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kupik [55]
3 years ago
8

a difference in density between gases creates buoyancy, which is the ability of an object to float. At 0 degrees celsius, helium

has a density of 0.18kg
Chemistry
1 answer:
GaryK [48]3 years ago
3 0

Answer: for an object or gas or liquid to float in another there must be a difference in density

Explanation:

Helium has a density of 0.18 kg/m³ and air has a density of 1.29 kg/m^³. If a balloon is filled with helium it will float in air due to density differences

Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces

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Help plzzzzz ASAP!!!!!!!!
SIZIF [17.4K]

1. NaF, Na₂S, Na₃P, Na₂O

2. MgF₂, MgS, Mg₃P₂, MgO

3. AlF₃, Al₂S₃, AlP, Al₂O₃

<h3>Further explanation</h3>

Given

Ionic charge

Required

The formula of binary ionic compounds

Solution

Ionic compounds consisting of cations (ions +) and anions (ions -)

Ionic compounds usually consist of metal cations and non-metal anions

Metal: cation, positively charged.

Nonmetal: negatively charged

The anion cation's charge is crossed

The ionic compounds :

1. NaF, Na₂S, Na₃P, Na₂O

2. MgF₂, MgS, Mg₃P₂, MgO

3. AlF₃, Al₂S₃, AlP, Al₂O₃

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3 years ago
A buffer solution contains 0.345 M acetic acid and 0.377 M sodium acetate . If 0.0613 moles of potassium hydroxide are added to
melamori03 [73]

Answer:

pH = 5.54

Explanation:

The pH of a buffer solution is given by the <em>Henderson-Hasselbach (H-H) equation</em>:

  • pH = pKa + log\frac{[CH_3COO^-]}{[CH_3COOH]}

For acetic acid, pKa = 4.75.

We <u>calculate the original number of moles for acetic acid and acetate</u>, using the <em>given concentrations and volume</em>:

  • CH₃COO⁻ ⇒ 0.377 M * 0.250 L = 0.0942 mol CH₃COO⁻
  • CH₃COOH ⇒ 0.345 M * 0.250 L = 0.0862 mol CH₃COOH

The number of CH₃COO⁻ moles will increase with the added moles of KOH while the number of CH₃COOH moles will decrease by the same amount.

Now we use the H-H equation to <u>calculate the new pH</u>, by using the <em>new concentrations</em>:

  • pH = 4.75 + log\frac{(0.0942+0.0613)mol/0.250L}{(0.0862-0.0613)mol/0.250L} = 5.54
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3 years ago
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