Answer : The final volume will be, 84.4 mL
Explanation :
Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

or,

where,
= initial pressure = 760 torr
= final pressure = 120 kPa = 900 torr (1 kPa = 7.5 torr)
= initial volume = 100 mL
= final volume = ?
Now put all the given values in the above equation, we get:


Therefore, the final volume will be, 84.4 mL
Answer:
0.7561 g.
Explanation:
- The hydrogen than can be prepared from Al according to the balanced equation:
<em>2Al + 6HCl → 2AlCl₃ + 3H₂,</em>
It is clear that 2.0 moles of Al react with 6.0 mole of HCl to produce 2.0 moles of AlCl₃ and 3.0 mole of H₂.
- Firstly, we need to calculate the no. of moles of (6.8 g) of Al:
no. of moles of Al = mass/atomic mass = (6.8 g)/(26.98 g/mol) = 0.252 mol.
<em>Using cross multiplication:</em>
2.0 mol of Al produce → 3.0 mol of H₂, from stichiometry.
0.252 mol of Al need to react → ??? mol of H₂.
∴ the no. of moles of H₂ that can be prepared from 6.80 g of aluminum = (3.0 mol)(0.252 mol)/(2.0 mol) = 0.3781 mol.
- Now, we can get the mass of H₂ that can be prepared from 6.80 g of aluminum:
mass of H₂ = (no. of moles)(molar mass) = (0.3781 mol)(2.0 g/mol) = 0.7561 g.
Answer:
nitrate and sodium are spectator ions .The reactions that form an aqueous solution are often a spectator ions.
Answer: The difference between an element and a compound is that an element is a pure substance that is made of only one element, but a compound is 2 or more elements together
Explanation:
160 g of SO3 are needed to make 400 g of 49% H2SO4.
<h3>How many grams of SO3 are required to prepare 400 g of 49% H2SO4?</h3>
The equation of the reaction for the formation of H2SO4 from SO3 is given below as follows:

1 mole of SO3 produces 1 mole of H2SO4
Molar mass of SO3 = 80 g/mol
Molar mass of H2SO4 = 98 g/mol
80 g of SO3 are required to produce 98 og 100%H2SO4
mass of SO3 required to produce 400 g of 100 %H2SO4 = 80/98 × 400 = 326.5 g of SO3
Mass of SO3 required to produce 49% of 400 g H2SO4 = 326.5 × 49% = 160 g
Therefore, 160 g of SO3 are needed to make 400 g of 49% H2SO4.
Learn more about mass and moles at: brainly.com/question/15374113
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