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3 years ago

A magnet is an example of a

Chemistry

1 answer:

statuscvo [17]3 years ago

7 0

The answer is distance force

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What’a the structure of 4-ethyl-2,2-dimethylheptane?

Slav-nsk [51]

PubChem CID: 142982

Chemical Names: 4-Ethyl-2,2-dimethylhexane; Hexane, 4-ethyl-2,2-dimethyl-; 52896-99-8; 2,2-dimethyl-4-ethylhexane; AC1L3M80; Hexane,4-ethyl-2,2-dimethyl- More...

Molecular Formula: C10H22

Molecular Weight: 142.286 g/mol

InChI Key: QHLDBFLIDFTHQI-UHFFFAOYSA-N

7 0

3 years ago

How many formula units make up 24.2 g of magnesium chloride (MgCl2)? Help!!

NNADVOKAT [17]

Answer:

Approximately $1.53 \times 10^{23}$ formula units ( $0.254\; \rm mol$ ).

Explanation:

Refer to a modern periodic table for the relative atomic mass of magnesium ( $\rm Mg$ ) and chlorine ( $\rm Cl$ ):

$\rm Mg$ : $24.305$ .
$\rm Cl$ : $35.45$ .

In other words, the mass of $1\; \rm mol$ of $\rm Mg$ atoms would be (approximately) $24.305\; \rm g$ .

Likewise, the mass of $1\; \rm mol$ of $\rm Cl$ atoms would be approximately $35.45\; \rm g$ .

One formula unit of the ionic compound $\rm MgCl_{2}$ includes exactly as many atoms as there are in the given formula. The formula mass of a compound is the mass of $1\; \rm mol$ of the formula units of this compound.

The formula $\rm MgCl_{2}$ includes one $\rm Mg$ atom and two $\rm Cl$ atoms.

Hence, every formula unit of $\rm MgCl_{2} \!$ would include the same number of atoms: one $\rm Mg\!$ atom and two $\rm Cl\!$ atoms. There would be $1\; \rm mol$ of $\rm Mg$ atoms and $2\; \rm mol$ of $\rm Cl$ atoms in $1\; \rm mol\!$ of $\rm MgCl_{2}$ formula units.

Thus, the mass of $1\; \rm mol\!$ of $\rm MgCl_{2}$ formula units would be equal to the mass of $1\; \rm mol$ of $\rm Mg$ atoms plus the mass of $2\; \rm mol$ of $\rm Cl$ atoms. (The mass of $1\; \rm mol\!\!$ of each atom could be found from the relative atomic mass of each element.)

$\begin{aligned}& M({\rm MgCl_{2}}) \\ =\; & 24.305\; {\rm g \cdot mol^{-1}} + 2\times {\rm 35.45 \; \rm g \cdot mol^{-1}} \\ =\; & 95.205\; \rm g \cdot mol^{-1}\end{aligned}$ .

In other words, the formula mass of $\rm MgCl_{2}$ is $95.205\; \rm g \cdot mol^{-1}$ .

Therefore, the number of formula units in $m = 24.2\; \rm g$ of $\rm MgCl_{2}$ would be:

$\begin{aligned}n &= \frac{m({\rm MgCl_{2}})}{M({\rm MgCl_{2}})} \\ &= \frac{24.2\; \rm g}{95.205\; \rm g\cdot mol^{-1}} \\ & \approx 0.254\; \rm mol\end{aligned}$ .

Multiple $n$ by Avogadro's Number $N_{A} \approx 6.022 \times 10^{23}\; \rm mol^{-1}$ to estimate the number of formula units in $0.254\; \rm mol$ :

$\begin{aligned}N &= n \cdot N_{A} \\ &\approx 0.254\; \rm mol \times 6.022 \times 10^{23}\; \rm mol^{-1} \\ &\approx 1.53\times 10^{23}\end{aligned}$ .

6 0

3 years ago

What mass of aluminum is needed to produce 0.500 mole of aluminum chloride?

3241004551 [841]

Answer: " 13.5 g Al " ;

→ that is: "13.5 grams of aluminum."

