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3 years ago

Which gas occupies the highest volume at STP?

2 answers:

7 0

The gas, 2 mol of H2, occupies the highest volume at STP since at STP the volume of this gas is approximately 44.8 mol as compared to other options this has the greatest amount.

Anna [14]3 years ago

4 0

Answer: D. 2 mol of H2

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How many grams of SO2 can be formed from 0.622 mole FeS2

Anna71 [15]

Send answer choices. Wanna make sure My answer is correct

8 0

3 years ago

A sample of TNT, C7H5N3O6 , has 8.94 × 1021 nitrogen atoms. How many hydrogen atoms are there in this sample of TNT?

Bess [88]

Answer:

1.49×10²² atoms of H are contained in the sample

Explanation:

TNT → C₇H₅N₃O₆

1 mol of this has 7 moles of C, 5 moles of H, 3 moles of N and 6 moles of O

Let's determine the mass of TNT.

Molar mass is = 227 g/mol

As 1 mol has (6.02×10²³) NA atoms, how many moles are 8.94×10²¹ atoms.

8.94×10²¹ atoms / NA = 0.0148 moles

So this would be the rule of three to determine the mass of TNT

3 moles of N are in 227 g of compound

0.0148 moles of N are contained in (0.0148 .227) / 3 = 1.12 g

Now we can work with the hydrogen.

227 grams of TNT contain 5 moles of H

1.12 grams of TNT would contain (1.12 .5) / 227 = 0.0247 moles

Finally let's convert this moles to atoms:

0.0247 mol . 6.02×10²³ atoms / 1 mol = 1.49×10²² atoms

8 0

3 years ago

balance the following reaction. a coefficient of "1" is understood. choose option "blank" for the correct answer if the coeffici

suter [353]

Balanced chemical reaction: 2KCl + Pb(NO₃)₂ → PbCl₂ + 2KNO₃.

According to principle of mass conservation, number of atoms must be equal on both side of balanced chemical reaction.

KCl is potassium chloride.

Pb(NO₃)₂ is lead(II) nitrate.

KNO₃ is potassium nitrate.

PbCl₂ is lead(II) chloride.

3 0

3 years ago

Which structure is located between the auricle and the eardrum? O stirrup O ear canal O auditory nerve Ocochlea

Damm [24]

Answer:

The structure that is located between the auricle and the eardrum is the ear canal.

Explanation:

The ear canal, or external ear canal, is a a tubular hole about 30 mm long that runs from the auricle to the eardrum, forming part of the external ear.

Its function is to conduct sound, in the form of vibrations, from the outside to the eardrum. It also has the function of producing a viscous secretion called cerumen, capable of trapping dust particles and small foreign bodies.

Other options are not correct because:

Stirrup is located in the middle ear, along with the anvil and hammer.
Cochlea is in the inner ear and continues with the auditory nerve.

3 0

3 years ago

Use this equation for the following problems: 2NaN3 --> 2Na+3N2

olchik [2.2K]

Answer:

1) 65.0

2) 16.434 L = 16434 mL.

Explanation:

2NaN₃ → 2Na + 3N₂,

It is clear from the balanced equation that 2.0 moles of NaN₃ are decomposed to 2.0 moles of Na and 3.0 moles of N₂.

Q1: How many grams of NaN₃ are needed to make 23.6L of N₂?

Density of N₂ = 0.92 g/L which means that every 1.0 L of N₂ contains 0.92 g of N₂.

Now, we can get the mass of N₂ in 23.6 L N₂ using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂.

23.6 L of N₂ contains → ??? g of N₂.

∴ The mass of N₂ in 23.6 L of N₂ = (23.6 L)(0.92 g)/(1.0 L) = 21.712 g.

We can get the no. of moles of 23.6 L of N₂ (21.712 g) using the relation:

n = mass/molar mass = (21.712 g)/(28.0 g/mol) = 0.775 mol.

We can get the no. of moles of NaN₃ needed to produce 0.775 mol of N₂:

using cross multiplication:

2.0 moles of NaN₃ produce → 3.0 moles of N₂, from the balanced equation.

??? mol of NaN₃ produce → 0.775 moles of N₂.

∴ The no. of moles of NaN₃ needed = (2.0 mol)(0.775 mol)/(3.0 mol) = 0.517 mol.

Finally, we can get the grams of NaN₃ needed:

mass = no. of moles x molar mass = (0.517 mol)(65.0 g/mol) = 33.6 g.

Q2: How many mL of N₂ result if 8.3 g Na are also produced?

We need to get the no. of moles of 8.3 g Na using the relation:

n = mass/atomic mass = (8.3 g)/(22.98 g/mol) = 0.36 mol.

We can get the no. of moles of N₂ produced with 0.36 mol of Na:

using cross multiplication:

2.0 moles of Na produced with → 3.0 moles of N₂, from the balanced equation.

0.36 moles of Na produced with → ??? moles of N₂.

∴ The no. of moles of N₂ needed = (3.0 mol)(0.36 mol)/(2.0 mol) = 0.54 mol.

We can get the mass of 0.54 mol of N₂:

mass = no. of moles x molar mass = (0.54 mol)(28.0 g/mol) = 15.12 g.

Now, we can get the mL of 15.12 g of N₂:

using cross multiplication:

1.0 L of N₂ contains → 0.92 g of N₂, from density of N₂ = 0.92 g/L.

??? L of N₂ contains → 15.12 g of N₂.

∴ The volume of N₂ result = (1.0 L)(15.12 g)/(0.92 g) = 16.434 L = 16434 mL.

4 0

3 years ago