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3 years ago

How many moles of carbon are in 3.7 moles of C 8 H11 NO2

1 answer:

6 0

In every molecule of $C_{8} H_{11}NO_2$ there is 8 atoms of Carbon.

IF we have 3.7 moles of $C_{8} H_{11}NO_2$ to find the number of moles of Carbon, just multiply by 8

3.7 * 8 = 29.6 mol Carbon

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Can any one help on this one

enyata [817]

0.004382166 Make sure to round to the right amount of Sig Figs

8 0

3 years ago

If the volume of wet gas collected over water is 85.0 mL at 20°C and 760 mm Hg , what is the volume of dry gas at STP conditions

dimulka [17.4K]

Answer: 77.4 mL

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of dry gas = (760 - 17.5) mmHg= 742.5 mm Hg

$P_2$ = final pressure of dry gas at STP = 760 mm Hg

$V_1$ = initial volume of dry gas = 85.0 mL

$V_2$ = final volume of dry gas at STP = ?

$T_1$ = initial temperature of dry gas = $20^oC=273+20=293K$

$T_2$ = final temperature of dry gas at STP = $0^oC=273+0=273K$

Now put all the given values in the above equation, we get the final volume of wet gas at STP

$\frac{742.5mmHg\times 85.0ml}{293K}=\frac{760mmHg\times V_2}{273K}$

$V_2=77.4mL$

Volume of dry gas at STP is 77.4 mL.

5 0

3 years ago

An ideal gas occupies a volume V at an absolute temperature T. If the volume is halved and the pressure kept constant, what will

Kruka [31]

Answer:

It will be halve of T

Explanation:

V1 = V

T1 = T

V2 = ½V

T2 = x

V1/T1 = V2/T2

V/T = ½V/x

Vx = ½VT

2Vx = VT

2x = T

x = ½T

6 0

3 years ago

How many moles are in 18.2 g of CO2?<br> 41.4 moles<br> 801 moles<br> 0.414 moles<br> 0 2.42 moles

mrs_skeptik [129]

Answer:

0.414 mole (3 sig. figs.)

Explanation:

Given grams, moles = mass/formula weight

moles in 18.2g CO₂(g) = 18.2g/44g/mole = 0.413636364 mole (calc. ans.)

≅ 0.414 mole (3 sig. figs.)

6 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