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lina2011 [118]
3 years ago
12

16. What are the reactants in the equation?

Chemistry
1 answer:
Dimas [21]3 years ago
4 0
5. Iron and chlorine
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What determines how large each population can grow
lana66690 [7]

Answer:

laws

Explanation:

in some areas and countries there is a limit on how many children someone can have, this would clearly regulate population growth.

hope I helped ;)

8 0
3 years ago
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Para el elemento de Z-56 b, que formula minima tiene el compuesto que forma con 17-cl¿ c, que tipo de union hay en dicho compues
inessss [21]

Answer:

ver explicacion

Explanation:

El elemento que tiene el número atómico 56 en la tabla periódica del bario. El bario es un elemento del grupo dos.

Se combina con el cloro para formar cloruro de bario. que tiene la fórmula BaCl2 de acuerdo con la valencia de ambos elementos.

El cloruro de bario es un compuesto iónico.

5 0
3 years ago
Cars run on gasoline, where octane (C8H18) is the principle component. This combustion reaction is responsible for generating en
Bezzdna [24]

Answer:

  • 10.19 g CO₂
  • 4.69 g H₂O

Explanation:

The combustion reaction of Octane is:

  • C₈H₁₈ → 8CO₂ + 9H₂O

To calculate the mass of CO₂ and H₂O produced, we need to know the mass of octane combusted.

We calculate the mass of Octane from the given volume and density, using the following <em>conversion factors</em>:

  • 1 gallon = 3.785 L
  • 1 L = 1000 mL

Now we<u> convert 1.24 gallons to mL</u>:

  • 1.24 gallon * \frac{3.785L}{1gallon} *\frac{1000mL}{1L} = 4693.4 mL

We <u>calculate the mass of Octane</u>:

  • 4693.4 mL * 0.703 g/mL = 3.30 g Octane

Now we use the <em>stoichiometric ratios</em> and <em>molecular weights</em> to <u>calculate the mass of CO₂ and H₂O</u>:

  • CO₂ ⇒ 3.30 g Octane ÷ 114g/mol * \frac{8molCO_{2}}{1molOctane} * 44 g/mol =  10.19 g CO₂
  • H₂O ⇒ 3.30 g Octane ÷ 114g/mol * \frac{9molH_{2}O}{1molOctane} * 18 g/mol = 4.69 g H₂O

7 0
4 years ago
The combustion of 0.25 mol of an unknown organic compound results in the release of 320 kJ of energy. Which of the compounds in
vichka [17]

Answer: C. ethanol

The enthalpy of combustion is the amount of heat produced when one mole of ethanol undergoes complete combustion at 25 ° C and 1 atmosphere pressure, yielding products also at 25 ° C and 1 atm.

<u>The enthalpy of combustion of the unknown compound is</u>

ΔH = - 320 kJ / 0.25 mol = - 1280 kJ / mol

<u>To choose a probable compound according to this combustion enthalpy, we must evaluate the deviation in relation to the values reported in the literature for the three probable compounds</u> (methane, ethylene and ethanol). The deviation (e%) will be calculated according to the following equation,

e% =  ( | ΔHx - ΔH | / ΔHx ) x 100%

where ΔHx is the enthalpy of combustion of the probable compound.

The following table shows the combustion enthalpies of the probable compounds and their deviation in relation to the enthalpy of ΔH = - 1280 kJ / mol

Compound Enthalpy of combustion (kJ/mol)   Deviation

Methane                        - 890.7                                 43.8%

Ehylene                         -1411.2                                   9.3%

Ethanol                        -1368.6                                    6.5%

According to the previous table, we can say that the most probable compound is ethanol, since it has the smallest deviation in relation to the experimental enthalpy value of combustion.

5 0
4 years ago
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Isoflavones, quercetin and anthocyanins are all what type of phytochemicals
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Isoflavones, quercetin, and anthocyanin are all Flavonoids.
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