Calculate the number of moles of NaOH contained in 250. mL of a 0.05M solution.

Answer:

B C4H10 will be the empirical formulae.

Answer:

The specific heat of water is more specific than the heat of sand, therefore, it will take more energy to raise the same amount of water with the same temperature.

Explanation:

The temp is 89 F so, 31.67° C

The specific heat of  the water is 4.184 J/g° C

The specific heat of the sand is 0.290 J/ g° C

The heat is the amount  of energy that is needed to raise the temp. of the substance

You will need a lot more energy to raise the temp of the water then of the same amount of sand, therefore, because of the specific lower heat of the sand it will raise it's temperature quicker compared to water.

Answer:

H2O2

Explanation:

There's hydrogen and ox in the formula to represent

Answer:

C) Na

Explanation:

The question tells us that all of the sodium (Na) is consumed and that water is left over. This means that sodium would be the limiting reagent.

Answer:

D. 0.50

Explanation:

Use avogadro number to find the whole work.

Answer:

There's no such thing as absolute zero

Answer:

I think number 1 but if wrong report me

Answer: 4.003 g/mol

Explanation:

So basically Molar mass is the atomic mass of an element and the atomic mass of Helium is 4.0026 or 4.003 g/mol

Answer:

A) barium hydroxide

B) ethanol  

C) glucose

D) silver nitrate  

E) dichloromethane

F) postassium chloride

The other compounds ethanol and glucose  are also soluble in water but do not dissociate when dissolved in water.

Explanation:

Please mark this Brainilest.

These compounds will dissociate in water:

A) barium hydroxide

D) silver nitrate

F) potassium chloride

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Answer:

The correct answer is - 66g.

Explanation:

Given:

molecules of CO2 = 9.03 x 10^23

We know:

1 mole of any substance = 6.02x10^23 molecules. (Avogadro's number)

M (CO2) = 12 + (2x16) = 12 + 32 = 44g

Solution:

The mass of CO2 with 9.03x10^23 molecules would be:

44g of CO2 = 6.02x10^23 molecules.

So, the mass in grams of CO2 (X) = 9.03x10^23 molecules

Xg of CO2 = (44x9.03x10^23)/6.02x10^23 = 66g

Thus, the correct answer would be - 66gm.

Answer:

1) The new pressure of the gas is 500 kilopascals.

2) The final volume is 1.44 liters.

3) Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

Explanation:

1) From the Equation of State for Ideal Gases we construct the following relationship:

[tex]\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}[/tex] (1)

Where:

[tex]P_{1}, P_{2}[/tex] - Initial and final pressure, measured in kPa.

[tex]V_{1}, V_{2}[/tex] - Initial and final pressure, measured in mililiters.

If we know that [tex]P_{1} = 100\,kPa[/tex], [tex]V_{1} = 500\,mL[/tex] and [tex]V_{2} = 1000\,mL[/tex], then the new pressure of the gas is:

[tex]P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}} \right)[/tex]

[tex]P_{2} = 500\,kPa[/tex]

The new pressure of the gas is 500 kilopascals.

2) Let suppose that gas experiments an isothermal process. From the Equation of State for Ideal Gases we construct the following relationship:

[tex]\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}[/tex] (1)

Where:

[tex]P_{1}, P_{2}[/tex] - Initial and final pressure, measured in kPa.

[tex]V_{1}, V_{2}[/tex] - Initial and final pressure, measured in mililiters.

If we know that [tex]V_{1} = 3.60\,L[/tex], [tex]P_{1} = 10\,kPa[/tex] and [tex]P_{2} = 25\,kPa[/tex] then the new volume of the gas is:

[tex]V_{2} = V_{1}\cdot \left(\frac{P_{1}}{P_{2}} \right)[/tex]

[tex]V_{2} = 1.44\,L[/tex]

The final volume is 1.44 liters.

3) From the Equation of State for Ideal Gases we construct the following relationship:

[tex]\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}}[/tex] (1)

Where:

[tex]P_{1}, P_{2}[/tex] - Initial and final pressure, measured in kPa.

[tex]V_{1}, V_{2}[/tex] - Initial and final pressure, measured in mililiters.

