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2 years ago

How would you know if this combination is likely to be found in some dirt?

Chemistry

1 answer:

Mkey [24]2 years ago

6 0

Answer:

very unlikely to be found in dirt

Explanation:

That combination has a ratio of ¹⁴N to ¹⁵N of 10:4 = 2.5: 1. The most ¹⁵N enriched natural material has a ratio of ¹⁴N to ¹⁵N on the order of 250 : 1 or higher. The combination shown is very unlikely to be found in dirt.

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Which of the following shows an accurate combustion reaction?

Jet001 [13]

Answer:

Cu + O₂ → CuO₂

Explanation:

A combustion reaction is the reaction with oxygen along with the release of energy in form of heat or light.

Organic compounds (like CH₄) undergo combustion forming water and CO₂.

The combustion reaction of CH₄ is:

CH₄ + 2O₂ → CO₂ + 2H₂O

Hence, the first equation from the choices is not showing the combustion reaction of CH₄.

Not only organic compounds can undergo combustion. Metals and no metals can undergo combustion, i.e. metals and no metals can react with oxygen releasing light or heat.

The reaction of copper and oxygen (second choice) is a combustion reaction:

Cu + O₂ → CuO₂

The formation of water (2H₂ + O₂ → 2H₂O) is other example of a combustion reaction where no organic compounds are involved.

On the other hand, the other two equations from the choice list are not reactions with oxygen, so they do not show combustion reactions.

5 0

2 years ago

An alkene with the molecular formula C8H16 undergoes ozonolysis to yield a mixture of (CH3)2C=O and (CH3)3CCHO. The alkene is:

aalyn [17]

Answer:

2,4,4-trimethyl-2-pentene yields mixture of $(CH_{3})_{2}C=O$ and $(CH_{3})_{3}CHO$

Explanation:

In ozonolysis (hydrolysis step involve a reducing agent such as Zn, $Me_{2}S$ etc.), a pi bond is broken to form ketone/aldehyde.

Ketone is formed from di-substituted side of double bond and aldehyde is formed from mono-substituted side of double bond.

Ozoznolysis involves two consecutive steps : (1) formation of ozonide, (2) hydrolysis of ozonide.

Hydrolysis can be done with/without using reducing agent. Carboxylic acid/carbon dioxide/ketone is produced when hydrolysis is done without using reducing agent.

Here, 2,4,4-trimethyl-2-pentene yields mixture of $(CH_{3})_{2}C=O$ and $(CH_{3})_{3}CHO$

Reaction steps are shown below.

8 0

2 years ago

Phosphorus has the molecular formula p4, and sulfur has the molecular formula s8. how many grams of phosphorus contain the same

Alinara [238K]

1) Find the number of molecules in 7.88 g of sulfur

molar mass of S8 = 8*atomic mass of S = 8 * 32.0 g / mol = 256.0 g/mol

Number of moles = mass in grams / atomic mass = 7.88 g / 256.0 g / mol = 0.0308 moles

2) Find the mass of 0.0308 moles of P4

mass = number of moles * molar mass

molar mass of P4 = 4 * atomic mass of P = 4 * 31 g/mol = 124 g/mol

mass of P4 = 0.0308 moles * 124 g/mol = 3.8192g ≈ 3.82 g.

Answer: 3.82 grams of P4 will have the same number of molecules as 7.88 g of S8 (that is 0.0308 moles of molecules)

6 0

2 years ago

Calculate the amount of work done against an atmospheric pressure of 1.00 atm when 500.0 g of zinc dissolves in excess acid at 3

ZanzabumX [31]

Answer:

19,26 kJ

Explanation:

The work done when a gas expand with a constant atmospheric pressure is:

W = PΔV

Where P is pressure and ΔV is the change in volume of gas.

Assuming the initial volume is 0, the reaction of 500g of Zn with H⁺ (Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)) produce:

500,0g Zn(s)× $\frac{1molZn}{65,38g}$ × $\frac{1molH_{2}(g)}{1molZn}$ = 7,648 moles of H₂

At 1,00atm and 303,15K (30°C), the volume of these moles of gas is:

V = nRT/P

V = 7,648mol×0,082atmL/molK×303,15K / 1,00atm

V = 190,1L

That means that ΔV is:

190,1L - 0L = 190,1L

And the work done is:

W = 1atm×190,1L = 190,1atmL.

In joules:

190,1 atmL× $\frac{101,325}{1atmL}$ = 19,26 kJ

I hope it helps!

5 0

2 years ago

One liter of ocean water contains 35.06 g of salt. What volume of ocean water would contain 1.00 kg of salt? Express your answer

sweet [91]

Answer:

28.52 L

Explanation:

First, let's calculate the density of the ocean, which is the mass divided by the volume:

d = m/V

d = 35.06/1

d = 35.06 g/L

So, for a mass of 1.00 kg = 1000.00 g

d = m/V

35.06 = 1000.00/V

V = 1000.00/35.06

V = 28.52 L

How all the data are expressed with two significant figures, the volume must also be expressed with two.

7 0

2 years ago