a 5.00 L sample of helium expands to 12.0 L at which point the pressure is measured to be 0.720atm. what was the original pressu

The visual representation is displayed in the following image.

For calculations, consider 100 grams of the compound:

ω(Cl) = 85.5% ÷ 100%.

ω(Cl) = 0.855; signifying the mass percentage of chlorine in the compound.

m(Cl) = 0.855 · 100 g.

m(Cl) = 85.5 g; this represents the mass of chlorine.

m(C) = 100 g - 85.5 g.

m(C) = 14.5 g; indicating the mass of carbon.

n(Cl) = m(Cl) ÷ M(Cl).

n(Cl) = 85.5 g ÷ 35.45 g/mol.

n(Cl) = 2.41 mol; this is the quantity of chlorine.

n(C) = 14.5 g ÷ 12 g/mol.

n(C) = 1.21 mol; this is the quantity of carbon.

n(Cl): n(C) = 2.41 mol: 1.21 mol = 2: 1.

The compound in question is identified as dichlorocarbene CCl₂.

Answer:

Amount of salt in 1 L seawater = 34 g

Explanation:

Based on Archimedes' principle, the mass of fresh water and the mass of the cup are equal to the mass of the same volume of seawater.

The mass of freshwater can be calculated using density times volume.

1 cm³ is equivalent to 1 mL.

The mass of freshwater is 0.999 g/cm³ multiplied by 735 cm³, which results in 734.265 g.

The total mass of the freshwater and cup combined is 734.265 g plus 25 g, equating to 759.265 g.

This means the mass for an equal volume of seawater is 759.265 g.

The volume of the seawater displaced is 735 mL, which is 0.735 L (assuming the cup's volume can be disregarded).

We know that 1 liter equals 1000 cm³ or 1000 mL.

The density of seawater can be determined as mass divided by volume.

The density of seawater becomes 759.265 g divided by 0.735 L, yielding 1033.01 g/L.

Conversely, the density of freshwater in g/L is calculated as 0.999 g/(1/1000) L, equating to 999 g/L.

The mass of salt dissolved in 1 liter of seawater is calculated as 1033.01 g - 999 g, which equals 34.01 g.

Thus, the amount of salt in 1 L of seawater is 34 g.

Answer:NH₃/NH₄Cl

Explanation:

The pH of a buffer can be determined using Henderson-Hasselbalch's equation.

When the concentration of acid equals that of the base, the pH aligns with the pKa of the buffer. The ideal pH range is pKa ± 1.

Below are the buffers and their corresponding pKa values:

• CH₃COONa/CH3COOH (pKa = 4.74)

• NH₃/NH₄Cl (pKa = 9.25)

• NaOCl/HOCl (pKa = 7.49)

• NaNO₂/HNO₂ (pKa = 3.35)

• NaCl/HCl Not a buffer

Thus, the ideal buffer is NH₃/NH₄Cl.

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