Answer:The new volume is 5mL
Explanation:
The formular for Boyles Law is; P1 V1 = P2 V2
Where P1 = 1st Pressure V1 = First Volume
P2 = 2nd Pressure V2 = Second Volume
From the question; P1 = 5atm, V1 = 10ml
P2 = 2 x P1 (2 x 5) = 10 atm V2 =?
Using the Boyles Law Formular; P1 V1 = P2 V2, we make V2 the subject of formular; P1 V1/ P2 = V2
∴ 5 x 10/ 10 = 5
∴ V2 = 5mL
One aluminum atom will cost 0.84 × 10⁻²³ cents.
Explanation:
Aluminium costs 85 cents per 453.592 grams.
number of moles = mass / atomic weight
number of moles of aluminium = 453.592 / 27 = 16.8 moles
To calculate the number of aluminum atoms we use the Avogadro's number:
if 1 mole of aluminum contains 6.022 × 10²³ aluminum atoms
then 16.8 moles of aluminum contains X aluminum atoms
X = (16.8 × 6.022 × 10²³) / 1 = 101.17 × 10²³ aluminum atoms
Now, taking in account the aluminium price, we devise the following reasoning:
if 101.17 × 10²³ aluminum atoms costs 85 cents
then 1 aluminum atom costs Y cents
Y = (1 × 85) / (101.17 × 10²³) = 0.84 × 10⁻²³ cents
Learn more about:
Avogadro's number
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Surface tension increases by increasing the intermolecular forces.
<h3>What is intermolecular forces?</h3>
The electromagnetic forces of attraction or repulsion that act between atoms and other types of nearby particles, such as atoms or ions, are examples of intermolecular forces (IMFs), also known as secondary forces.
Between molecules, intermolecular forces are at work. In contrast, molecules themselves exert intramolecular pressures. In comparison to intramolecular forces, intermolecular forces are weaker. Intermolecular forces include things like the London dispersion force, dipole-dipole interaction, ion-dipole interaction, and van der Waals forces.
Intermolecular forces come in five flavors: ion-induced dipole forces, dipole-induced dipole forces, induced dipole forces, and dipole-dipole forces. Ions and polar (dipole) molecules are held together by ion-dipole forces.
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Answer:
The mass of NaBr needed is 0.22969 g.
Explanation:
1 mole of NaBr contains 22.4 dm^3 of NaBr
Therefore, 0.05 dm^3 (50/1000 = 0.05 dm^3) of NaBr = 0.05/22.4 = 0.00223 mol of NaBr
MW of NaBr = 23 + 80 = 103 g/mol
Mass of NaBr = number of moles × MW = 0.00223 × 103 = 0.22969 g
