Answer number six please

The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 3.001 g sample of B

loris [4]

I believe here is the right question, so will just ignore the rest of the junk information from the previous message

The compound known as butylated hydroxytoluene, abbreviated as BHT, contains carbon, hydrogen, and oxygen. A 3.001 g sample of BHT was combusted in an oxygen rich environment to produce 8.990 g of CO2(g) and 2.944 g of H2O(g). Insert subscripts below to appropriately display the empirical formula of BHT

Answer:

$C_{15}H_{24}0$

Explanation:

A 3.001 g sample of BHT was combusted in an oxygen rich environment to produce 8.990 g of CO2(g) and 2.994 g of H2O(g).

If all the carbon in BHT is present in $CO_2$ and also, all the hydrogen in BHT is present in $H_2O$ , Then we can determine for the corresponding numbers of moles of Carbon(C) and Hydrogen (H) respectively as:

moles of $CO_2$ = $8.990 g*(\frac{1mole}{44.01g})$

= 0.2043 moles

∴ moles of C = 0.2043 moles

moles of $H_2O$ = $2.944 g *(\frac{1mole}{18.01g} )$

= 0.1635 moles

∴ moles of H = 2 × 0.1635 moles

= 0.327 moles

Since number of moles= $\frac{mass}{molarmass}$

number of moles of H = 0.327 moles

molar mass of H = 1.008 g/mol

∴ mass of H in the sample = 0.327 moles × 1.008 g/mol

= 0.329616g

mass of C in the sample can be calculated as = 0.2043 moles × $(\frac{12.01g}{1 mole} )$

= 2.453643 g

mass of C+H in the sample = 2.453643g + 0.329616g

= 2.783259 g

mass of O can be calculated as = 3.001 g - 2.783259 g

= 0.217741 g

∴ moles of O = 0.217741g × $(\frac{1mole}{16.0g})$

= 0.0136 moles

Now, since; we've gotten our data, we can now proceed to calculate for the empirical formula.

C H O

0.2043 0.327 0.0136

Dividing by the least number (0.0136) , we have :

$\frac{0.2043}{0.0136}$ $\frac{0.327}{0.0136}$ $\frac{0.0136}{0.0136}$

15.02 24.04 1

≅

15 24 1

Therefore, the empirical formula would be : $C_{15}H_{24}0$

7 0

3 years ago

If each NADHNADH generates 3 ATPATP molecules and each FADH2FADH2 generates 2 ATPATP molecules, calculate the number of ATPATP m

Nadusha1986 [10]

Answer:

$128~ATP$

Explanation:

The metabolic pathway by which energy can be obtained from a fatty acid is called "beta-oxidation". In this route, acetyl-Coa is produced by removing 2 carbons from the fatty acid for each acetyl-Coa produced. In other words, for each round, 1 acetyl Coa is produced and for each round 2 carbons are removed from the initial fatty acid. Therefore, the first step is to calculate the number of rounds that will take place for an 18-carbon fatty acid using the following equation:

$Number~of~Rounds=\frac{n}{2}-1$

Where "n" is the number of carbons, in this case "18", so:

$Number~of~Rounds=\frac{18}{2}-1~=~8$

We also have to calculate the amount of Acetyl-Coa produced:

$Number~of~Acetyl-Coa=\frac{18}{2}~=~9$

Now, we have to keep in mind that in each round in the beta-oxidation we will have the production of 1 $FADH_2$ and 1 $NADH$ . So, if we have 8 rounds we will have 8 $FADH_2$ and 8 $NADH$ .

Finally, for the total calculation of ATP. We have to remember the yield for each compound:

-) $1~FADH_2~=~2~ATP$

-) $1~NADH~=~3~ATP$

-) $Acetyl~CoA~=~10~ATP$

Now we can do the total calculation:

$(8*2)~+~(8*3)~+~(9*10)=130~ATP$

We have to subtract "2 ATP" molecules that correspond to the activation of the fatty acid, so:

$130-2=128~ATP$

In total, we will have 128 ATP.

I hope it helps!

6 0

3 years ago

Write the number of sig. fig. in four numbers given in the sentence below. An (one) octopus has 8 legs. 13 octopi have 104 legs.

Marrrta [24]

Answer:

1, 1, 2, 3

Explanation:

The numbers 1 and 8 both have 1 sig. fig.

The number 13 has 2 sig. figs.

The number 104 has 3 sig. figs.

6 0

3 years ago

8. Thermal Energy is?

Burka [1]

Thermal energy is the sum of the kinetic and potential energy of all the particles in an object. The figure shows that if either potential or kinetic energy increases, thermal energy increases.

hope it really helps...!!!

4 0

2 years ago

Which will contain the greater number of moles of potassium ion: 30.0 ml of 0.15 m k2cro4 or 25.0 ml of 0.080 m k3po4?

Dahasolnce [82]

30.0 ml of 0.15 m K2CrO4 solution will have more potassium ions. Let's see the relative number of potassium ions for each solution. Since all the measurements are the same, the real difference is the K2CrO4 will only have 2 potassium ions per molecule while the K3PO4 solution will have 3 potassium ions per molecule. K2CrO4 solution 30.0 * 0.15 * 2 = 9 K3PO4 solution 25.0 * 0.080 * 3 = 6 Since 9 is greater than 6, the K2CrO4 solution will have more potassium ions.

3 0

3 years ago