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3 years ago

Cual es la finalidad de las citas en el texto argumentativo

Chemistry

1 answer:

Genrish500 [490]3 years ago

5 0

Answer:

Dar crédito a la persona u organización cuyas ideas está usando (y evitar acusaciones de plagio). Mostrarle a su audiencia que su argumento es bueno porque consultó a expertos y también pensó en el tema usted mismo.

Explanation:

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Which is a nonpolar molecule?

Colt1911 [192]

The answer is C- sulfur hexachlorine (SF6)

SF6 is the only molecule here that is non-polar. That's due to having the fluorine atoms arranged in a way that, in pairs, they lie opposite to each other. Also, these pairs are perpendicular to each other on three different axis.

7 0

3 years ago

Read 2 more answers

Ethanol (C2H5OH) melts a - 144 oC and boils at 78 °C. The enthalpy of fusion of ethanol is 5.02 kj/mol, and its enthalpy of vapo

hammer [34]

Answer:

For a: The total heat required is 36621.5 J

For b: The total heat required is 58944.5 J

Explanation:

For a:

To calculate the heat required at different temperature, we use the equation:

$q=mc\Delta T$ .........(1)

where,

q = heat absorbed

m = mass of substance

$c$ = specific heat capacity of substance

$\Delta T$ = change in temperature

To calculate the amount of heat required at same temperature, we use the equation:

$q=m\times \Delta H$ ........(2)

where,

q = heat absorbed

m = mass of substance

$\Delta H$ = enthalpy of the reaction

The processes involved in the given problem are:

$1.)C_2H_5OH(l)(35^oC)\rightarrow C_2H_5OH(l)(78^oC)\\2.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$

For process 1:

We are given:

Change in temperature remains the same.

$m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=35^oC\\\Delta T=[T_2-T_1]=[78-35]^oC=43^oC=43K$

Putting values in equation 1, we get:

$q_1=42.0g\times 2.3J/g.K\times 43K\\\\q_1=4153.8J$

For process 2:

We are given:

Conversion factor: 1 kJ = 1000 J

Molar mass of ethanol = 46 g/mol

$m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{35.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g$

Putting values in equation 2, we get:

$q_2=42.0g\times 773.04J/g\\\\q_2=32467.7J$

Total heat required = $[q_1+q_2]$

Total heat required = $[4153.8J+32467.7J]=36621.5J$

Hence, the total heat required is 36621.5 J

For b:

The processes involved in the given problem are:

$1.)C_2H_5OH(s)(-155^oC)\rightarrow C_2H_5OH(s)(-144^oC)\\2.)C_2H_5OH(s)(-144^oC)\rightarrow C_2H_5OH(l)(-144^oC)\\3.)C_2H_5OH(l)(-144^oC)\rightarrow C_2H_5OH(l)(78^oC)\\4.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$

For process 1:

We are given:

Change in temperature remains the same.

$m=42.0g\\c_s=0.97J/g.K\\T_2=-144^oC\\T_1=-155^oC\\\Delta T=[T_2-T_1]=[-144-(-155)]^oC=11^oC=11K$

Putting values in equation 1, we get:

$q_1=42.0g\times 0.97J/g.K\times 11K\\\\q_1=448.14J$

For process 2:

We are given:

$m=42.0g\\\Delta H_{fusion}=5.02kJ/mol=\frac{5.02kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=109.13J/g$

Putting values in equation 2, we get:

$q_2=42.0g\times 109.13J/g\\\\q_2=4583.5J$

For process 3:

We are given:

Change in temperature remains the same.

$m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=-144^oC\\\Delta T=[T_2-T_1]=[78-(-144)]^oC=222^oC=222K$

Putting values in equation 1, we get:

$q_3=42.0g\times 2.3J/g.K\times 222K\\\\q_3=21445.2J$

For process 4:

We are given:

$m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{38.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g$

Putting values in equation 2, we get:

$q_4=42.0g\times 773.04J/g\\\\q_4=32467.7J$

Total heat required = $[q_1+q_2+q_3+q_4]$

Total heat required = $[448.14+4583.5+21445.2+32467.7]J=58944.5J$

Hence, the total heat required is 58944.5 J

8 0

3 years ago

PLEASE PEOPLE I NEED HELP PLEASE IF YOU SEE THIS HELP ME ! :,( !!!!! PLEASEEEEEEEEE

Tems11 [23]

Answer:

year 1 is 5.5%

year 2 is 7.5%

year 3 is 10.2%

Explanation:

since,

length of the transect covered in seaweed / total lenth of transect x 100

then,

0.55 / 10.0 x 100 = 5.5

and

0.75 / 10.0 x 100 = 7.5

and

1.02 / 10.0 x 100 = 10.2

you could also just move the decimal to the right once

4 0

2 years ago

5. The reaction of magnesium oxide with hydrochloric acid carried out in a calorimeter caused the

saul85 [17]

Answer:

kek w

Explanation:

kek w

5 0

2 years ago

the gas left in an used aerosol can is at a pressure of 103 kPa at 25 degrees celsius if this can be thrown into fire what is th

Rainbow [258]

Hello!

The pressure of the gas when it's temperature reaches 928 °C is 3823,36 kPa

To solve that we need to apply Gay-Lussac's Law. It states that the pressure of a gas when the volume is left constant (like in the case of a sealed container like an aerosol can) is proportional to temperature. This is the relationship derived from this law that we use to solve this problem:

$P2= \frac{P1}{T1}*T2= \frac{103 kPa}{25}*928=3823,36 kPa$

Have a nice day!

5 0

3 years ago

Read 2 more answers

Cual es la finalidad de las citas en el texto argumentativo​

Cual es la finalidad de las citas en el texto argumentativo