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2 years ago

1.- Ilumina de amarillo los verbos en pretérito y de verde los verbos en copretérito.

Chemistry

1 answer:

ivanzaharov [21]2 years ago

3 0

Answer:

Pretérito (Pretérito perfecto):

<em>estiró, sudó, peleó, compartí</em>

Copretérito (Pretérito imperfecto):

<em>saludaba, pedias, leia, compraba, barrias, cargaba, estrenabas, escribian</em>

Explanation:

Los verbos en pretérito, también llamado pretérito perfecto, son los siguientes:

<em>estiró, sudó, peleó, compartí</em>

Como se aprecia, los verbos conjugados en pretérito perfecto terminan la última sílaba en vocal con acento, es decir, <em>-á</em>, <em>-é</em>, <em>-í</em> u <em>-ó</em>.

Los verbos en copretérito, también llamado pretérito imperfecto, son los siguientes:

<em>saludaba, pedias, leia, compraba, barrias, cargaba, estrenabas, escribian</em>

Como se aprecia, los verbos conjugados en pretérito imperfecto terminan la última sílaba en <em>-ía</em>, <em>ías</em>, <em>-ían</em>, <em>-aba</em>, <em>-abas</em> o <em>-aban</em>.

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None of these are correct, because there is no way to balance this equation, but I hope these steps help you figure out your answer.

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Because they have similar attractive forces in their molecules

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There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In

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Answer:

The net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide is 523.2 kJ.

Explanation:

Step 1:

$CaC_2(s) + 2H_2O(g)\rightarrow C_2H_2(g) + Ca(OH)_2(s),\Delta H_1=414.0 kJ$ ...[1]

Step 2 :

$6C_2H_2(g) + 3CO_2(g) + 4H_2O(g)\rightarrow 5CH_2CHCO_2H(g) \Delta H_2=132.0kJ$ ..[2]

Adding 6 × [1] and [2]:

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$6C_2H_2(g)+3CO_2(g)+16H_2O(g)\rightarrow 5CH_2CHCO_2H(g)$

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$6CaC_2(s) + 8H_2O(g)+3CO_2(g)\rightarrow 5CH_2CHCO_2H(g)+ 6Ca(OH)_2(s),\Delta H'=?$

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san4es73 [151]

The number of moles of the magnesium (mg) is 0.00067 mol.

The number of moles of hydrogen gas is 0.0008 mol.

The volume of 1 more hydrogen gas (mL) at STP is 22.4 L.

<h3>Number of moles of the magnesium (mg)</h3>

The number of moles of the magnesium (mg) is calculated as follows;

number of moles = reacting mass / molar mass

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<h3>Number of moles of hydrogen gas</h3>

PV = nRT

n = PV/RT

Apply Boyle's law to determine the change in volume.

P1V1 = P2V2

V2 = (P1V1)/P2

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Now determine the number of moles using the following value of ideal constant.

R = 8.314 LkPa/mol.K

n = (15.15 kPa x 0.127 L)/(8.314 x 290.95)

n = 0.0008

<h3>Volume of 1 mole of hydrogen gas at STP</h3>

V = nRT/P

V = (1 x 8.314 x 273) / (101.325)

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