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dolphi86 [110]
3 years ago
9

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

Chemistry
1 answer:
Dafna1 [17]3 years ago
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
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