Composición centecimal de H2SO4

When 100 mL of 0.200 M NaCl(aq) and 100 mL of 0.200 M AgNO3(aq), both at 21.9 °C, are mixed in a coffee cup calorimeter, the tem

masya89 [10]

Answer:

There is 1.3 kJ heat produced(released)

Explanation:

Step 1: Data given

Volume of a 0.200 M Nacl solution = 100 mL = 0.1 L

Volume of a 0.200 M AgNO3 solution = 100 mL = 0.1 L

Initial temperature = 21.9 °C

Final temperature = 23.5 °C

Solid AgCl will be formed

Step 2: The balanced equation:

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

AgCl(s) + NaNO3(aq) → Na+(aq) + NO3-(aq) + AgCl(s)

Step 3: Define the formula

Pressure is constant. → the heat evolved from the reaction is equivalent to the enthalpy of reaction.

Q=m*c*ΔT

⇒ Q = the heat transfer (in joule)

⇒ m =the mass (in grams)

⇒ c= the heat capacity (J/g°C)

⇒ ΔT = Change in temperature = T2- T1

Step 4: Calculate heat

Let's vonsider the density the same as the density of water (1g/mL)

Mass = volume * density

Mass = 200 mL * 1g/mL

Mass = 200 grams

Q= m*c*ΔT

⇒ m = 200 grams

⇒ c = the heat capacity (let's consider the heat capacity of water) = 4.184 J/g°C

⇒ ΔT = 23.5 -21.9 = 1.6°C

Q = 200 * 4.184 * 1.6 = 1338 .9 J = 1.3 kJ

There is 1.3 kJ heat produced(released)

Therefore, we assumed no heat is absorbed by the calorimeter, no heat is exchanged between the calorimeter and its surroundings, and the specific heat and mass of the solution are the same as those for water (1g/mL and 4.184 J/g°C)

7 0

3 years ago

A disk of radius 2.0 cm has a surface charge density of 6.3 μC/m2 on its upper face. What is the magnitude of the electric field

maksim [4K]

Answer:

the electric field at Z = 12 cm is E = 9.68 × 10³ N/C = 9.68 kN/C

Explanation:

Given: radius of disk, R = 2.0 cm = 2 × 10⁻² cm, surface charge density,σ = 6.3 μC/m² = 6.3 × 10⁻⁶ C/m², distance on central axis, z = 12 cm = 12 × 10⁻² cm.

The electric field, E at a point on the central axis of a charged disk is given by E = σ/ε₀( $1 - \frac{z}{\sqrt{z^{2} + R^{2} } }$ )

Substituting the values into the equation, it becomes

E = σ/ε₀( $1 - \frac{z}{\sqrt{z^{2} + R^{2} } }$ ) = 6.3 × 10⁻⁶/8.854 × 10⁻¹²( $1 - \frac{0.12}{\sqrt{0.12^{2} + 0.02^{2} } }$ ) = 7.12 × 10⁵( $1 - \frac{0.12}{0.1216}$ ) = 7.12 × 10⁵(1 - 0.9864) = 7.12 × 10⁵ × 0.0136 = 0.0968 × 10⁵ = 9.68 × 10³ N/C = 9.68 kN/C

Therefore, the electric field at Z = 12 cm is E = 9.68 × 10³ N/C = 9.68 kN/C

7 0

3 years ago

Consider the following intermediate chemical equations.

QveST [7]

Answer: 250 kJ

Explanation: According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to Hess’s law, the chemical equation can be treated as algebraic expressions and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

$P_4(s)+6Cl_2\rightarrow 4PCl_3$ $\Delta H_1=-2439kJ$ (1)

$4PCl_5(g)\rightarrow P_4(s)+10Cl_2(g)$ $\Delta H_2=3438kJ$ (2)

Net chemical equation:

$PCl_5(g)\rightarrow PCl_3(g)+Cl_2(g)$ $\Delta H=?$ (3)

Adding 1 and 2 we get,

$4PCl_5(g)\rightarrow 4PCl_3(g)+4Cl_2$ $\Delta H_3=\Delta H_1+\Delta H_2=-2439+3438=1000kJ$ (4)

Now dividing equation (4) by 4, we get

$PCl_5(g)\rightarrow PCl_3(g)+Cl_2$

$\Delta H=\frac{\Delta H_3}{4}=\frac{1000kJ}{4}=250kJ$ (4)

8 0

3 years ago

Which subatomic particle gives off visible light when it drops back down to a lower energy state?

icang [17]

Answer:

Option C = electron

Explanation:

Electrons are responsible for the production of colored light.

Electron:

The electron is subatomic particle that revolve around outside the nucleus and has negligible mass. It has a negative charge.

Symbol= e-

Mass= 9.10938356×10⁻³¹ Kg

It was discovered by j. j. Thomson in 1897 during the study of cathode ray properties.

How electrons produce the colored light:

Excitation:

When the energy is provided to the atom the electrons by absorbing the energy jump to the higher energy levels. This process is called excitation. The amount of energy absorbed by the electron is exactly equal to the energy difference of orbits.

De-excitation:

When the excited electron fall back to the lower energy levels the energy is released in the form of radiations. this energy is exactly equal to the energy difference between the orbits. The characteristics bright colors are due to the these emitted radiations. These emitted radiations can be seen if they are fall in the visible region of spectrum.

Other process may involve,

Fluorescence:

In fluorescence the energy is absorbed by the electron having shorter wavelength and high energy usually of U.V region. The process of absorbing the light occur in a very short period of time i.e. 10 ∧-15 sec. During the fluorescence the spin of electron not changed.

The electron is then de-excited by emitting the light in visible and IR region. This process of de-excitation occur in a time period of 10∧-9 sec.

Phosphorescence:

In phosphorescence the electron also goes to the excitation to the higher level by absorbing the U.V radiations. In case of Phosphorescence the transition back to the lower energy level occur very slowly and the spin pf electron also change.

5 0

3 years ago

Determine the mass of sodium carbonate required to produce 23.4g of sodium chloride when it reacts with excess hydrochloric acid

Leni [432]

The balanced chemical reaction for the described reaction above is,
Na2CO3 + 2HCl ---> 2NaCl + H2CO3
From the reaction, 1 mole of Na2CO3 is needed to produce 2 moles of NaCl. In terms of mass, 106 grams of Na2CO3 are needed to produce 116.9 grams of NaCl. From this,
(23.4 g NaCl) x (106 g Na2CO3 / 116.9 NaCl = 21.22 g Na2CO3
Thus, approximately 21.22 g Na2CO3 is needed for the desired reaction.

8 0

3 years ago