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marta [7]
3 years ago
5

An atom has 8 protons, 8 neutrons and 9 electrons. is thus a cation, anion, or neutral atom?​

Chemistry
1 answer:
Black_prince [1.1K]3 years ago
7 0

Answer:

It is an <u>Anion</u> because there are more electrons than protons making it negative

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State the type of bonding—ionic, covalent, or metallic—you would expect in each: (c) Na(s).
miv72 [106K]

Na(s) forms an ionic bond.

<h3>What is ionic bond?</h3>

The main interaction in ionic compounds is ionic bonding, a type of chemical bonding that involves the electrostatic attraction between two atoms or ions with dramatically differing electronegativities. Along with metallic and covalent bonds, it is one of the most common types of bonds. Atoms (or collections of atoms) possessing an electrical charge are known as ions. Ions with negative charges are created when atoms gain electrons (called anions). Positively charged ions are produced when atoms lose electrons (called cations). In contrast to covalence, this electron transfer is referred to as electrovalence.

Ionic chemicals normally do not conduct electricity when solid, only when molten or in solution. Depending on the charge of the ions they are made of, ionic compounds typically have a high melting point.

To learn more about ionic bond from the given link:

brainly.com/question/13526463

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5 0
1 year ago
The temperature of evaporation is much higher for water than for alcohol. Without knowing more about the chemistry of alcohol, w
3241004551 [841]

Answer:

C

Explanation:

Alcohols are organic molecules characterized majorly by the presence of the OH group in their molecule. The OH group is majorly responsible for several of their characteristics. This include the formation of hydrogen bonds between alcohol molecules. While this makes them more inorganic than most organic compounds, comparatively the hydrogen bonding formed in alcohols is not as strong as that which is present in water.

The higher strength of the hydrogen bonding is responsible for some comparable properties. While water boils at a temperature of 100 degrees Celsius, alcohol boils at a temperature of 78 degrees Celsius. This is an evidence to the fact that hydrogen bonding in alcohol is less stronger that that in water.

4 0
3 years ago
What is the major product of this reaction? which reaction sequence would provide the highest yield of 1 -bromo-2-mcihylcyclohex
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Of which reaction may I ask? And do you mean 1-bromo-2-methylcyclohexane??
3 0
3 years ago
Which statement correctly describes diamond and graphite, which are different forms of solid carbon?
mart [117]
They differ in their molecular structures and properties.
5 0
3 years ago
Read 2 more answers
Calculate the unit cell edge length for an 85 wt% fe-15 wt% v alloy. All of the vanadium is in solid solution, and, at room temp
Lady bird [3.3K]

Answer is 0.289nm.

Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.

wt % of Fe in Fe-V alloy = 85%

wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}

For calculating edge length,

a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}

For calculating M_{ave}, we use the formula

M_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{V}}{M_V}}

Similarly for calculating (\rho)_{ave}, we use the formula

\rho_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{\rho_{Fe}}+\frac{(wt\%)_{V}}{\rho_V}}

From the periodic table, masses of the two elements can be written

M_{Fe}= 55.85g/mol

M_{V}=50.941g/mol

Specific density of both the elements are

(\rho)_{Fe}=7.874g/cm^3\\(\rho)_{V}=6.10g/cm^3

Putting  M_{ave} and \rho_{ave} formula's in edge length formula, we get

a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}}  \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}}  \right )}  \right ]^{1/3}

a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}}  \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}}  \right )}  \right ]^{1/3}

By calculating, we get

a=2.89\times10^{-8}cm=0.289nm

7 0
3 years ago
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