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Alexxx [7]
3 years ago
7

PLEASE HELP HELP HELP

Chemistry
1 answer:
mihalych1998 [28]3 years ago
8 0

11.nicotinamid dinucleotid

12.nicotinamid dinucleotid phosphate

13.nicotinamid dinucleotid phosphate hydrogen

14.adenosine diphosphate

15.adenoside triphosphate

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Sulfuric acid is formed when sulfur dioxide reacts with oxygen and water. Write a balanced chemical equation for the reaction. (
Nesterboy [21]

Balanced chemical equation for the reaction is:

2SO_{2} (g) + O_{2} (g)+ 2H_{2}O (l) ⇒H_{2} S_{} O_{4}

Moles of H_{2} S_{} O_{4} formed is 5.75 moles.

Moles of oxygen used is 5.75 moles in the reaction.

Explanation:

Data given:

moles of SO_{2} = 11.5 moles

moles of H_{2} S_{} O_{4} = ?

Moles of O_{2} needed =?

balanced equation with states of matter =?

Balanced chemical reaction under STP condition is given as:

2SO_{2}(g) + O_{2} (g) + 2H_{2}O (l) ⇒H_{2} S_{} O_{4}

From the balanced reaction 2 moles of sulphur dioxide reacted to form 1 mole of sulphuric acid:

so, from 11.5 moles of SO_{2}, x moles of H_{2} S_{} O_{4} is formed

\frac{1}{2}  =\frac{x}{11.5}

2x = 11.5

x = 5.75 moles of sulphuric acid formed.

From the balanced reaction 1 mole of oxygen reacted to form 1  mole of sulphuric acid.

when 11.5 moles of Sulphur dioxide reacted then oxygen in the reaction is 5.75 moles.

7 0
3 years ago
Any one know the answer?
quester [9]
The last one, answer is D
8 0
4 years ago
Sodium-potassium pumps are examples of what type of cellular transport?
Dmitrij [34]

Answer:

Active transport

Explanation:

Sodium-potassium pumps are examples of Active type of cellular transport. Sodium potassium pump exchanges sodium ions from potassium ions through the plasma membrane of animal cells.

Whereas Active transport can be defined as movement of ions and molecules across a cell membrane to the region  of higher concentration with the help of enzymes and energy.

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4 years ago
What is the percent yield if 107.50 g NH3 reacts with excess O2 according to the
Maksim231197 [3]

Answer:

81.59%

Explanation:

  • 4NH₃ + 5O₂ → 4NO + 6H₂O

First we <u>convert 107.50 g of NH₃ into moles</u>, using its <em>molar mass</em>:

  • 107.50 g NH₃ ÷ 17 g/mol = 6.32 mol NH₃

Now we <u>calculate how many moles of NO would have been formed by the complete reaction of 6.32 moles of NH₃</u>:

  • 6.32 mol NH₃ * \frac{4molNO}{4molNH_3} = 6.32 mol NO

Then we <u>convert 6.32 moles of NO to grams</u>, using its <em>molar mass</em>:

  • 6.32 mol NO * 30 g/mol = 189.60 g NO

Finally we <u>calculate the percent yield</u>:

  • 154.70 g / 189.60 g * 100% = 81.59%

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A. 1.4 m<br> B.1.9m<br> C. 2.4m<br> D. 3.6m
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