution: approximately $1.05\; \rm mol \cdot dm^{-3}$ .

Explanation:

Notice that the concentration of the $\rm HCl$ solution is in the unit $\rm mol\cdot dm^{-3}$ . However, the unit of the two volumes is $\rm cm^{3}$ . Convert the unit of the two volumes to $\rm dm^{3}$ to match the unit of concentration.

$\begin{aligned} V(\mathrm{NaOH}) &= 25.0\; \rm cm^{3} \\ &= 25.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0250\; \rm dm^{3} \end{aligned}$ .

$\begin{aligned} V(\mathrm{HCl}) &= 35.0\; \rm cm^{3} \\ &= 35.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0350\; \rm dm^{3} \end{aligned}$ .

Calculate the number of moles of $\rm HCl$ molecules in that $0.0350\; \rm dm^{3}$ of $0.75\; \rm mol \cdot dm^{3}$ $\rm HCl\!$ solution:

$\begin{aligned}n(\mathrm{HCl}) &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.00350\; \rm dm^{3} \times 0.75\; \rm mol \cdot dm^{3} \\ &\approx 0.02625\; \rm mol\end{aligned}$ .

$\rm HCl$ is a monoprotic acid. In other words, each $\rm HCl\!$ would release up to one proton $\rm H^{+}$ .

On the other hand, $\rm KOH$ is a monoprotic base. Each $\rm KOH\!$ formula unit would react with up to one $\rm H^{+}$ .

Hence, $\rm HCl$ molecules and $\rm KOH\!$ formula units would react at a one-to-one ratio.

${\rm HCl}\, (aq) + {\rm NaOH}\, (aq) \to {\rm NaCl}\, (aq) + {\rm H_2O}\, (l)$ .

Therefore, that $0.02625\; \rm mol$ of $\rm HCl$ molecules would neutralize exactly the same number of $\rm NaOH$ formula units. That is: $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ .

Calculate the concentration of a $\rm NaOH$ solution where $V(\mathrm{NaOH}) = 0.0250\; \rm dm^{3}$ and $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ :

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} \\ &= \frac{0.02625\; \rm mol}{0.0250\; \rm dm^{3}}\approx 1.05\; \rm mol \cdot dm^{-3}\end{aligned}$ .

7 0

3 years ago