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kaheart [24]
3 years ago
7

Calculate the volume of dry co2 produced at body temperature (37 ∘c) and 0.960 atm when 24.0 g of glucose is consumed in this re

action.
Chemistry
1 answer:
weqwewe [10]3 years ago
6 0
The balanced equation for the above reaction is as follows;
C₆H₁₂O₆ + 6O₂ --> 6CO₂ + 6H₂O
molar ratio of glucose to CO₂ is 1:6
the mass of glucose reacted - 24.0 g
Number of glucose moles reacted - 24.0 g /180 g/mol = 0.133 mol
Number of CO₂ formed when 0.133 mol of glucose reacted -0.133 x 6 = 0.798 mol
since CO₂ is a gas we can use the ideal gas law equation to find the volume occupied 
PV = nRT
where P - pressure - 0.960 atm x 101 325 Pa / atm = 97 272 Pa
n - number of moles - 0.798 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 273 + 37 °C = 310 K
substituting the values in the equation
97 272 Pa x V= 0.798 mol x 8.314 Jmol⁻¹K⁻¹ x 310 K
V = 21.1 dm³
volume occupied by CO₂ is 21.1 dm³
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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
A sulfuric acid solution containing 571.3 g of h2so4 per liter of aqueous solution has a density of 1.329 g/cm3. Part a calculat
loris [4]

Mass percentage of a solution is the amount of solute present in 100 g of the solution.

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Therefore we have:

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Answer:

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We have, Billy the friendly Robot uses 50 N of force to lift a box 2 meters in the air.

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So, the work performed by Billy is 100 J.

4 0
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