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jonny [76]
3 years ago
14

How many particles are in 7.07x10^-6 moles of a substance?

Chemistry
1 answer:
Butoxors [25]3 years ago
5 0

Answer:

4.25*10^18

Explanation:

1 mole =6.023*10^23 particles

so 7.07*10^-6 mole=6.023*10^23*7.07*10^-6 particles=4.25*10^18 particles

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If four hydrogen atoms react with two hydrogen atoms, how many atoms<br> will be in the product?
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Answer:6

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An aqueous solution at 25 °C has a pOH of 12.42. Calculate the pH. Round your answer to 2 decimal places
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Answer:

pH=1.58

Explanation:

As we know that

pH+pOH=14

Given that pOH=12.42

we get pH=14-pOH

or pH=14-12.42

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Pls help this is due in 10 minutes can someone tell me question 4
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Then the person will only have the dominant trait. There needs to be two recessive genes in order for that gene to appear.

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The combustion of ethane ( C 2 H 6 ) produces carbon dioxide and steam. 2 C 2 H 6 ( g ) + 7 O 2 ( g ) ⟶ 4 CO 2 ( g ) + 6 H 2 O (
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10.4 moles of CO2 are produced

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The carbon-14 content of a wooden harpoon handle found in an Inuit archaeological site was found to be 61.9% of the carbon-14 co
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3,964 years.

Explanation:

  • It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
  • Half-life time is the time needed for the reactants to be in its half concentration.
  • If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
  • Also, it is clear that in first order decay the half-life time is independent of the initial concentration.

  • The half-life of the element is 5,730 years.

  • For, first order reactions:

<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>

Where, k is the rate constant of the reaction.

t1/2 is the half-life of the reaction.

∴ k =0.693/(t1/2) = 0.693/(5,730 years) = 1.21 x 10⁻⁴ year⁻¹.

  • Also, we have the integral law of first order reaction:

<em>kt = ln([A₀]/[A]),</em>

where, k is the rate constant of the reaction (k = 1.21 x 10⁻⁴ year⁻¹).

t is the time of the reaction (t = ??? year).

[A₀] is the initial concentration of the sample ([A₀] = 100%).

[A] is the remaining concentration of the sample ([A] = 61.9%).

∴ t = (1/k) ln([A₀]/[A]) = (1/1.21 x 10⁻⁴ year⁻¹) ln(100%/61.9%) = 3,964 years.

7 0
4 years ago
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