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guapka [62]
3 years ago
15

If 4.0 mol of NO and 4.0 mol of O2 are combined, how many moles

Chemistry
1 answer:
Masja [62]3 years ago
4 0
4.0


i think it has something to do with molar ratios and finding the limiting reactant

4.0 mol NO * 2 mol NO2/2 mol NO = 4.0 moles of NO2

4.0 mol O2 * 2 mol NO2/1 mol O2 = 8.0 moles of NO2

so the limiting reactant (the reactant that runs out the quickest leaving an excess) is NO

once the limiting reactant is found, we can use that data for that substance to calculate the amount of product

4.0 mol NO * 2 mol NO2/2 mole NO = 4.0 moles of NO2

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Answer:

43.868 J

Explanation:

Kinetic energy of a body is the amount of energy possessed by a moving body. The SI unit of kinetic energy is the joule (kg⋅m²⋅s⁻²).

According to classical mechanics, kinetic energy = 1/2 m·v²

Where, m= mass in kg and v= velocity in m/s

Given: m = 19.2 lb and v = 7.10 miles/h

Since, 1 lb= 0.453592 kg

∴ m = 19.2 lb = 19.2 × 0.453592 kg = 8.709 kg

Also, 1 mi = 1609.34 m and 1 h = 3600 sec

∴ v = 7.10 mi/h = 7.10 × 1609.34 m ÷ 3600 sec = 3.174 m/sec

Therefore, <u>kinetic energy of the goose</u> = 1/2 m·v² = 1/2 × (8.709 kg)× (3.174 m/sec)² = 43.868 J

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3 years ago
A nucleotide consists of a phosphate group, a pentose sugar, and a __________________, all linked together by covalent bonds. po
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1. The reaction below describes cellular respiration. Which compounds are the reactants and which are the
swat32

Answer:

C₆H₁₂O₆ and O₂ are reactant.

CO₂ and H₂O are products.

C₆H₁₂O₆ + 6O₂  →  6CO₂ + 6H₂O + ATP

Explanation:

There are two types of respiration:

1. Aerobic respiration  

2. Anaerobic respiration

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Glucose + oxygen → carbon dioxide + water + 38ATP

Anaerobic Respiration

It is the breakdown of glucose molecule in the absence of oxygen and produce small amount of energy. Alcohol or lactic acid and carbon dioxide are also produced as byproducts.  

Glucose→ lactic acid/alcohol + 2ATP + carbon dioxide

This process use respiratory electron transport chain as electron acceptor instead of oxygen. It is mostly occur in prokaryotes. Its main advantage is that it produce energy (ATP) very quickly as compared to aerobic respiration.

Steps involve in anaerobic respiration are:

Glycolysis

Glycolysis is the first step of both aerobic and anaerobic respiration. It involve the breakdown of one glucose molecule into pyruvate and 2ATP.

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Read 2 more answers
Determine the pHpH of an HFHF solution of each of the following concentrations. In which cases can you not make the simplifying
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The question is incomplete, complete question is :

Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? (K_a for HF is 6.8\times 10^{-4}.)

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Express your answer to two decimal places.

Answer:

The pH of an 0.280 M HF solution is 1.87.

Explanation:3

Initial concentration if HF = c = 0.280 M

Dissociation constant of the HF = K_a=6.8\times 10^{-4}

HF\rightleftharpoons H^++F^-

Initially

c          0            0

At equilibrium :

(c-x)      x             x

The expression of disassociation constant is given as:

K_a=\frac{[H^+][F^-]}{[HF]}

K_a=\frac{x\times x}{(c-x)}

6.8\times 10^{-4}=\frac{x^2}{(0.280 M-x)}

Solving for x, we get:

x = 0.01346 M

So, the concentration of hydrogen ion at equilibrium is :

[H^+]=x=0.01346 M

The pH of the solution is ;

pH=-\log[H^+]=-\log[0.01346 M]=1.87

The pH of an 0.280 M HF solution is 1.87.

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