How many moles are in 12.0 grams of O2

Answers

Answer 1
Answer:

Answer:

Moles = 0.375

Explanation:

Moles= m/M

= 12/32 = 0.375mol


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Identify the true statements about introns.a- they code for polypeptide proteinsb- they have a branch site located 20 to 50 nucleotides upstream of the 3' splice sitec- they end with the nucleotides AG at the 3' endd- they begin with the nucleotides GU at the 5' ende- they tend to be common in bacterial genes

Answers

Answer:

The answer is "Option b, c, and d".

Explanation:

In such a gene, Autosomes are also the sequence for code and transposable elements, not the series of encoding. Through the expression of genes, such fragments of its introns are split through protein complexes throughout the translation process. There has been no kenaf fiber in the genomes of prokaryotic cells.

For each set of dilutions (bleach and mouthwash), the first tube contained a 1:11 dilution (0.5 mL of agent was added to 5.0 mL of nutrient broth). What was the dilution of the MIC for the mouthwash and the bleach? Hint: You will have to calculate the dilution of each tube 2 – 4.

Answers

The dilution of each tube are as follows;

  • Tube 2; 8.26 × 10-³
  • Tube 3; 7.5 × 10-⁴
  • Tube 4; 6.83 × 10-⁵

For each time a dilution is further diluted;

The dilution ratio is; 1 : 11; In essence, 0.5 mL of agent was added to 5.0 mL of nutrient broth.

  • For the first tube; dilution factor is; 1 : 11 = 9.1× 10-²

  • For the second tube; 0.5mL of the first tube was added to 5.0mL of the nutrient broth; The dilution factor is then; (1/11) × (1/11) = 1/121 = 8.26 × 10-³.

  • For the third tube; 0.5mL of the second tube was added to 5.0mL of the nutrient broth; The dilution factor is then; (1/11) × (1/11) ×(1/11) = 1/121 = 7.5 × 10-⁴.

  • For the fourth tube; 0.5mL of the third tube was added to 5.0mL of the nutrient broth; The dilution factor is then; (1/11) × (1/11) × (1/11) × (1/11) = 1/14641 = 6.83 × 10-⁵.

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Answer:

Tube 2: 8.26 * 10^-3; Tube 4: 6.83 * 10^-5

Explanation:

In the serial dilutions for MIC test, the volume of nutrient broth in each tube should be equal: 5.0 mL. And the volume of agent in each dilution should also be similar: 0.5 mL.

The serial dilutions was as following:

  • Tube 1: 0.5/5.5
  • Tube 2: 0.5 mL of tube 1 was diluted with 5.0 mL broth. Then, the dilution of tube 2 is (1:11) * (1:11) = (0.5/5.5) * (0.5/5.5) = 1:121 = 8.26 * 10^-3
  • Tube 3: We perform the similar calculation. Thus, the result is 1:1331 = 7.51 * 10^-4
  • Tube 4: It is 1:14641 = 6.83 * 10^-5.

Copper and iron(III) nitrate
Does it have a reaction?

Answers

Yes it does have a reaction

A chemist titrates 160.0mL of a 0.3403M aniline C6H5NH2 solution with 0.0501M HNO3 solution at 25°C . Calculate the pH at equivalence. The pKb of aniline is 4.87 . Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HNO3 solution added.

Answers

Answer : The pH of the solution is, 5.24

Explanation :

First we have to calculate the volume of HNO_3

Formula used :

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the initial molarity and volume of C_6H_5NH_2.

M_2\text{ and }V_2 are the final molarity and volume of HNO_3.

We are given:

M_1=0.3403M\nV_1=160.0mL\nM_2=0.0501M\nV_2=?

Putting values in above equation, we get:

0.3403M* 160.0mL=0.0501M* V_2\n\nV_2=1086.79mL

Now we have to calculate the total volume of solution.

Total volume of solution = Volume of C_6H_5NH_2 +  Volume of HNO_3

Total volume of solution = 160.0 mL + 1086.79 mL

Total volume of solution = 1246.79 mL

Now we have to calculate the Concentration of salt.

\text{Concentration of salt}=(0.3403M)/(1246.79mL)* 160.0mL=0.0437M

Now we have to calculate the pH of the solution.

At equivalence point,

pOH=(1)/(2)[pK_w+pK_b+\log C]

pOH=(1)/(2)[14+4.87+\log (0.0437)]

pOH=8.76

pH+pOH=14\n\npH=14-pOH\n\npH=14-8.76\n\npH=5.24

Thus, the pH of the solution is, 5.24

Testbank, Question 098 In the reaction between an alkyne and Na metal in liquid ammonia, the role of Na is as a(n) ___________.a. catalyst
b. electrophile
c. Brønsted base
d. reducing agent
e. Bronsted acid

Answers

Answer:

d. reducing agent

Explanation:

Na acts as a  reducing agent. A reducing agent is a substance whose function is to reduce or donate electrons to another, and by doing so it becomes oxidized.  We can see the mechanism of the reaction in the image attached below.

Final answer:

In the chemical reaction between an alkyne and Na in liquid ammonia, Na acts as a reducing agent, donating electrons and facilitating the reduction of the alkyne to a trans-alkene.

Explanation:

In the reaction between an alkyne and Na metal in liquid ammonia, the role of Na is as a reducing agent. A reducing agent is a substance that donates electrons in a chemical reaction, facilitating the process of reduction. In this specific reaction, Na donates its outer shell electron to the alkyne, facilitating its reduction to a trans-alkene. Therefore, answer d. reducing agent is the correct choice among the given options.

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If you put a strong base in water what will it produce ?

Answers

Conversely, strong bases react completely with water to produce the hydroxide ion, whereas weak bases react only partially with water to form hydroxide ions. The reaction of a strong acid with a strong base is a neutralization reaction, which produces water plus a salt.