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Olegator [25]
3 years ago
13

Identify the oxidizing and reducing agents in this reaction: C7H12(I) + 10O2(g) → 7CO2(g) + 6H2O(I)

Chemistry
2 answers:
kiruha [24]3 years ago
5 0
H has a charge of +1, and O has a charge of -2. In this case, C in C7H12 would have a charge of -12/7 to balance it out. Meanwhile, C in CO2 would have a charge of +4. Since the charge of C went from negative to positive, this implies that C lost electrons, which is indicative of oxidation (and thus it was the reducing agent). Therefore, C7H12 was the reducing agent, while O2 was the oxidizing agent.
polet [3.4K]3 years ago
4 0

Answer:

Oxidation agent:O

Reducing agent:C

Explanation:

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dmitriy555 [2]

Answer:

The flow rate of a tube is the volume of fluid flowing through the tube per unit time. The flowrate is proportional to the product of the velocity of the fluid through the tube, and the cross-sectional area of the tube.

That is

Q = AV

where

A is the area of the tube

V is the velocity of the tube

The cross-sectional area of the tube is proportional to the radius of the tube. From the above equation, we can deduce that if the velocity of the fluid flowing through the tube is held constant, the flowrate of the fluid through the tube will increase with an increase in the radius of the tube, and it will decrease with a decrease in the radius of the tube.

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Why does warm air raise and cold air sink
ki77a [65]
Warm is less heaver then cold air so warm air rise and cold air sinks
  
3 0
3 years ago
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Calculate the volume of dry co2 produced at body temperature (37 ∘c) and 0.970 atm when 25.5 g of glucose is consumed in this re
katrin2010 [14]

I believe the balanced chemical equation is:

C6H12O6 (aq) + 6O2(g) ------> 6CO2(g) + 6H2O(l) 

 

First calculate the moles of CO2 produced:

moles CO2 = 25.5 g C6H12O6 * (1 mol C6H12O6 / 180.15 g) * (6 mol CO2 / 1 mol C6H12O6)

moles CO2 = 0.8493 mol

 

Using PV = nRT from the ideal gas law:

<span>V = nRT  / P</span>

V = 0.8493 mol * 0.08205746 L atm / mol K * (37 + 273.15 K) / 0.970 atm

<span>V = 22.28 L</span>

6 0
3 years ago
The following reaction shows sodium carbonate reacting with calcium hydroxide.
velikii [3]

Ans: 15.1 grams

Given reaction:

Na2CO3 + Ca(OH)2 → 2NaOH + CaCO3

Mass of Na2CO3 = 20.0 g

Molar mass of Na2CO3 = 105.985 g/mol

# moles of Na2CO3 = 20/105.985 = 0.1887 moles

Based on the reaction stoichiometry: 1 mole of Na2CO3 produces 2 moles of NaOH

# moles of NaOH produced = 0.1887*2 = 0.3774 moles

Molar mass of NaOH = 22.989 + 15.999 + 1.008 = 39.996 g/mol

Mass of NaOH produced = 0.3774*39.996 = 15.09 grams


5 0
3 years ago
A solution contains 15.27 grams of NaCl in 0.670 kg water at 25 °C. What is the vapor pressure of the solution?
Anvisha [2.4K]

Answer:

The vapor pressure of the solution is 23.636 torr

Explanation:

P_{solution} = X_{solvent}*P_{solvent}

Where;

P_{solution is the vapor pressure of the solution

X_{solvent is the mole fraction of the solvent

P_{solvent is the vapor pressure of the pure solvent

Thus,

15.27 g of NaCl = [(15.27)/(58.5)]moles = 0.261 moles of NaCl

0.67 kg of water =  [(0.67*1000)/(18)]moles = 37.222 moles of H₂O

Mole fraction of solvent (water) = (number of moles of water)/(total number of moles present in solution)

Mole fraction of solvent (water) = (37.222)/(37.222+0.261)

Mole fraction of solvent (water) = 0.993

<u>Note:</u> the vapor pressure of water at 25°C is 0.0313 atm

Therefore, the vapor pressure of the solution = 0.993 * 0.0313 atm

the vapor pressure of the solution = 0.0311 atm = 23.636 torr

5 0
4 years ago
Read 2 more answers
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