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3 years ago

Is there a new powerful weapons that stronger than nuclear weapons. If yes, explain why and how.

Chemistry

1 answer:

ahrayia [7]3 years ago

7 0

No. There are stronger and more powerful nuclear weapons, but nothing is stronger than a nuclear weapon.

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During a storm, a tree fell over into a river. What might happen to this tree?

mina [271]

Break down in to tiny prices as the water hit the tree

6 0

3 years ago

Over long time scales the solubility-temperature feedback (below), can affect climate. An increase in atmospheric CO2 concentrat

Mars2501 [29]

The fact that CO2 is released from oceans due to further rise in temperature is an example of a negative feedback.

A negative feedback is one in which the process that produces the feedback is interrupted. That is, the process is stopped as a result of the feedback received.

In this case, CO2 which leads to global warming dissolves in the ocean which serves a large sink for the gas. However, as the increase in ocean temperatures causes decrease in solubility of CO2, more CO2 is released leading to further temperature rise. This is an example of a negative feedback loop.

Learn more: brainly.com/question/13440572

5 0

3 years ago

How many mol of sodium corresponds to 1.0 x 10^15 atoms of sodium?

andrew-mc [135]

The Avogadro number represents the number of units in one mole of a chemical substance.
So to find the mole number of a chemical element, you divide its atom number of the Avogadro number which Na = 6.02*10^23 approx.
So n=N/Na (n=mole number, N=number of atoms, Na=Avogadro number)
n=1.0*10^15/6.02*10^23
n=1.6 * 10^-9

So 1.0*10^15 atoms of Sodium represent 1.6*10^-9 mol.

Hope this Helps! :)

3 0

3 years ago

Read 2 more answers

The most common form of elemental sulfur is S8, in which eight sulfur atoms linked in a ring of single bonds. At high temperatur

aliina [53]

Answer:

Explanation:

In S₈ , there are 8 single bonds which breaks up first . Energy absorbed

= 8 x 240 = 1920 kJ

In S₈ four double bonds of S₂ are formed . Let bond energy be x . In this process energy will be released . energy released in four S₂ molecules formed = 4 x

Given

1920 + 4x = 239

4x = 239 - 1920

x = - 420.25 kJ .

So bond energy of S₂ = 420.25 kJ .

8 0

3 years ago

Write packing efficiency of fcc ,BCC, SCC and formula also . ?

Minchanka [31]

❖ Packing Efficiency ❖

➪ The percentage of total space occupied by particles is called packing efficiency.

$\\ \qquad{\rule{200pt}{3pt}}$

$\bigstar \boxed{ \sf \: \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}$

❖ Packing efficiency of simple cubic structure (SCC).❖

$\rm \: {Let \: } \bf{r } \: \rm{ be \: the \: radius \: of \: a \: sphere }\: \\ \rm and \: \bf{a} \: \rm {be \: the \: edge \: of \: unit \: cell.} \\ \\ \; \bigstar \boxed{ \sf{Volume_{(sphere)} = \dfrac{4}{3} \pi r^3}} \bigstar$

❒ Since, simple cubic unit cell contain 1 atom (sphere). So, the total volume of sphere will be :

$: : \implies \sf \: 1 \times \dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi r^3$

Volume of unit cell = a³

Volume of unit cell = (2r)³

Volume of unit cell = 8r³

❒ Now, we know that,❒

$\bigstar \boxed{ \sf \: \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}$

➪ Substituting the known values in the formula, we get the following results:

$: : \implies \rm \: {\dfrac{\dfrac{4}{3}\pi r^3}{8r^3} \times 100} \\ \\ : : \implies \dfrac{\pi}{6} \times 100 \\\\\ : : \implies \bf {52.4 \%}$

❒ Hence, the packing efficiency of simple cubic structure is 52.4%.

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➪ Packing efficiency of cubic close packing (SCC)/face centred cubic structure (FCC).

❒ Since, simple cubic unit cell contain 2 atom (sphere). So, the total volume of sphere will be:

$: : \implies \rm 4 \times \dfrac{4}{3}\pi r^3 = \dfrac{16}{3}\pi r^3$

Volume of unit cell = a³

Volume of unit cell = (2√2r)³

Volume of unit cell = 16√2r³

❒ Now, we know that,❒

$\bigstar \boxed{ \sf \: \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}$

➪Substituting the known values in the formula, we get the following results:

$: : \implies \rm {\dfrac{\dfrac{16}{3}\pi r^3}{16\sqrt{2}r^3} \times 100 }\\\\ \implies \rm \dfrac{\pi}{3\sqrt{2}} \times 100 \\\\\ \implies\bf52.4 \%$

❖ Hence, the packing efficiency of face centred cubic structure is 74%.

$\\ \qquad{\rule{200pt}{3pt}}$

❖ Packing efficiency of body cubic structure (BCC).

❒ Since, simple cubic unit cell contain 2 atom (sphere). So, the total volume of sphere will be:

$: : \implies \rm \: 2 \times \dfrac{4}{3}\pi r^3 = \dfrac{8}{3}\pi r^3$

Volume of unit cell = a³

Volume of unit cell = (4r/√3)³

Volume of unit cell = 64r³/3√3

❒ Now, we know that,❒

$\bigstar \boxed{ \sf \: \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}$

➪Substituting the known values in the formula, we get the following results:

$: : \implies \rm {\dfrac{\dfrac{8}{3}\pi r^3}{\dfrac{64r^3}{3\sqrt{3}}} \times 100} \\\\ \implies \dfrac{3\pi}{8} \times 100 \\\ \\ \bf \implies \: 68 \%$

❒ Hence, the packing efficiency of body centred cubic structure is 68%.

$\\ \qquad{\rule{200pt}{3pt}}$

6 0

2 years ago