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3 years ago

Why does tin oxide weigh more than tin?

Chemistry

1 answer:

Oxana [17]3 years ago

4 0

Answer:

Explanation:

The mass of the actual tin will remain constant; however I assume you are comparing the mass of the reactant, tin, with the mass of the products. The increase in mass stems from the oxides of tin present due to its reaction with oxygen in the air:

Sn(s) + O2(g) → SnO2(s)

By working out the atomic mass of the tin and the relative formula mass of tin oxide, you can see why there is a mass increase; 119:151. As you can see here, for every 119g of tin reacted, there will be 151g of tin oxide, hence why you think the mass of the tin increased (by 32g).

Hope this helped!

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How do toothpaste prevent tooth decay?

padilas [110]

Answer:Flouride it creates enamel.

Explanation:

enamel (that's the tooth's hard outer surface) more resistant to acids produced by the bacteria living on the teeth and gums.

3 0

3 years ago

A cup of gold colored metal beads was measured to have a mass 425 grams. By water displacement, the volume of the beads was calc

Butoxors [25]

If the volume of 425 grams was 48.0 cm³, simply divide

g/cm³ = 425 g/48 cm³ = 8.85 g/cm³

If using water in water displacement, 1 mL = 1 cm³

8.85 g/cm³ = 8.85 g/mL

This density is most closely aligned with that of B. Copper

Hope I helped!

4 0

3 years ago

Q1. Briefly explain why the Zeff experienced by a valence electron in Cl is larger than in Mg. Q2. Briefly explain why the Zeff

Inessa05 [86]

Answer:

See explanation

Explanation:

Q1:

Chlorine has 17 protons while magnesium has only 12 protons. Recall that the Zeff depends on the size of the nuclear charge. The greater the size of the nuclear charge, the larger the Zeff experienced by a valence electron.

Q2:

The larger the Zeff, the smaller the atomic radius. Since the valence electrons of Cl experience a greater Zeff than those of Mg due to greater size of the nuclear charge, the atomic radius of chlorine will be smaller than that of Mg.

Q3:

The radius of an atom increases as the value of the principal quantum number (n) increases down the group due to addition of more shells. The greater the number of shells added, the greater the principal quantum number (n) and the greater the atomic radius, hence the answer.

5 0

3 years ago

How do electrons differ from protons and neutrons? A. They do not move. B. They are larger. C. They are not in a fixed position.

nikdorinn [45]

Hi!

Electrons are particles which basically 'orbit' around the nucleus. Protons and neutrons are condensed, in a fixed position inside the nucleus.

With this in mind, the answer will be C.

Hopefully, this helps! =)

4 0

4 years ago

Read 2 more answers

Help Please! Will give brainlyest for correct answers! 99 points!!

lapo4ka [179]

Answer:

1- 1.54 mol.

2- 271.9 kPa.

3- Yes, the tires will burst.

4- 235.67 kPa.

5- As, the temperature increased, the no. of molecules that has minimum kinetic energy increases as shown in image 1 that represents the Maxwell’s Distribution of Speeds of molecules. "Kindly, see the explanation and the attached images".

Explanation:

Q1- How many moles of nitrogen gas are in each tire?

To calculate the no. of moles of nitrogen gas in each tire, we can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the nitrogen gas (P = 247.0 kPa/101.325 = 2.44 atm),

V is the volume of the nitrogen gas (V = 15.2 L),

n is the no. of moles of the nitrogen gas (n = ??? mole),

R is the general gas constant (R = 0.082 L.atm/mol.K),

T is the temperature of the nitrogen gas (T = 21°C + 273 = 294 K).

∴ n = PV/RT = (2.44 atm)(15.2 L)/(0.082 L/atm/mol.K)(294.0 K) = 1.54 mol.

Q2: What would the maximum tire pressure be at 50 degrees C?

Now, the temperature is raised to be 50°C (T = 50°C + 273 = 323 K).

The pressure can be calculated using the general gas law: PV = nRT.

∴ P = nRT/V = (1.54 atm)(0.082 L/atm/mol.K)(323.0 K)/(15.2 L) = 2.68 atm = 271.9 kPa.

Q3: Will the tires burst in Spokane? Explain.

Yes, the tires will burst because the internal pressure be 271.9 kPa that exceeds 270 kPa, the pressure above which the tires will burst.

Q4: If you must let nitrogen gas out of the tire before you go, to what pressure must you reduce the tires before you start your trip? (Assume no significant change in tire volume.)

To get the pressure that we must begin with:

Firstly, we should calculate the no. of moles at:

T = 55°C + 273 = 328 K,

Pressure = 270 kPa (the pressure above which the tires will burst). (P =270 kPa/101.325 = 2.66 atm).

V = 15.2 L, as there is no significant change in tire volume.

∴ n = PV/RT = (2.66 atm)(15.2 L)/(0.082 L.atm/mol.K)(328 K) = 1.5 mol.

1.5562 moles of N₂ in the tires will give a pressure of 270 kPa at 55°C, so this is the minimum moles of N₂ that will make the tires burst.

Now, we can enter this number of moles into the original starting conditions to tell us what pressure the tires will be at if we start with this number of moles of N₂.

P = ???

V = 15.6 L.

n = 1.5 mol

T = 21°C + 273 = 294.0 K

R = 0.0821 L.atm/mol.K.

∴ P = nRT/V = (1.5 mol x 0.082 x 294.0 K) / (15.6 L) = 2.2325 atm = 235.67 kPa.

So, the starting pressure needs to be 235.67 kPa or just under in order for the tires not to burst.

Q5: Create a drawing of the tire and show a molecular view of the air molecules in the tire at 247 kpa vs the molecular view of the air molecules after the tires have been heated. Be mindful of the number of molecules that you use in your drawing in the before and after scenarios. Use a caption to describe the average kinetic energy of the molecules in both scenarios.

As, the temperature increased, the no. of molecules that has minimum kinetic energy increases as shown in “image 1” that represents the Maxwell’s Distribution of Speeds of molecules.

The no. of molecules that possess a critical K.E. of molecules increases due to increasing the temperature activate the motion of molecules with high velocity as

(K.E. = 3RT/2), K.E. directly proportional to the temperature of the molecules (see image 2).

Also, the average speed of molecules increases as the K.E of the molecules increases (see image 3).

5 0

3 years ago