If we know that [tex]\frac{P_{2}}{P_{1}} = 3[/tex], then the volume ratio is:

[tex]\frac{V_{1}}{V_{2}} = 3[/tex]

[tex]\frac{V_{2}}{V_{1}} = \frac{1}{3}[/tex]

Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

Answer:

Found in every living cell

Answer:

True

Explanation:

Just like the Earth, half of the Moon is lit by the Sun while the other half is in darkness.

Answer:

Pretty sure it's true.

On Earth, we only see part of the moon that is illuminated or not, since it revolves around the Earth. However, the sun is stationary and doesn't move, meaning that one side of the moon is always lit by the sun as it revolves around the Earth.

Answer: 25.13 g of [tex]N_2O[/tex]  and 20.56 g of [tex]H_2O[/tex] will be produced from 45.70 g of [tex]NH_4NO_3[/tex]

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} NH_4NO_3=\frac{45.70g}{80.04g/mol}=0.571moles[/tex]

The balanced chemical equation is:

[tex]NH_4NO_3\rightarrow N_2O+2H_2O[/tex]  

According to stoichiometry :

1 mole of [tex]NH_4NO_3[/tex] produce = 1 mole of [tex]N_2O[/tex]

Thus 0.571 moles of [tex]NH_4NO_3[/tex] will require=[tex]\frac{1}{1}\times 0.571=0.571moles[/tex]  of [tex]N_2O[/tex]  

Mass of [tex]N_2O=moles\times {\text {Molar mass}}=0.571moles\times 44.01g/mol=25.13g[/tex]

1 mole of [tex]NH_4NO_3[/tex] produce = 2 moles of [tex]H_2O[/tex]

Thus 0.571 moles of [tex]NH_4NO_3[/tex] will require=[tex]\frac{2}{1}\times 0.571=1.142moles[/tex]  of [tex]H_2O[/tex]  

Mass of [tex]H_2O=moles\times {\text {Molar mass}}=1.142moles\times 18g/mol=20.56g[/tex]

Thus 25.13 g of [tex]N_2O[/tex]  and 20.56 g of [tex]H_2O[/tex] will be produced from 45.70 g of [tex]NH_4NO_3[/tex]

Answer:

78

Explanation:

Answer:

The OH would be .7825 and the basic solution is a strong base.

Explanation:

What you would do is -log(0.165) in your calculator which would give you 0.7825160065 as an answer. Im not sure what the significant figure is so I will not be rounding to that, but that is your answer for the first part.

The second part: because your pOH is a .7825, this would be consiered a strong base in the pOH because it is closer to 1 which is your base.

Answer:

okay but give me brainliest

Explanation:

A volcano is formed when hot molten rock, ash and gases escape from an opening in the Earth's surface. The molten rock and ash solidify as they cool, forming the distinctive volcano shape shown here. As a volcano erupts, it spills lava that flows downslope. Volcanic flows are called lahars.

The percentage yield of the reaction has been 46.74%.

Percentage yield can be given as the ratio of the actual yield to the theoretical yield.

For the theoretical yield, the balanced equation will be:

[tex]\rm 2\;ZnS\;+\;3\;O_2\;\rightarrow\;2\;ZnO\;+\;2\;SO_2[/tex]

Thus, 2 moles of ZnS will give 2 moles of ZnO.

The moles of 46.5 g ZnS:

Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]

ZnS = [tex]\rm \dfrac{46.5}{97.474}[/tex]

ZnS = 0.477 mol

The moles of 13.3 grams Oxygen:

Oxygen = [tex]\rm \dfrac{13.3}{32}[/tex]

Oxygen = 0.415 mol.

The moles of 18.14 grams ZnO:

ZnO = [tex]\rm \dfrac{18.14}{81.38}[/tex]

ZnO = 0.222 mol

According to the balanced chemical equation,

2 moles ZnO = 2 moles ZnS

0.222 mol ZnO = 0.222 mol ZnS

2 moles ZnO = 3 moles Oxygen

0.222 mol ZnO = 0.334 mol of Oxygen

Since, both the reactants are in enough concentration,

The theoretical yield can be calculated with any of the reactants. The theoretical yield of ZnO from 46.5 grams of ZnS can be given as:

1 mole ZnS = 1 mol ZnO

0.477 mol of ZnS = 0.477 mol of ZnO.

The mass of 0.477 mol of ZnO:

Mass = moles [tex]\times[/tex] molecular weight

Mass of ZnO = 0.477 [tex]\times[/tex] 81.38

Mass of ZnO = 38.818 grams.

