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3 years ago

CAN SOMEONE PLEASE HELP ME

Chemistry

2 answers:

sammy [17]3 years ago

3 0

Answer:

Condensation

Explanation:

son4ous [18]3 years ago

3 0

try condensation ...............

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What are the characteristics of Mercury? (Select all that apply.)

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less gravity, closest to the sun

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3 years ago

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This image shows electronegativities of elements on the periodic table. Based on this information, which element is most likely

andrezito [222]

Answer:

helium

Explanation:

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3 years ago

Draw the best Lewis structure for NH3 by filling in the bonds, lone pairs, and formal charges. (Assign bonds, lone pairs, radica

kiruha [24]

Answer : The Lewis-dot structure of $NH_3$ is shown below.

Explanation :

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The given molecule is, $NH_3$

As we know that hydrogen has '1' valence electron and nitrogen has '5' valence electrons.

Therefore, the total number of valence electrons in $NH_3$ = 5 + 3(1) = 8

According to Lewis-dot structure, there are 6 number of bonding electrons and 2 number of non-bonding electrons.

Now we have to determine the formal charge for each atom.

Formula for formal charge :

$\text{Formal charge}=\text{Valence electrons}-\text{Non-bonding electrons}-\frac{\text{Bonding electrons}}{2}$

$\text{Formal charge on N}=5-2-\frac{6}{2}=0$

$\text{Formal charge on }H_1=1-0-\frac{2}{2}=0$

$\text{Formal charge on }H_2=1-0-\frac{2}{2}=0$

$\text{Formal charge on }H_3=1-0-\frac{2}{2}=0$

Hence, the Lewis-dot structure of $NH_3$ is shown below.

3 0

3 years ago

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

How many salt and water would you use to prepare 10% salt solution?

tatuchka [14]

Answer:

We can make 10 percent solution by volume or by mass. A 10% of NaCl solution by mass has ten grams of sodium chloride dissolved in 100 ml of solution. Weigh 10g of sodium chloride. Pour it into a graduated cylinder or volumetric flask containing about 80ml of water.

Explanation:

have a good day

4 0

3 years ago