How many grams of h3po3 would be produced from the complete reaction of 93.2 g p2o3

A box has a volume of 45m3 and is filled with air held at 25∘C and 3.65atm. What will be the pressure (in atmospheres) if the sa

Marina CMI [18]

Answer:

Given:

Initial pressure: $3.65\; \rm atm$ .
Volume was reduced from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$ .
Temperature was raised from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$ .

New pressure: approximately $3.4\times 10\; \rm atm$ ( $34\; \rm atm$ .) (Assuming that the gas is an ideal gas.)

Explanation:

Both the volume and the temperature of this gas has changed. Consider the two changes in two separate steps:

Reduce the volume of the gas from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$ . Calculate the new pressure, $P_1$ .
Raise the temperature of the gas from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$ . Calculate the final pressure, $P_2$ .

By Boyle's Law, the pressure of an ideal gas is inversely proportional to the volume of this gas (assuming constant temperature and that no gas particles escaped or was added.)

For this gas, $V_0 = 45\; \rm m^{3}$ while $V_1 = 5.0\; \rm m^{3}$ .

Let $P_0$ denote the pressure of this gas before the volume change ( $P_0 = 3.65\; \rm atm$ .) Let $P_1$ denote the pressure of this gas after the volume change (but before changing the temperature.) Apply Boyle's Law to find the ratio between $P_1\!$ and $P_0\!$ :

$\displaystyle \frac{P_1}{P_0} = \frac{V_0}{V_1} = \frac{45\; \rm m^{3}}{5.0\; \rm m^{3}} = 9.0$ .

In other words, because the final volume is $(1/9)$ of the initial volume, the final pressure is $9$ times the initial pressure. Therefore:

$\displaystyle P_1 = 9.0\times P_0 = 32.85\; \rm atm$ .

On the other hand, by Amonton's Law, the pressure of an ideal gas is directly proportional to the temperature (in degrees Kelvins) of this gas (assuming constant volume and that no gas particle escaped or was added.)

Convert the unit of the temperature of this gas to degrees Kelvins:

$T_1 = (25 + 273.15)\; \rm K = 298.15\; \rm K$ .

$T_2 = (35 + 273.15)\; \rm K = 308.15\; \rm K$ .

Let $P_1$ denote the pressure of this gas before this temperature change ( $P_1 = 32.85\; \rm atm$ .) Let $P_2$ denote the pressure of this gas after the temperature change. The volume of this gas is kept constant at $V_2 = V_1 = 5.0\; \rm m^{3}$ .

Apply Amonton's Law to find the ratio between $P_2$ and $P_1$ :

$\displaystyle \frac{P_2}{P_1} = \frac{T_2}{T_1} = \frac{308.16\; \rm K}{298.15\; \rm K}$ .

Calculate $P_2$ , the final pressure of this gas:

$\begin{aligned} P_2 &= \frac{308.15\; \rm K}{298.15\; \rm K} \times P_1 \\ &= \frac{308.15\; \rm K}{298.15\; \rm K} \times 32.85\; \rm atm \approx 3.4 \times 10\; \rm atm\end{aligned}$ .

In other words, the pressure of this gas after the volume and the temperature changes would be approximately $3.4\times 10\; \rm atm$ .

8 0

3 years ago

Aqueous potassium phosphate was mixed with aqueous magnesium chloride, and a crystallized magnesium phosphate product was formed

scoray [572]

Answer: The other product formed is potassium chloride.

Explanation:

Precipitation reaction is defined as the chemical reaction in which an insoluble salt is formed when two solutions are mixed containing soluble substances. The insoluble salt settles down at the bottom of the reaction mixture.

The chemical equation for the reaction of potassium phosphate and magnesium chloride follows:

$2K_3PO_4(aq.)+3MgCl_2(aq.)\rightarrow Mg_3(PO_4)_2(s)+6KCl(aq.)$

By Stoichiometry of the reaction:

2 moles of aqueous solution of potassium phosphate reacts with 3 moles of aqueous solution of magnesium chloride to produce 1 mole of solid magnesium phosphate and 6 moles of aqueous solution of potassium chloride.

Hence, the other product formed is potassium chloride.

4 0

4 years ago

2 reasons for chemical reactivity of nitrogen

erma4kov [3.2K]

Answer:

Due to presence of a triple bond between the two N−atoms, the bond dissociation enthalpy (941.4 kJ mol

−1

) is very high. Hence, N

2

is the least reactive.

6 0

3 years ago

An atom has 25 protons, 30 neutrons, and 25 electrons. What is the charge of the nucleus?

DochEvi [55]

Answer:

25

Explanation:

With Protons being the only charged particles in the nucleus, their overall charge is zero. This is because electrons and protons have opposite charges. The nuclear charge of ANY atom is given by its atomic number.

I am joyous to assist you anytime.

4 0

3 years ago

Consider the following reaction where Kc = 1.29×10-2 at 600 K: COCl2 (g) CO (g) + Cl2 (g) A reaction mixture was found to contai

vaieri [72.5K]

Answer:

1. In order to reach equilibrium COCl₂(g) must be consumed.

B. False

2. In order to reach equilibrium Kc must increase.

B. False .

3. In order to reach equilibrium CO must be consumed.

A. True.

4. Qc is greater than Kc.

A. True

5. The reaction is at equilibrium. No further reaction will occur.

B. False.

Explanation:

Based on the reaction:

COCl₂(g) → CO (g) + Cl₂(g)

And Kc is defined as:

Kc = 1.29x10⁻² = [CO] [Cl₂] / [COCl₂]

Molar concentrations of each species are:

[COCl₂] = 0.104 moles of COCl₂ / 1L = 0.104M

[CO] = 4.66×10⁻² moles of CO / 1L = 4.66×10⁻²M

[Cl₂] = 3.76×10⁻² moles of Cl₂ / 1L = 3.76×10⁻²M

Replacing in Kc formula:

4.66×10⁻²M × 3.76×10⁻²M / 0.104M = 1.68x10⁻²

As the concentrations are not in equilibrium, 1.68x10⁻² is defined as the reaction quotient, Qc.

As Qc > Kc, the reaction will shift to the left producing more COCl₂ and consuming CO and Cl₂. Thus

1. In order to reach equilibrium COCl₂(g) must be consumed.

B. False

2. In order to reach equilibrium Kc must increase.

B. False . Kc is a constant that never change.

3. In order to reach equilibrium CO must be consumed.

A. True.

4. Qc is greater than Kc.

A. True

5. The reaction is at equilibrium. No further reaction will occur.

B. False. The reaction is in equilibrium when Qc = Kc

6 0

4 years ago