The theoretical yield of ZnO = 38.818 g.

The actual yield of ZnO = 18.14 g.

Percentage yield = [tex]\rm \dfrac{18.14}{38.81}\;\times\;100[/tex]

Percentage yield = 46.74 %

The percentage yield of the reaction has been 46.74%.

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Answer:

Well it depends, There are THOUSANDS of inactive satellites. Even scientists don't know how many as for active ones there is about 1 957 according to nasa

Explanation:

Answer:

Answer is A. Ethyl.........

Answer:

D. ethane

Explanation:

Answer:

None of these cause the correct answer is 9/-5

Answer: Answer:

Explanation:

Apparent magnitude does not take into account the fact that not all stars are the same distance from Earth.

Apparent magnitude is only on a comparative scale that determines how bright a star looks from a theoretical distance of 32 light-years from Earth. Therefore, the correct option is option D.

Any large, gaseous celestial body that is capable of producing its own light and illuminates through radiation. Only a very small portion of the tens of milliards of milliards of stars that make up the observable cosmos are readily visible to the eye.

When measuring the brightness of a star, what is a limitation of using apparent magnitude rather than absolute brightness. Apparent magnitude is only on a comparative scale that determines how bright a star looks from a theoretical distance of 32 light-years from Earth.

Therefore, the correct option is option D.

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Answer:

It will strengthen Electromagnetism

Answer:

true

Answer:

Assignment Summary

For this assignment, you will develop a model that shows a roller coaster cart in four different positions on a track. You will then use this model to discuss the changes in potential and kinetic energy of the cart as it moves along the track.

Background Information

The two most common forms of energy are potential energy and kinetic energy. Potential energy is the stored energy an object has due to its position. Kinetic energy is the energy an object has due to its motion. An object’s kinetic energy changes with its motion, while its potential energy changes with its position, but the total energy stays the same. If potential energy increases, then kinetic energy decreases. If potential energy decreases, then kinetic energy increases.

Potential energy related to the height of an object is called gravitational potential energy. Gravitational potential energy is directly related to an object’s mass, the acceleration due to gravity, and an object’s height.

Materials

 One poster board per student  Drawing utensils

Assignment Instructions

Step 1: Prepare for the project.

a) Read the entire Student Guide before you begin this project.

b) If anything is unclear, be sure to ask your teacher for assistance before you begin.

c) Gather the materials you will need to complete this project.

Step 2: Create your poster.

a) On the poster board, draw a roller coaster track that starts with one large hill, then is followed by a valley and another, smaller hill.

b) Draw a cart in four positions on the track as outlined below.

i. Draw the first cart at the top of the first hill. Label it A.

ii. Draw the second cart going down the first hill into the valley. Label it B.

iii. Draw the third cart at the bottom of the valley. Assume that the height of the cart in this position is zero. Label it C.

iv. Draw the last cart at the top of the second, smaller hill. Label it D.

c) Make sure that your name is on the poster. Step 3: Type one to two paragraphs that describe the energy of the cart.

a) Type one to two paragraphs describing the changes in potential and kinetic energy of the cart. Be sure to discuss how the potential and kinetic energy of the cart changes at each of the four positions along the track, and explain why these changes occur.

b) Make sure your name is on the document.

c) Later, you will submit this document through the virtual classroom.

Step 4: Evaluate your project using this checklist.

If you can check each criterion below, you are ready to submit your project.

 Did you draw a model of a roller coaster track with one large hill, a valley, and a smaller hill?

 Did you draw a cart on the track in the four required positions A–D? Did you label the cart at each of the four positions?

 Did you type a paragraph describing the changes in potential and kinetic energy of the cart at each of the four positions on the roller coaster track? Did you explain why the changes in potential and kinetic energy occur?

Step 5: Revise and submit your project.

a) If you were unable to check off all of the requirements on the checklist, go back and make sure that your project is complete.

b) When you have completed your project, submit your poster to your teacher for grading. Be sure that your name is on it.

c) Submit the typewritten document through the virtual classroom. Be sure that your name is on it.

Step 6: Clean up your work space.

a) Clean up your work space. Return any reusable materials to your teacher and throw away any trash.

b) Congratulations! You have completed your project.

Explanation:

Answer:

20009 is answer is right 1345678899444

Explanation:

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