Anyone help please more & you get extra points

Answer:

true

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:

Zr(Cr2O7)2

Explanation:

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:

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:

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:

It will strengthen Electromagnetism

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: 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:

D. 0.50

Explanation:

Use avogadro number to find the whole work.

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:

Q58. number of moles(n) and relative formula mass(Mr)

Q59. a)   2 H2S +  O2 --> 2 H2O + 2 S

         b)   After the balanced equation you should able to draw it yourself - I                              cannot draw it on here :)

Explanation:

Answer:

A solid metal bead of the same size is dropped into the same cup and sinks to the bottom of the cup. How do the pebble and the metal bead compare? They both have the same mass. They are both denser than water.

Explanation:

nun

Answer:

Both the pebble and metal bead of the same mass will be equally accelerated  if they are at the same place, in this case, they are in the same place meaning they will be accelerated at the same speed towards the earth. Thus even if they are in the water they will sink at the same speed and finally will reach the bottom of the cup.

Answer:

20009 is answer is right 1345678899444

Explanation:

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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:

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

Answer:

D. ethane

Explanation:

Answer:

To back up any statements I make.

To refute any errors presented.

Explanation:

During a group discussion, you look at your notes to point out an error or when you want to flesh out a point you made.

The term group discussion has to do with learners talking about a topic in a manner ion which each learner takes a turn to explain a portion of the topic.

During a group discussion, you look at your notes to point out an error or when you want to flesh out a point you made.

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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:

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:

Found in every living cell

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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