Ask question

Ask question

All categories

erma4kov [3.2K]

3 years ago

8

A sample of gas has a volume of 4.4 liters when the pressure is 3.3atm. What is the volume when the pressure is reduced to 2.2at

m, if all other remain constant

1 answer:

oee [108]3 years ago

4 0

You'll remain with 2.9 liter

You might be interested in

Bess [88]

Answer:

Explination:

Given Data:

Trail 1 Trial 2 Trial 3

Student A 448.0 cm 485.6 cm 463.4 cm

Student B 450.5 cm 441.3 cm 446.8 cm

Student C 422.6 cm 445.2 cm 432.7 cm

Accepted Value = 435.0 cm

Required:

A: Accurate measurement =?

B: Reason for the answer =?

C: Precise measurement =?

D: Reason for the answer =?

Solution:

Student A:

Trail 1: 435.0 cm – 448.0 cm = (13.0 cm greater than accepted value)

Trail 2: 435.0 cm – 485.6 cm = (50.6 cm greater than accepted value)

Trial 3: 435.0 cm – 463.4 cm = (28.4 cm greater than accepted value)

Student B:

Trail 1: 435.0 cm – 450.5 cm = (14.5 cm greater than accepted value)

Trail 2: 435.0 cm – 441.3 cm = (6.3 cm greater than accepted value)

Trial 3: 435.0 cm – 446.8 cm = (11.8 cm greater than accepted value)

Student C:

Trial 1: 435.0 cm – 422.6 cm = (12.4 cm less than accepted value)

Trial 2: 435.0 cm – 445.2 cm = (10.2 cm greater than accepted value)

Trial 3: 435.0 cm – 432.7 cm = (1.3 cm less than accepted value)

A: The 3rd trial of the student C is accurate measurement = 432.7 cm.

B: The 3rd value of students’ C measurement is accurate because it is quite near to the accepted value, i.e. 435.0 cm.

As we know that Accuracy refers to the closeness of a measured value to a standard or known value.

This value has only the difference of 1.3 cm.

435.0 cm – 432.7 cm = (1.3 cm less than accepted value)

All the other have larger difference which is as above.

___________________

Student A:

1. 448.0 cm – 485.6 cm = (37.6 cm far)

2. 485.6 cm – 463.4cm = (22.2 cm far)

Student B:

1. 450.5 cm – 441.3 cm = (9.2 cm far)

2. 441.3 cm – 446.8 cm = (5.5 cm far)

Student C:

1. 422.6 cm – 445.2 cm = (22.6 cm far)

2. 445.2 cm – 432.7cm = (12.5 cm far)

So,

C: The values of Student B are more precise.

D: As we know that Precision refers to the closeness of two or more measurements to each other.

The measurements of student C are more close to each other. The values are only 9.2 cm and 5.5 cm far from each other.

5 0

3 years ago

How are we made from star debris? What does this mean? Explain

4vir4ik [10]

Answer:

How star stuff got to Earth

When it has exhausted its supply of hydrogen, it can die in a violent explostion, called a nova. The explosion of a massive star, called a supernova, can be billions of times as bright as the Sun , according to "Supernova," (World Book, Inc., 2005). Such a stellar explosion throws a large cloud of dust and gas into space, with the amount and composition of the material expelled varying depending on the type of supernova.

A supernova reaches its peak brightness a few days after it first occurred, during which time it may outshine an entire galaxy of stars. The dead star then continues to shine intensely for several weeks before gradually fading from view, according to "Supernova."

The material from a supernova eventually disperses throughout interstellar space. The oldest stars almost exclusively consisted of hydrogen and helium, with oxygen and the rest of the heavy elements in the universe later coming from supernova explosions, according to "Cosmic Collisions: The Hubble Atlas of Merging Galaxies," (Springer, 2009).

"It's a well-tested theory," Impey said. "We know that stars make heavy elements, and late in their lives, they eject gas into the medium between stars so it can be part of subsequent stars and planets (and people)."

Cosmic connections

So, all life on Earth and the atoms in our bodies were created in the furnace of now-long-dead stars, he said.

In 2002, music artist Moby released "We Are All Made of Stars," explaining during a press interview that his lyrics were inspired by quantum physics. "On a basic quantum level, all the matter in the universe is essentially made up of stardust," he said.

Explanation:

7 0

3 years ago

Mendeleev created this table as he noticed that a

Nimfa-mama [501]

Answer:

Mendeleev had left the noble gases out of his periodic table.

Explanation:

Mendeleev's periodic table is pictured in the image attached to the question.

Mendeleev's table obviously lacked the noble gases. The reason for this grave omission is simple; the noble gases were not known as at the time when he formulated his periodic table. There weren't any known elements whose properties were similar to the properties of the noble gases. This would have lead him to suspect their existence.

4 0

2 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

3 0

2 years ago

73.4 km to m then answer in correct number of significant numbers

Fittoniya [83]

Answer:

<em><u>To determine the number of significant figures in a number use the following 3 rules:</u></em>

<em><u>To determine the number of significant figures in a number use the following 3 rules:Non-zero digits are always significant.</u></em>

<em><u>To determine the number of significant figures in a number use the following 3 rules:Non-zero digits are always significant.Any zeros between two significant digits are significant.</u></em>

<em><u>To determine the number of significant figures in a number use the following 3 rules:Non-zero digits are always significant.Any zeros between two significant digits are significant.A final zero or trailing zeros in the decimal portion ONLY are significant.</u></em>

8 0

2 years ago