____________________________

Explanation:

____________________________

Note: What is missing from the question is the "balanced chemical equation" for the "chemical reaction" that contains:

The reactants: "aluminum (Al) " ; and "chlorine (Cl) " ; and:

The product: "aluminum choloride (AlCl₃) " .

____________________________

The "balanced chemical equation" is:

____________________________

2 Al + 3 Cl₂ → 2 AlCl₃ ;

_____________________________

Note: The molecular weight of "aluminum (Al)" is: " 26.98 g /mol " .

____________________________

So: We call solve using a technique known as: "dimensional analysis" :

____________________________

0.500 mol AlCl₃ * $(\frac{2mol Al}{2mol AlCl_{3} }) * (\frac{26.98g Al}{1 mol Al}) = ?$

____________________________

Note: The units of "mol AlCl₃" cancel out to "1' ; and:

The units of "mol Al" cancel out to "1" ; and we are left with:

____________________________

" $\frac{(0.500 * 2 * 26.98)}{2}$ g Al ["grams of aluminum"] ;

____________________________

Note: We can "cancel out the "2's" ; since "2/2 = 1 " ; and we have:

→ (0.500 * 26.98) g Al ;

= 13.49 g Al ;

→ Round to 3 (Three) significant figures;

→ Since: "0.500" has 3 (Three) significant figures:

____________________________

= 13.5 g Al ; that is: "13.5 grams of aluminum."

____________________________

Hope this is helpful!

Best wishes to you in your academic pursuits—and within the "Brainly" community!

____________________________

3 0

3 years ago

Which piece of the planetary object data could be used to decide if there actually is a solid surface to land on?

zloy xaker [14]

A planetary surface is where the solid (or liquid) material of the outer crust on certain types of astronomical objects contacts the atmosphere or outer space. Planetary surfaces are found on solid objects of planetary mass, including terrestrial planets (including Earth), dwarf planets, natural satellites, planetesimals and many other small Solar System bodies (SSSBs).[1][2][3] The study of planetary surfaces is a field of planetary geology known as surface geology, but also a focus of a number of fields including planetary cartography, topography, geomorphology, atmospheric sciences, and astronomy. Land (or ground) is the term given to non-liquid planetary surfaces. The term landing is used to describe the collision of an object with a planetary surface and is usually at a velocity in which the object can remain intact and remain attached.

In differentiated bodies, the surface is where the crust meets the planetary boundary layer. Anything below this is regarded as being sub-surface or sub-marine. Most bodies more massive than super-Earths, including stars and gas giants, as well as smaller gas dwarfs, transition contiguously between phases, including gas, liquid, and solid. As such, they are generally regarded as lacking surfaces.

Planetary surfaces and surface life are of particular interest to humans as it is the primary habitat of the species, which has evolved to move over land and breathe air. Human space exploration and space colonization therefore focuses heavily on them. Humans have only directly explored the surface of Earth and the Moon. The vast distances and complexities of space makes direct exploration of even near-Earth objects dangerous and expensive. As such, all other exploration has been indirect via space probes.

Indirect observations by flyby or orbit currently provide insufficient information to confirm the composition and properties of planetary surfaces. Much of what is known is from the use of techniques such as astronomical spectroscopy and sample return. Lander spacecraft have explored the surfaces of planets Mars and Venus. Mars is the only other planet to have had its surface explored by a mobile surface probe (rover). Titan is the only non-planetary object of planetary mass to have been explored by lander. Landers have explored several smaller bodies including 433 Eros (2001), 25143 Itokawa (2005), Tempel 1 (2005), 67P/Churyumov–Gerasimenko (2014), 162173 Ryugu (2018) and 101955 Bennu (2020). Surface samples have been collected from the Moon (returned 1969), 25143 Itokawa (returned 2010), 162173 Ryugu and 101955 Bennu.

3 0

3 years ago

How does increasing pressure affect the rate of reaction between gases?

Elina [12.6K]

Answer:

B, increases rate of collisions

Explanation:

7 0

2 years ago