when we convert 32.5 lb/in² to atmosphere, the result obtained is 2.21 atm
<h3>Conversion scale</h3>
14.6959 lb/in² = 1 atm
<h3>Data obtained from the question</h3>
- Pressure (in lb/in²) = 32.5 lb/in²
- Pressure (in ATM) =?
<h3>How to convert 32.5 lb/in² to atm</h3>
14.6959 lb/in² = 1 atm
Therefore
32.5 lb/in² = 32.5 / 14.6959
32.5 lb/in² = 2.21 atm
Thus, 32.5 lb/in² is equivalent to 2.21 atm
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Answer:
As you move from left to right, the nucleus gains protons. This increases the positive charge of the nucleus and its attractive force on the electrons. At the same time, electrons are added to the atoms as you move from left to right across a period.
Explanation:
Answer:
V = 85.2
Explanation:
STP = 273K and 1 atm
Considering what we know about STP, we get the moles, temperature, and pressure. Using the ideal gas law we can find the volume (PV = nRT). Plug in our variables: (1 * V = 3.80 * R * 273). Since we are dealing with atm and not kPA or mmHg, we use the constant for atm (0.0821) which we use for R. (So.. now our equation is 1 * V = 3.80 * 0.0821 * 273). We now multiply the right side to get 85.17054. So... V = 85.2 considering sigificant figures (this is the part where I am the least sure of, since I havent done sig figs in a while)
A. Magnesium is an element and after it is burned it is a compound
The number of grams of Cl2 formed when 0.385 mol HCl reacts with an excess of O2 is 13.6675 g.
<h3>What are moles?</h3>
A mole is defined as 6.02214076 ×
of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.
Given data:
Moles of hydrochloric acid = 0.385 mol
Mass of chlorine gas =?
Chemical equation:
4HCl + O₂ → 2Cl₂ + 2H₂O
Now we will compare the moles of Cl₂ with HCl.
HCl : Cl₂
4 : 2
0.385 : 2÷4× 0.385 = 0.1925 mol
Oxygen is present in excess that's why the mass of chlorine produced depends upon the available amount of HCl.
Mass of Cl₂ :
Mass of Cl₂ = moles × molar mass
Mass of Cl₂ =0.1925 mol × 71 g/mol
Mass of Cl₂ = 13.6675 g
Hence, the number of grams of Cl2 formed when 0.385 mol HCl reacts with an excess of O2 is 13.6675 g.
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