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3 years ago

) Experimental evidence indicates that the nucleus of an atom

Chemistry

1 answer:

geniusboy [140]3 years ago

5 0

Answer:

A (contains most of the mass of the atom)

Evidence has it that a proton is about 2000 times as massive as an electron.

And there is usually multiple protons and neutrons in the nucleus

From what I just said, you can say that B is wrong

C however is also wrong because protons have a +charge and neutrons are neutrle which means you always have a charge > (greater than) 0

And D is wrong because electrons (which are not in the nucleus) have a neg charge. and protons have a + charge and are in the nucleus

So your answer is A

Hope it helped

Spiky Bob

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The area that includes everything between the cell membrane and the nucleus of a cell is called the ___________.

m_a_m_a [10]

The answer is C. Cytoplasm

happy to help!!

7 0

2 years ago

Read 2 more answers

Use the unbalanced equation NH3+O2=NO+H2O to find the mole ratio between NO and H2O

Leona [35]

Answer:

For every 4 moles of NO created, 6 moles of H2O are created so the ratio is 4:6

Explanation:

You just need to balance the equation.

NH3 + O2 -> NO + H2O

1. I started with hydrogen; there's 3 on the left and 2 on the right. Multiply them together to find a number they both go into (3×2=6, but in this case 6 hydrogen on each side does not work so I doubled it so there is 12 hydrogen on each side).

This will bring you to this:

4NH3 + O2 -> NO + 6H2O

2. Now get equal amounts of nitrogen on each side. There's 4 nitrogen on the left side, and 1 on the right. Multiply the right by 4. Then you will have this:

4NH3 + O2 -> 4NO + 6H2O

3. Last thing you need to do is have the same amount of oxygen on both sides. On the left you have 2 and on the right you have 10. Get the left to 10 by multiplying it by 5.

Balanced: 4NH3 + 5O2 -> 4NO + 6H2O

In word form, for every reaction between 4 moles of ammonia and 5 moles of oxygen, 4 moles of nitric oxide and 6 moles of water will be created.

I hope this helps!

5 0

3 years ago

How do you write 561983 in scientific notation?

Artist 52 [7]

Answer:

5.61983 × 10^5

Explanation:

Move the decimal forward 5 spaces, each time doing so you add 10^(# of spaces moved, in this case 5)

7 0

3 years ago

You mix 285.0 mL of 1.20 M lead(II) nitrate with 300.0 mL of 1.60 M potassium iodide. The lead(II) iodide is insoluble. Which of

SIZIF [17.4K]

Answer:

D. The final concentration of NO3– is 0.821 M.

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For potassium iodide :

Molarity = 1.60 M

Volume = 300.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 300.0×10⁻³ L

Thus, moles of potassium iodide :

$Moles=1.60 \times {300.0\times 10^{-3}}\ moles$

Moles of potassium iodide = 0.48 moles

For lead(II) nitrate :

Molarity = 1.20 M

Volume = 285 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 285×10⁻³ L

Thus, moles of lead(II) nitrate :

$Moles=1.20\times {285\times 10^{-3}}\ moles$

Moles of lead(II) nitrate = 0.342 moles

According to the given reaction:

$2KI_{(aq)}+Pb(NO_3)_2_{(aq)}\rightarrow PbI_2_{(s)}+2KNO_3_{(aq)}$

2 moles of potassium iodide react with 1 mole of lead(II) nitrate

1 mole of potassium iodide react with 1/2 mole of lead(II) nitrate

0.48 moles potassium iodide react with 0.48/2 mole of lead(II) nitrate

Moles of lead(II) nitrate = 0.24 moles

Available moles of lead(II) nitrate = 0.342 moles

Limiting reagent is the one which is present in small amount. Thus, potassium iodide is limiting reagent.

Also, consumed lead(II) nitrate = 0.24 moles (lead ions precipitate with iodide ions)

Left over moles = 0.342 - 0.24 moles = 0.102 moles

Total volume = 300 + 285 mL = 585 mL = 0.585 L

So, Concentration = 0.102/0.585 M = 1.174 M

Statement A is correct.

The formation of the product is governed by the limiting reagent. So,

2 moles of potassium iodide gives 1 mole of lead(II) iodide

1 mole of potassium iodide gives 1/2 mole of lead(II) iodide

0.48 mole of potassium iodide gives 0.48/2 mole of lead(II) iodide

Mole of lead(II) iodide = 0.24 moles

Molar mass of lead(II) iodide = 461.01 g/mol

Mass of lead(II) chloride = Moles × Molar mass = 0.24 × 461.01 g = 111 g

Statement B is correct.

Potassium iodide is the limiting reagent. So all the potassium ion is with potassium nitrate . Thus,

2 moles of Potassium iodide on reaction forms 2 moles of potassium ion

0.48 moles of Potassium iodide on reaction forms 0.48 moles of potassium ion

Total volume = 300 + 285 mL = 585 mL = 0.585 L

So, Concentration = 0.48/0.585 M = 0.821 M

Statement C is correct.

Nitrate ions are furnished by lead(II) nitrate . So,

1 mole of lead(II) nitrate produces 2 moles of nitrate ions

0.342 mole of lead(II) nitrate produces 2*0.342 moles of nitrate ions

Moles of nitrate ions = 0.684 moles

So, Concentration = 0.684/0.585 M = 1.169 M

Statement D is incorrect.

4 0

3 years ago

Determine whether each of the properties described applies to volumetric or graduated glassware. 1. Used for applications in whi

kogti [31]

Answer:

1) volumetric

2) graduated

3) volumetric

Explanation:

A volumetric glassware is a glassware that is marked at a particular point. A typical example of a volumetric glassware is the volumetric flask. A volumetric glassware is capable of measuring only a specific volume of a liquid.

On the other hand, graduated glassware can measure a range of volumes of liquid. However, a volumetric glassware is still required where a high degree of accuracy is important.

5 0

3 years ago