What role do glucose and oxygen play in cellular respiration?

Suppose a student started with 142.0 mg of trans-cinnamic acid, 412 mg of pyridinium tribromide, and 2.30 mL of glacial acetic a

nirvana33 [79]

Answer: Theoretical Yield = 0.2952 g

Percentage Yield = 75.3%

Explanation:

Calculation of limiting reactant:

n-trans-cinnamic acid moles = (142mg/1000) / 148.16 = 9.584*10⁻⁴ mol

pyridium tribromide moles = (412mg/1000) / 319.82= 1.288*10⁻³ mol

n-trans-cinnamic acid is the limiting reactant

The molar ratio according to the equation mentioned is equals to 1:1

The brominated product moles is also = 9.584*10⁻⁴ mol

Theoretical yield = (9.584*10⁻⁴ mol) * (Mr of brominated product)

= (9.584*10⁻⁴ mol) * (307.97) = 0.2952 g

Percentage Yield is : Actual Yield / Theoretical Yield = 0.2223/0.2952

= 75.3%

4 0

3 years ago

Crushing large chunks of sodium chloride: What lab equipment

finlep [7]

Goggles Gloves materials

4 0

2 years ago

Problem Page Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 70. g of

shusha [124]

Answer:

The maximum mass of carbon dioxide that could be produced by the chemical reaction is 70.6gCO_{2}

Explanation:

1. Write down the balanced chemical reaction:

$2C_{6}H_{14}_{(l)}+19O_{2}_{(g)}=12CO_{2}_{(g)}+14H_{2}O_{(g)}$

2. Find the limiting reagent:

- First calculate the number of moles of hexane and oxygen with the mass given by the problem.

For the hexane:

$70.0gC_{6}H_{14}*\frac{1molC_{6}H_{14}}{86.2gC_{6}H_{14}}=0.81molesC_{6}H_{14}$

For the oxygen:

$81.3gO_{2}*\frac{1molO_{2}}{32.0gO_{2}}=2.54molesO_{2}$

- Then divide the number of moles between the stoichiometric coefficient:

For the hexane:

$\frac{0.81}{2}=0.41$

For the oxygen:

$\frac{2.54}{19}=0.13$

- As the fraction for the oxygen is the smallest, the oxygen is the limiting reagent.

3. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction:

The calculations must be done with the limiting reagent, that is the oxygen.

$81.3gO_{2}*\frac{1molO_{2}}{32gO_{2}}*\frac{12molesCO_{2}}{19molesO_{2}}*\frac{44.0gCO_{2}}{1molCO_{2}}=70.6gCO_{2}$

7 0

3 years ago

2.3.12mL = L* O a. .312 O b.31.2 O C..00312 O d..0312

Anna11 [10]

Answer:

a..312 I guesss kkkkkkkkkkkkkkkkkk

6 0

2 years ago

3. The following data of decomposition reaction of thionyl chloride (SO2Cl2) were collected at a certain temperature and the con

KonstantinChe [14]

Answer:

a) First-order.

b) 0.013 min⁻¹

c) 53.3 min.

d) 0.0142M

Explanation:

Hello,

In this case, on the attached document, we can notice the corresponding plot for each possible order of reaction. Thus, we should remember that in zeroth-order we plot the concentration of the reactant (SO2Cl2 ) versus the time, in first-order the natural logarithm of the concentration of the reactant (SO2Cl2 ) versus the time and in second-order reactions the inverse of the concentration of the reactant (SO2Cl2 ) versus the time.

a) In such a way, we realize the best fit is exhibited by the first-order model which shows a straight line (R=1) which has a slope of -0.0013 and an intercept of -2.3025 (natural logarithm of 0.1 which corresponds to the initial concentration). Therefore, the reaction has a first-order kinetics.

b) Since the slope is -0.0013 (take two random values), the rate constant is 0.013 min⁻¹:

$m=\frac{ln(0.0768)-ln(0.0876)}{200min-100min} =-0.0013min^{-1}$

c) Half life for first-order kinetics is computed by:

$t_{1/2}=\frac{ln(2)}{k}=\frac{ln(2)}{0.013min^{-1}} =53.3min$

d) Here, we compute the concentration via the integrated rate law once 1500 minutes have passed:

$C=C_0exp(-kt)=0.1Mexp(-0.013min^{-1}*1500min)\\\\C=0.0142M$

Best regards.

6 0

3 years ago