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Tomtit [17]
3 years ago
7

How many atoms are there in one molecule of C22H44O?

Chemistry
2 answers:
Alik [6]3 years ago
8 0

Answer:

64

Explanation:

LUCKY_DIMON [66]3 years ago
4 0
Answer
64 atoms

Hope it helps
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In an investigation with density, Marcia’s teacher measures the mass and volume of 10 different samples of a substance. The samp
valina [46]

Answer:

Mass  

Step-by-step explanation:

Usually, you plot the independent variable along the horizontal (x) axis and the dependent variable along the vertical (y) axis.

Marcia's teacher plotted the mass of the sample along the x-axis and volume along the y-axis.

The mass is the independent variable, because that is <em>what the teacher varied</em>.

The volume is the <em>dependent variable</em>, because it <em>depends</em> on the mass.

Sample number is <em>wrong</em>, because it is not a variable.

Substance is <em>wrong</em>, because all samples consist of the same substance.

Density is <em>wrong</em>, because it is constant. It is the slope of the graph.

3 0
3 years ago
Read 2 more answers
A 50-mL beaker only has a scale that measures 10, 20, 30, and 40 mL. What is the uncertainty associated with the 50 mL beaker.
Juliette [100K]
If the uncertainty of a certain measurement instrument is not given, then it is assumed to be equal to half of the least count of that instrument. In this case, the least count is 10 ml, so half of this is 5 ml. Therefore, the graduated cylinder has an uncertainty of +/- 5 ml
8 0
3 years ago
Calculate total ATP produced from a fatty acid of 32 carbons
emmasim [6.3K]

Answer:

Total number of ATP molecules generated from a 32-carbon fatty acid = 206 ATP molecules

Explanation:

A 32 carbon fatty acid which undergoes complete beta-oxidation assuming that the fatty acid is fully saturated will pass through the beta-oxidation cycle 14 times to produce the following:

15 molecules of acetylCoA, 14 molecules of FADH₂, and 14 molecules of NADH.

Each of the 15 acetylCoA molecules can be further oxidized in the citric acid cycle to yield the following: 15 × 3 NADH; 15 × 1 FADH₂, and 15 ATP molecules from the substrate level phosphorylation occuring at the succinylCoA synthetase catalyzed-reaction.

Total FADH₂ produced = 15 + 14 = 29 molecules of FADH₂

Total NADH produced = 45 + 14 = 59 molecules of NADH

The FADH₂ and NADH will each donate a pair of electrons to the electron transfer flavoprotein and mitochondrial NADH dehydrogenase respectively of the electron transport chain, and about 1.5 and 2.5 molecules of ATP are generated respectively when these electrons are transfered to molecular oxygen.

Thus, number of molecules of ATP generated by 29 molecules of FADH₂ = 1.5 × 29 = 43.5 molecules of ATP.

Number of molecules of ATP generated by 59 molecules of NADH = 2.5 × 59 = 147.5

Sum of ATP generated from FADH₂ and NADH = 43.5 + 147.5 = 191 ATP molecules

Total number of ATP molecules generated = 191 + 15 = 206 ATP molecules

Total number of ATP molecules generated from a 32-carbon fatty acid = 206 ATP molecules

7 0
3 years ago
A 20.0 mL 0.100 M solution of lactic acid is titrated with 0.100 M NaOH.
yan [13]

Answer:

(a) See explanation below

(b) 0.002 mol

(c) (i) pH = 2.4

(ii) pH = 3.4

(iii) pH = 3.9

(iv) pH = 8.3

(v) pH = 12.0

Explanation:

(a) A buffer solution exits after addition of 5 mL of NaOH  since after reaction we will have  both the conjugate base lactate anion and unreacted weak  lactic acid present in solution.

