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3 years ago

13

A metal oxide reacts chemically with water to form a base in the reaction shown below. If the temperature of the metal oxide and

water solution is increased, this will most likely

1 answer:

Alla [95]3 years ago

6 0

increase the rate of chemical change.

Explanation:

The reaction of the metal oxide with water to form a base in the presence of a spike in temperature will lead to an increase in the rate of chemical change.

Temperature change has considerable effect on reaction rates.

Temperature is directly proportional to the average kinetic energy of reacting particles.
Reaction rates varies directly with a spike in temperature.
It has been known that for every 10°C rise in temperature, above the room temperature, reaction rates become double or tripled.
Temperature increases the kinetic energy of each of the reacting particles.
Many of the reacting particles also acquires an energy greater than or equal to the activation energy of the reaction.
The frequency of ordinary collisions and effective collisions per unit time increases.

Learn more;

Activation energy brainly.com/question/3930233

#learnwithBrainly

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BY ANSWERING THIS QUESTION UR PUTTING IT ON UR MOM's LIFE THAT U WON'T STEAL MY POINTS.

Yakvenalex [24]

Answer:

$T_2=-125.58\°C$

Explanation:

Hello!

In this case, considering the Gay-Lussac's law which describes the pressure-temperature behavior as a directly proportional relationship by holding the volume as constant, we write:

$\frac{T_1}{P_1} =\frac{T_2}{P_2}$

Whereas solving for the final temperature T2, we get:

$T_2=\frac{T_1P_2}{P_1}$

Thus, we plug in the given data (temperature in Kelvins) to obtain:

$T_2=\frac{(22+273.15)K*1.75atm}{3.50atm} \\\\T_2=147.58K-273.15\\\\T_2=-125.58\°C$

Best regards!

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3 years ago

Olivia, a Latina student, is told that she can check only two books out of the library at a time, but Leann, a white student, is

Sliva [168]

Answer:

a

Explanation:

the others are rude, and rather support this, while a helps to support the ending of white privlige

4 0

3 years ago

Read 2 more answers

As you add protons to the nucleus and you add more what happen

Lubov Fominskaja [6]

The more protons you add, the more positively charged the atom becomes
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3 years ago

At 85°C, the vapor pressure of A is 566 torr and that of B is 250 torr. Calculate the composition of a mixture of A and B that b

Phantasy [73]

Answer:

Composition of the mixture:

$x_{A} =0.652=65.2$ %

$x_{B} =0.348=34.8$ %

Composition of the vapor mixture:

$y_{A} =0.809=80.9$ %

$y_{B} =0.191=19.1$ %

Explanation:

If the ideal solution model is assumed, and the vapor phase is modeled as an ideal gas, the vapor pressure of a binary mixture with $x_{A}$ and $x_{B}$ molar fractions can be calculated as:

$P_{vap}=x_{A}P_{A}+x_{B}P_{B}$

Where $P_{A}$ and $P_{B}$ are the vapor pressures of the pure compounds. A substance boils when its vapor pressure is equal to the pressure under it is; so it boils when $P_{vap}=P$ . When the pressure is 0.60 atm, the vapor pressure has to be the same if the mixture is boiling, so:

$0.60*760=P_{vap}=x_{A}P_{A}+x_{B}P_{B}\\456=x_{A}P_{A}+(1-x_{A})P_{B}\\456=x_{A}*(P_{A}-P_{B})+P_{B}\\\frac{456-P_{B}}{P_{A}-P_{B}}=x_{A}\\\\\frac{456-250}{566-250}=x_{A}=0.652$

With the same assumptions, the vapor mixture may obey to the equation:

$x_{A}P_{A}=y_{A}P$ , where P is the total pressure and y is the fraction in the vapor phase, so:

$y_{A} =\frac{x_{A}P_{A}}{P}=\frac{0.652*566}{456} =0.809=80.9$ %

The fractions of B can be calculated according to the fact that the sum of the molar fractions is equal to 1.

7 0

3 years ago

An air tight freezer measures 4 mx 5 m x 2.5 m high. With the door open, it fills with 22 °C air at 1 atm pressure.

____ [38]

Answer:

(a) Density of the air = 1.204 kg/m3

(b) Pressure = 93772 Pa or 0.703 mmHg

(c) Force needed to open the door = 15106 N

Explanation:

(a) The density of the air at 22 deg C and 1 atm can be calculated using the Ideal Gas Law:

$\rho_{air}=\frac{P}{R*T}$

First, we change the units of P to Pa and T to deg K:

$P=1 atm * \frac{101,325Pa}{1atm}=101,325 Pa\\\\ T=22+273.15=293.15^{\circ}K$

Then we have

$\rho_{air}=\frac{P}{R*T}=\frac{101325Pa}{287.05 J/(kg*K)*293.15K} =1.204 \frac{kg}{m3}$

(b) To calculate the change in pressure, we use again the Ideal Gas law, expressed in another way:

$PV=nRT\\P/T=nR/V=constant\Rightarrow P_{1}/T_{1}=P_{2}/T_{2}\\\\P_{2}=P_{1}*\frac{T_{2}}{T_{1}}=101325Pa*\frac{7+273.15}{22+273.15}=101,325Pa*0.9254=93,772Pa\\\\P2=93,772 Pa*\frac{1mmHg}{133,322Pa}= 0.703 mmHg$

(c) To calculate the force needed to open we have to multiply the difference of pressure between the inside of the freezer and the outside and the surface of the door. We also take into account that Pa = N/m2.

$F=S_{door}*\Delta P=2m^{2} *(101325Pa - 93772Pa)=2m^{2} *7553N/m2=15106N$

7 0

3 years ago