Convert this temperature from F to C 26.6°c 93.600 62.2°C 5.7°C

What is the theory that can explain the model of Pangaea?

NISA [10]

The bones of the same animal found out continents far away from each other

4 0

3 years ago

The positive sign of the energy difference in a chemical reaction would indicate that ..........

trasher [3.6K]

Answer:

B) the chemicals are gaining energy from the surroundings.

Explanation:

The positive sign of the energy difference in a chemical reaction would indicate that the chemicals are gaining energy from the surroundings. This is what happens in an endothermic reaction.

In an endothermic reaction, heat is absorbed from the surroundings hence the surrounding becomes colder at the end of the changes.

Here the energy change is assigned a positive value.
This is because the heat energy level of the final state is higher than that of the initial state.
So, the difference gives a positive value.

3 0

2 years ago

A 265-mL flask contains pure helium at a pressure of 751 torrs. A second flask with a volume of 465 mL contains pure argon at a

Nadya [2.5K]

Answer:

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar = 472.8 torr

Explanation:

Step 1: Data given

Volume of the flask helium = 265 mL

Pressure in the helium flask = 751 torr = 751/760 atm

Volume of the flask argon = 465 mL

Pressure in the argon flask = 727 torr = 727/760 atm

The total pressure exerted by a gaseous mixture is equal to the sum of the partial pressures of each individual component in a gas mixture.

Step 2: Calculate total volume

Total volume = 265 mL + 465 mL = 730 mL = 0.730 L

Step 3: Boyle's Law:

P1V1=P2V2

⇒ with P1 = total pressure gas exerts in its own flask

⇒ with V1 = volume of flask with stopcock valve closed

⇒ with P2 = partial pressure of gas exerts on total volume of both flasks when stopcock valve is opened

⇒ with V2 = total volume of both flasks with stopcock valve opened

Helium using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of helium = 751 /760 = 0.98816 atm

⇒ with V1 = volume of helium = 0.265 L

⇒ with P2 = The new partial pressure of helium

⇒ with V2 = total volume = 0.730 L

(0.98816 atm)(0.265L)=P2(0.730L)

P2=0.359 atm

Argon using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of argon = 727/760 = 0.95658 atm

⇒ with V1 = volume of argon = 0.465 L

⇒ with P2 = The new partial pressure of argon

⇒ with V2 = total volume = 0.730 L

(0.95658 atm)(0.465L)=P2(0.730L)

P2=0.609 atm

Step 4: Convert pressure in atm to torr

Pressure helium = 0.359 atm = 272.8 torr

Pressure argon = 0.609 atm = 472.8 torr

Step 5: Calculate Total pressure

Ptotal = P(He)+P(Ar)

⇒ Pt = total pressure of the gas mixture

⇒ P(He) = partial pressure of Helium

⇒ P(Ar) = partial pressure of Argon

Pt = 272.8 torr + 472.8 torr

Pt = 745.6 torr

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar = 472.8 torr

5 0

3 years ago

Design a test to determine whether thorium-234 also emits particles. First, explain how Rutherford’s experiment measured positiv

liubo4ka [24]

The characteristics of the α and β particles allow to find the design of an experiment to measure the ²³⁴Th particles is:

On a screen, measure the emission as a function of distance and when the value reaches a constant, there is the beta particle emission from ²³⁴Th.

The neutrons cannot be detected in this experiment because they have no electrical charge.

In Rutherford's experiment, the positive particles directed to the gold film were measured on a phosphorescent screen that with each arriving particle a luminous point is seen.

The particles in this experiment are α particles that have two positive charge and two no charged is a helium nucleus.

The test that can be carried out is to place a small ours of Thorium in front of a phosphorescent screen and see if it has flashes, with the amount of them we can determine the amount of particle emitted per unit of time.

Thorium has several isotopes, with different rates and types of emission:

²³²Th emits α particles, it is the most abundant 99.9%

²³⁴Th emits β particles, exists in small traces.

In this case they indicate that the material used is ²³⁴Th, which emits β particles that are electrons, the detection of these particles is more difficult since it has one negative charge, it has much lower mass, but they can travel further than the particles α, therefore, for what type of isotope we have, we can start measuring at a small distance and increase the distance until the reading is constant. At this point all the particles that arrive are β, which correspond to ²³⁴Th.

Neutron detection is much more difficult since these particles have no charge and therefore do not interact with electrons and no flashing on the screen is varied.

In conclusion with the characteristics of the α and β particles we can find the design of an experiment to measure the ²³⁴Th particles is:

On a screen, measure the emission as a function of distance and when the value reaches a constant, there is the β particle emission from ²³⁴Th.

The neutrons cannot be detected in this experiment because they have no electrical charge.

Learn more about radioactive emission here: brainly.com/question/15176980

7 0

3 years ago

if the observed test value of a hypothesis test is outside of the established critical value(s), a researcher would __________.

Makovka662 [10]

The answer to this question is:

If the observed test value of a hypothesis test is outside of the established critical value(s), a researcher would __________.
"Claim significant support for the hypothesis"

Hoped This Helped, Itsalishamariee
Your Welcome :)

6 0

3 years ago

Read 2 more answers