The reaction of propyl magnesium bromide with formaldehyde

Joe is concerned about the carbon dioxide emissions from his daily commute and the impact that it is having on global warming In

Vesna [10]

Amount of CO₂ emission per day is 11,356.23 g.

<u>Explanation:</u>

Joe travelling distance per day = 60 miles

Carbon dioxide emission per day = 20 mpg

Now we have to find the amount of carbon dioxide emitted per day by dividing the distance by the emission per day given in gallons.

Amount of Carbon dioxide emission = $$\frac{distance}{emission amount}$

Amount of CO₂ emission in gallons = $$\frac{ 60 miles}{20 mpg}$

= 3 gallons

Now we have to convert the gallons to grams as,

1 gallon = 3,785.41 g

3 gallons = 3 × 3785.41 g = 11,356.23 g

So the emission of CO₂ per day is 11,356.23 g.

6 0

3 years ago

What is the molarity of a solution containing 23 g of NaCl in 500 mL of solution?

trapecia [35]

molarity of a solution means mols per liter.

First, you need to convert 23 grams on NaCl into mols. 23g divided by molar mass (58.44g/mol) which gives you .394 mols.

Now, you need to convert 500ml to L which moves the decimal three places to the left, giving you .500L of solution.

Finally, divide the mols over solution to get .787M

7 0

3 years ago

Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy for the following redox

stealth61 [152]

Answer:

-973 KJ

Explanation:

The balanced reaction equation is;

N2H4(aq) + 2Cl2(g) + 4OH^-(aq)---------> 4Cl-(aq) + 4H ^+(aq) + 4OH^-(aq) + N2(g)

Reduction potential of hydrazine = -1.16 V

Reduction potential of chlorine = 1.36 V

From;

E°cell= E°cathode - E°anode

E°cell= 1.36 - (-1.16)

E°cell= 2.52 V

∆G°=- nFE°cell

n= number of moles of electrons = 4

F= Faraday's constant = 96500 C

E°cell = 2.52 V

∆G°=- (4 × 96500 × 2.52)

∆G°= -972720 J

∆G°= -972.72 KJ

7 0

4 years ago

Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water according to the following equation:

rusak2 [61]

Answer:

60.42% is the percent yield of the reaction.

Explanation:

Moles of methane gas at 734 Torr and a temperature of 25 °C.

Volume of methane gas = V = 26.0 L

Pressure of the methane gas = P = 734 Torr = 0.9542 atm

Temperature of the methane gas = T = 25 °C = 298.15 K

Moles of methane gas = n

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.9542 atm\times 26.0L}{0.0821 atm L/mol K\times 298.15 K}=1.0135 mol$

Moles of water vapors at 700 Torr and a temperature of 125 °C.

Volume of water vapor = V' = 23.0 L

Pressure of water vapor = P' = 700 Torr = 0.9100 atm

Temperature of water vapor = T' = 125 °C = 398.15 K

Moles of water vapor gas = n'

$P'V'=n'RT'$

$n'=\frac{PV}{RT}=\frac{0.9100 atm\times 23.0L}{0.0821 atm L/mol K\times 398.15 K}=0.6402 mol$

$CH_4(g)+H_2O(g)\rightarrow CO(g)+3H_2(g)$

According to reaction , 1 mol of methane reacts with 1 mol of water vapor. As we can see that moles of water vapors are in lessor amount which means it is a limiting reagent and formation of hydrogen gas will depend upon moles of water vapors.

According to reaction 1 mol of water vapor gives 3 moles of hydrogen gas.

Then 0.6402 moles of water vapor will give:

$\frac{3}{1}\times 0.6402 mol=1.9208 mol$ of hydrogen gas

Moles of hydrogen gas obtained theoretically = 1.9208 mol

The reaction produces 26.0 L of hydrogen gas measured at STP.

At STP, 1 mole of gas occupies 22.4 L of volume.

Then 26 L of volume of gas will be occupied by:

$\frac{1}{22.4 L}\times 26 L= 1.1607 mol$

Moles of hydrogen gas obtained experimentally = 1.1607 mol

Percentage yield of hydrogen gas of the reaction:

$\frac{Experimental}{Theoretical}\times 100$

$\%=\frac{ 1.1607 mol}{1.9208 mol}\times 100=60.42\%$

60.42% is the percent yield of the reaction.

8 0

3 years ago

What fuel source is Jan using if she exercises at 85% of her maximum aerobic capacity?

Marina86 [1]

Answer:

Carbohydrates

Explanation:

Increased exercise intensity means the overall need for energy increases. As we increase exercise intensity we increase our glucose uptake and oxidation which far exceeds uptake, indicating that muscle stores of glycogen are being used. At moderate intensities (65%) there is an increased need for muscle glycogen and muscle triglycerides which is fat. At higher levels of intensities (85%) there is an even greater need for energy, and this is met almost solely by an increased uptake of glucose from the blood and from muscle glycogen.

In the case of fats as an energy fuel source at high intensities, increasing levels of intensity increases fat oxidation but once we get into higher levels of intensity, we return to levels of fat oxidation similar to very low intensities.

4 0

3 years ago

The reaction of propyl magnesium bromide with formaldehyde​

The reaction of propyl magnesium bromide with formaldehyde