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2 years ago

How many grams of steam at 100 °C would be required to raise the temperature

Chemistry

1 answer:

Alex777 [14]2 years ago

3 0

The mass of steam required to raise the temperature of water is 3.5 g.

The given parameters;

mass of the benzene, = 47.6
initial temperature of the benzene, = 5.5 ⁰C
final temperature of the benzene = 30 ⁰C

The molar mass of Benzene = 78.11 g/mol

The molar mass of water = 18 g/mol

The number of moles of the Benzene is calculated as follows;

$n = \frac{47.6}{78.11} = 0.61 \ mole$

The mass of steam required is calculated as follows;

heat lost by steam = heat absorbed by benzene

$\frac{m}{18} \times 40.7 \times 10^3 = 47.6(1.63)(30-5.5) \ + \ 0.61 \times 9.87 \times 10^3\\\\2261.11 m = 7921.61\\\\m = \frac{7921.61}{2261.11} \\\\m = 3.5 \ g$

Thus, the mass of steam required to raise the temperature of water is 3.5 g.

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a steel cylinder is left out in the sun the pressure at 2:00 pm was 1014 kpa at 33.0 c at 7:00 am at a temperature of 21.4c what

e-lub [12.9K]

The easiest way is to use the Law of Gay-Lussac. This law states that there is a direct relation between the temperature in Kelvin of a gas and the pressure.

Then, namig p the pressure and T the temperature in Kelvin and using subscripts for every state:

p/T is constant ==> p_1 / T_1 = p_2/T_2

From which you obtain:

p_2 = [p_1 / T_1] * T_2

T_1 = 33.0 + 273.15 = 306.15 K
T _2 = 21.4 + 273.15 = 294.55 K

p_1 = 1014 kPa

p_2 = 1014 kPa * 294.55 K / 306.15 K = 975.6 kPa

5 0

3 years ago

How many grams of aluminum chloride are needed to react completely with 1.084g lithium sulfide?

Oksi-84 [34.3K]

Molar mass :

Li₂S = 45.947 g/mol

AlCl₃ = 133.34 g/mol

3 Li₂S + 2 AlCl₃ = 6 LiCl + Al₂S₃

3 * 45.947 g Li₂S ----------> 2 * 133.34 g AlCl₃
1.084 g Li₂S ----------------> ?

Mass Li₂S = 1.084 * 2 * 133.34 / 3 * 45.947

Mass Li₂S = 289.08112 / 137.841

Mass Li₂S = 2.0972 g

hope this helps!

5 0

2 years ago

A 44.0 g sample of an unknown metal at 99.0 oC was placed in a constant-pressure calorimeter of negligible heat capacity contain

tatiyna

Answer:

$C_m=0.474\frac{J}{g\°C}$

Explanation:

Hello.

In this case, since this is a system in which the water is heated up and the metal is cooled down in a calorimeter which is not affected by the heat lose-gain process, we can infer that the heat lost by the metal is gained be water, it means that we can write:

$Q_m=-Q_w$

Thus, in terms of masses, specific heats and temperatures we can write:

$m_mC_m(T_{eq}-T_m)=-m_wC_w(T_{eq}-T_w)$

Whereas the equilibrium temperature is the given final temperature of 28.4 °C and we can compute the specific heat of the metal as shown below:

$C_m=\frac{-m_wC_w(T_{eq}-T_w)}{m_m(T_{eq}-T_m)}$

Plugging the values in and since the density of water is 1.00 g/mL so the mass is 80.0g, we obtain:

$C_m=\frac{-80.0g*4.184\frac{J}{g\°C} (28.4\°C-24.0\°C)}{44.0g(28.4\°C-99.0\°C)}\\\\C_m=0.474\frac{J}{g\°C}$

Best regards!

6 0

2 years ago

True or False: Electrons in each lower level are filled before electrons fill

ale4655 [162]

the answer is 1....true

5 0

3 years ago

Helpppp pleaseee ill give brainliest

Studentka2010 [4]

Answer:

The answers are in the explanation.

Explanation:

The energy required to convert 10g of ice at -10°C to water vapor at 120°C is obtained per stages as follows:

Increasing temperature of ice from -10°C - 0°C:

Q = S*ΔT*m

Q is energy, S specific heat of ice = 2.06J/g°C, ΔT is change in temperature = 0°C - -10°C = 10°C and m is mass of ice = 10g

Q = 2.06J/g°C*10°C*10g

Q = 206J

Change from solid to liquid:

The heat of fusion of water is 333.55J/g. That means 1g of ice requires 333.55J to be converted in liquid. 10g requires:

Q = 333.55J/g*10g

Q = 3335.5J

Increasing temperature of liquid water from 0°C - 100°C:

Q = S*ΔT*m

Q is energy, S specific heat of ice = 4.18J/g°C, ΔT is change in temperature = 100°C - 0°C = 100°C and m is mass of water = 10g

Q = 4.18J/g°C*100°C*10g

Q = 4180J

Change from liquid to gas:

The heat of vaporization of water is 2260J/g. That means 1g of liquid water requires 2260J to be converted in gas. 10g requires:

Q = 2260J/g*10g

Q = 22600J

Increasing temperature of gas water from 100°C - 120°C:

Q = S*ΔT*m

Q is energy, S specific heat of gaseous water = 1.87J/g°C, ΔT is change in temperature = 20°C and m is mass of water = 10g

Q = 1.87J/g°C*20°C*10g

Q = 374J

Total Energy:

206J + 3335.5 J + 4180J + 22600J + 374J =

30695.5J =

30.7kJ

5 0

2 years ago