Lets call lactic acid HA, and A⁻ the lactate conjugate base. The reaction is:

HA + NaOH ⇒ A⁻ + H₂O

Some unreacted HA will remain in solution, and since HA is a weak acid , we will have the followin equilibrium:

HA  + H₂O ⇆ H₃O⁺ + A⁻

Since we are going to have unreacted acid, and some conjugate base, the buffer has the capacity of maintaining the pH in a narrow range if we add acid or base within certain limits.

An added acid will be consumed by the conjugate base A⁻ , thus keeping the pH more or less equal:

A⁻ + H⁺ ⇄ HA

On the contrary, if we add extra base it will be consumed by the unreacted lactic acid, again maintaining the pH more or less constant.

H₃O⁺ + B ⇆ BH⁺

b) Again letting HA stand for lactic acid:

mol HA =  (20.0 mL x  1 L/1000 mL) x 0.100 mol/L = 0.002 mol

c)

i) After 0.00 mL of NaOH have been added

In this case we just have to determine the pH of a weak acid, and we know for a monopric acid:

pH = - log [H₃O⁺] where  [H₃O⁺] = √( Ka [HA])

Ka for lactic acid = 1.4 x 10⁻⁴  ( from reference tables)

[H₃O⁺] = √( Ka [HA]) = √(1.4 x 10⁻⁴ x 0.100) = 3.7 x 10⁻³

pH = - log(3.7 x 10⁻³) = 2.4

ii) After 5.00 mL of NaOH have been added ( 5x 10⁻³ L x 0.1 = 0.005 mol NaOH)

Now we have a buffer solution and must use the Henderson-Hasselbach equation.

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.0005                0

after rxn    0.002-0.0005                  0                  0.0005

                        0.0015

Using Henderson-Hasselbach equation :

pH = pKa + log [A⁻]/[HA]

pKa HA = -log (1.4 x 10⁻⁴) = 3.85

pH = 3.85 + log(0.0005/0.0015)

pH = 3.4

iii) After 10.0 mL of NaOH have been ( 0.010 L x 0.1 mol/L = 0.001 mol)

                             HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.001               0

after rxn        0.002-0.001                  0                  0.001

                        0.001

pH = 3.85 + log(0.001/0.001)  = 3.85

iv) After 20.0 mL of NaOH have been added ( 0.002 mol )

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.002                 0

after rxn                 0                         0                   0.002

We are at the neutralization point and  we do not have a buffer anymore, instead we just have  a weak base A⁻ to which we can determine its pOH as follows:

pOH = √Kb x [A⁻]

We need to determine the concentration of the weak base which is the mol per volume in liters.

At this stage of the titration we added 20 mL of lactic acid and 20 mL of NaOH, hence the volume of solution is 40 mL (0.04 L).

The molarity of A⁻ is then

[A⁻] = 0.002 mol / 0.04 L = 0.05 M

Kb is equal to

Ka x Kb = Kw ⇒ Kb = 10⁻¹⁴/ 1.4 x 10⁻⁴ = 7.1 x 10⁻¹¹

pOH is then:

[OH⁻] = √Kb x [A⁻]  = √( 7.1 x 10⁻¹¹ x 0.05) = 1.88 x 10⁻⁶

pOH = - log (  1.88 x 10⁻⁶ ) = 5.7

pH = 14 - pOH = 14 - 5.7 = 8.3

v) After 25.0 mL of NaOH have been added (

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn           0.002                  0.0025              0

after rxn                0                         0.0005              0.0005

Now here what we have is  the strong base sodium hydroxide and A⁻ but the strong base NaOH will predominate and drive the pH over the weak base A⁻.

So we treat this part as the determination of the pH of a strong base.

V= (20 mL + 25 mL) x 1 L /1000 mL = 0.045 L

[OH⁻] = 0.0005 mol / 0.045 L = 0.011 M

pOH = - log (0.011) = 2

pH = 14 - 1.95 = 12

7 0
3 years ago
Compare la configuración realizada con la que aparece en la tabla periódica y confirma si esta en
mart [117]

Answer:

?

Explanation:

4 0
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