Why would it have been wrong to write Li2SO4 as lithium sulfur oxide?

     Because they often have long half-lives, their radioactive persistence in the environment is lengthy.

Number 4

If you spot any errors in my English, please tell me, since I am not a native speaker.

The element with atomic number 58 is Cerium, meaning its symbol should be Ce rather than Co, which belongs to Cobalt with atomic number 27. Therefore, the notation for isotopes consists of the element's symbol accompanied by a superscript and a subscript, properly aligned. The superscript indicates the mass number.

Mass number = protons + neutrons = 58 + 33 = 91

The subscript denotes the atomic number, which is 58. This notation is illustrated in the attached image.

Based on the titration results, the adult would need to consume 85.7 mL to meet the recommended daily intake of 60 mg of Vitamin C. This is calculated from the average total volume of DCPIP used during trials.

Answer:

0.400 moles of Oxygen

Explanation:

By employing the equation PV = nRT, the initial pressure of the flask can be calculated prior to the reaction, which leads to:

P = nRT/V

Where:

n signifies moles (4,000 moles: 2,000 moles of CO and 2,000 moles of H₂O)

R represents the gas constant (0.082 atm·L/mol·K)

T is the temperature (300.0 K)

V denotes volume (0.2000 L)

Substituting values results in P = 492.0 atm

To achieve a pressure reduction of 10.00%, the resulting pressure should be:

492.0 atm - 49.2 atm = 442.8 atm

Calculating with the new pressure under the same conditions gives the moles as:

n = PV/RT

n = 3,600 total moles

In the reaction:

2CO(g) + O₂(g) ⟶ 2CO₂(g)

The resulting moles are:

CO: 2,000 moles - 2X

O₂: 2,000 moles - X

CO₂: 2X

Where X accounts for the moles that react

Consequently, the total moles are:

4,000 moles - X = 3,600 moles

X = 0.400 moles

This indicates that the amount of oxygen needed for the reaction is 0.400 moles of Oxygen

I hope this is useful!

Context:

175 kilograms of methane (CH4) is to be converted into hydrogen cyanide (HCN)

The equation that balances this reaction is listed here:

2 CH4<span> + 2 NH</span>3<span> + 3 O</span>2<span> → 2 HCN + 6 H</span>2<span>O
</span>
To find the quantities of ammonia and oxygen needed, we will use 175 kg of CH4 as our reference.

Molar masses are as follows:
CH4 = 16 kg/kmol
NH3 = 17 kg/kmol
O2 = 32 kg/kmol

For ammonia: mass of NH3 = 175 kg CH4 / 16 kg/kmol * (2/2) * 17 kg/kmol  
This results in 185.94 kg of NH3 required

For oxygen: mass of O2 = 175 kg CH4 / 16 kg/kmol * (3/2) * 32 kg/kmol
So the mass of O2 needed equals 525 kg

To derive the mass of oxygen: mass of O = 525 kg / 32 kg/kmol * (1/2) * 16 kg/kmol
This gives a mass of O equal to 131.25 kg O