All categories

3 years ago

HELP ME PLEASE!!!

Chemistry

1 answer:

777dan777 [17]3 years ago

8 0

Answer:

A bonding that occurs between high electronegative atoms such are N, F, O and H atoms, is called a hydrogen bond. Hydrogen bond is a very strong bond. (C)

If hydrogen bonds are not formed between H atoms and N, F, O atom, then the atoms interact through dispersion forces (also known as london dispersion forces). Dispersion forces are weak and they are temporary forces formed by overlapping of orbitals. (B)

You might be interested in

Moises performs a chemical reaction according to the reaction

taurus [48]

Answer:

Explanation:

mass of the reactant = mass of the product

15.31 + 1.50 = 16.81g

4 0

3 years ago

Mass number is equal to

ValentinkaMS [17]

The answer : protons and neutrons

Mark me as brainliest ?❤️

4 0

4 years ago

When heat move in a gas or liquid, warm matter rise and cold matter sinks, creating currents. which type of heat transfer is thi

ch4aika [34]

I believe it is convection

6 0

3 years ago

the atomic mass of N is 14.01 g/mol and the atomic mass of H is 1.008 g/mol. what is the molecular mass of NH3

nikdorinn [45]

Molecular mass= (14.01∗1)+(1.008∗3)
14.01+3.024=17.03g/mol

7 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