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ser-zykov [4K]
3 years ago
13

How do the particles of a gas differ from the particles of a solid?

Chemistry
1 answer:
musickatia [10]3 years ago
5 0

Answer:

Gas is more open particles

Explanation: gas can flow

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If the pOH of vinegar is 9.45, what is its [OH− ]?
goldfiish [28.3K]

3.55 × 10n M n = -10

6 0
4 years ago
Read 2 more answers
Calculate the molality of acetone in an aqueous solution with a mole fraction for acetone of 0.241. Answer in units of m.
Anit [1.1K]

Answer: The molality of solution is 17.6 mole/kg

Explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

Molarity=\frac{n}{W_s}

where,

n = moles of solute

W_s = weight of solvent in kg

moles of acetone (solute) = 0.241

moles of water (solvent )= (1-0.241) = 0.759

mass of water (solvent )= moles\times {\text {Molar Mass}}=0.759\times 18=13.7g=0.0137kg

Now put all the given values in the formula of molality, we get

Molality=\frac{0.241}{0.0137kg}=17.6mole/kg

Therefore, the molality of solution is 17.6 mole/kg

3 0
3 years ago
Consider the following reaction:
adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

8 0
3 years ago
Nvm i got it lol how do i delete
Lady_Fox [76]

Answer:

delete what?

Explanation:

3 0
3 years ago
You have 47.0 mL of a 2.00 M concentrated or "stock" solution that must be diluted to 0.500 M. How much water should you add?
Inessa [10]

The required volume of water to make the dilute solution of 0.5 M is 188 mL.

<h3>How do we calculate the required volume?</h3>

Required volume of water to dilute the stock solution will be calculated by using the below equation as:

M₁V₁ = M₂V₂, where

  • M₁ & V₁ are the molarity and volume of stock solution.
  • M₂ & V₂ are the molarity and volume of dilute solution.

On putting values from the question to the above equation, we get

V₂ = (2)(47) / (0.5) = 188mL

Hence required volume of water is 188 mL.

To know more about volume & concentration, visit the below link:
brainly.com/question/7208546

#SPJ1

6 0
2 years ago
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