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Anna [14]
1 year ago
6

Adding 1.56 g of K2SO4 to 6.00 mL of water at 16.2ºC causes the temperature of the solution to drop by 7.70ºC.

Chemistry
1 answer:
DanielleElmas [232]1 year ago
7 0

Answer:

You need to add 0.243g of NaOH to raise the temperature back to 16.2°C

Explanation:

Using the equation:

Q = C*m*ΔT

<em>Where Q is heat</em>

<em>C is specific heat</em>

<em>m is mass</em>

<em>and ΔT is change in temperature</em>

<em />

We can find the heat required to increase the temperature of the solution back to 16.2°C:

Assuming specific heat of the solution of water + K2SO4 = Specific heat of water:

C = 4.184J/g°C

m = 1.56g + 6.00g = 7.56g

ΔT = 16.2°C - 7.70°C = 8.50°C

Q = 4.184J/g°C * 7.56g * 8.50°C

Q = 268.86J = 0.269kJ of heat are required

As this heat is obtained from the dissolution of NaOH:

0.269kJ * (1mol NaOH / 44.3kJ) = 0.00607 moles of NaOH are required

In grams -Molar mass NaOH: 40g/mol-:

0.00607 moles NaOH * (40g / mol) =

<h3>You need to add 0.243g of NaOH to raise the temperature back to 16.2°C</h3>

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Steam initially at 0.3 MPa, 2500 C is cooled at constant volume. (a) At what temperature will steam become saturated vapour? [12
Scilla [17]

Answer:

a. 123.9°C

b.

c.

Explanation:

Hello, I'm attaching a picture with the numerical development of this exercise.

a. Since the steam is overheated vapour, the specific volume is gotten from the corresponding table. Then, as it became a saturated vapour, we look for the interval in which the same volume of state 1 is, then we interpolate and get the temperature.

b. Now, at 80°C, since it is about a rigid tank (constant volume for every thermodynamic process), the specific volume of the mixture is 0.79645 m^3/kg as well, so the specific volume for the liquid and the vapour are taken into account to get the quality of 0.234.

c. Now,since this is an isocoric process, the heat transfer per kg of steam is computed as the difference in the internal energy, considering the initial condition (showed in a. part) and the final one computed here.

** The thermodynamic data were obtained from Cengel's thermodynamics book 7th edition.

Best regards.

7 0
2 years ago
The Lewis structure for CO2 has a central The Lewis structure for C O 2 has a central blank atom attached to blank atoms.
kirza4 [7]

Answer:

See the explanation

Explanation:

1) The Lewis structure for  CO_2 has a central Carbon<em> </em>atom attached to Oxygen atoms.

In the CO_2  we will have a structure:  O=C=O the <u>central atom</u> "carbon" we will have <u>2 sigma bonds and 2 pi bonds</u>, therefore, we have an <u>Sp hybridization</u>. For O we have <u>1 pi and 1 sigma bond</u>, therefore, we have an <u>Sp2 hybridization</u>.

2) These atoms are held together by <u>double bonds.</u>

<u></u>

Again in the structure of CO_2: O=C=O we only have double bonds.

3. Carbon dioxide has a Carbon dioxide has a <u>Linear</u> electron geometry.

Due to the double bonds we have to have a linear structure because in this geometry the atoms will be further apart from each other.

4. The carbon atom is <u>Sp</u> hybridized.

We will have for carbon 2 pi bonds, so we will have an <u>Sp</u> hybridization.

5. Carbon dioxide has two Carbon dioxide has two C(p) - O(p) π bonds and two C(sp) - O(Sp2) σ bonds.

(See figures)

Figure 1: Carbon hybridization

Figure 2: Oxygen hybridization

6 0
2 years ago
If 15.6 g of hydrate are heated and only 11.7 g of anhydrous salt remain, calculate the % of water lost.
Elodia [21]

Answer:

Percent loss of water = 25%

Explanation:

Given data:

Mass of hydrated salt = 15.6 g

Mass of anhydrous salt = 11.7 g

Percentage of water lost = ?

Solution:

First of all we will calculate the mass of water in hydrated salt.

Mass of water =  Mass of hydrated salt - Mass of anhydrous salt

Mass of water = 15.6 g - 11.7 g

Mass of water = 3.9 g

Now we will calculate the percentage.

Percent loss of water = mass of water / total mass × 100

Percent loss of water = 3.9 g/ 15.6 g × 100

Percent loss of water = 25%

8 0
2 years ago
En una determinación cuantitativa se utilizan 17.1 mL de Na2S2O3 0.1N para que reaccione todo el yodo que se encuentra en una mu
lozanna [386]

Answer:

La cantidad de yodo en la muestra es 0.217 g

Explanation:

Los parámetros dados son;

Normalidad de la solución de Na₂S₂O₃ = 0.1 N

Volumen de la solución de Na₂S₂O₃ = 17.1 mL

Masa de muestra = 0.376 g

La ecuación de reacción química se da de la siguiente manera;

I₂ + 2Na₂S₂O₃ → 2 · NaI + Na₂S₄O₆

Por lo tanto, el número de moles de sodio por 1 mol de Na₂S₂O₃ en la reacción = 1 mol

Por lo tanto, la normalidad por mol = 1 M × 1 átomo de Na = 1 N

Por lo tanto, 0.1 N = 0.1 M

El número de moles de Na₂S₂O₃ en 17,1 ml de solución 0,1 M de Na₂S₂O₃ se da de la siguiente manera;

Número de moles de Na₂S₂O₃ = 17.1 / 1000 × 0.1 = 0.00171 moles

Lo que da;

Un mol de yodo, I₂, reacciona con dos moles de Na₂S₂O₃

Por lo tanto;

0,000855 moles de yodo, I₂, reaccionan con 0,00171 moles de Na₂S₂O₃

La masa molar de yodo = 253.8089 g / mol

La masa de yodo en la muestra = 253.8089 × 0.000855 = 0.217 g.

5 0
2 years ago
If 29.4 mL of ethanol is dissolved in water to make 359 mL of solution, what is the concentration expressed in volume/volume % o
Sveta_85 [38]

Answer: 8.2\%

Explanation:- Volume percentage is the ratio of volume of solute to the volume of solution defined in terms of percentage.

{\text {volume percentage}}=\frac{\text {volume of solute}}{\text {volume of solution}}\times 100\%

Given: volume of solute = 29.4 ml

Volume of solution= 359 ml

{\text {volume percentage}=\frac{29.4}{359}\times 100\%=8.2\%


6 0
2 years ago
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