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Elodia [21]
3 years ago
8

Ag, CO, Find out the stock system thank you

Chemistry
1 answer:
Yuri [45]3 years ago
5 0

Answer:

a system in chemical nomenclature and notation of indicating the oxidation state of the significant element in a compound or ion by means of a Roman numeral that is used in parentheses after the name or part of the name designating this element and ending invariably in -ate in the case of an anion and that is placed. Stock nomenclature for inorganic compounds is a widely used system of chemical nomenclature developed by the German chemist Alfred Stock and first published in 1919. In the "Stock system", the oxidation states of some or all of the elements in a compound are indicated in parentheses by Roman numerals.

Explanation:

Brainliest please?

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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
astraxan [27]

Answer:

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
3 years ago
Determination of the amount of patassium hydroxide in patassium hydroxide/Sodium carbonate mixture
Eva8 [605]

Answer:

do you still need this answer?

Explanation:

6 0
3 years ago
What element is least likely to be found in nature
Doss [256]

Answer:

uranium, caesium, potassium, beryllium,

Explanation:

5 0
2 years ago
There are some data that suggest that zinc lozenges can significantly shorten the duration of a cold. If the solubility of zinc
zhuklara [117]

Answer:

K_{sp} of Zn(CH_{3}COO)_{2} is 0.0513

Explanation:

Solubility equilibrium of Zn(CH_{3}COO)_{2}:

Zn(CH_{3}COO)_{2}\rightleftharpoons Zn^{2+}+2CH_{3}COO^{-}

Solubility product of Zn(CH_{3}COO)_{2} (K_{sp}) is written as-            K_{sp}=[Zn^{2+}][CH_{3}COO^{-}]^{2}

Where [Zn^{2+}] and [CH_{3}COO^{-}] represents equilibrium concentration (in molarity) of Zn^{2+} and CH_{3}COO^{-} respectively.

Molar mass of Zn(CH_{3}COO)_{2} = 183.48 g/mol

So, solubility of Zn(CH_{3}COO)_{2} = \frac{43.0}{183.48}M = 0.234M

1 mol of Zn(CH_{3}COO)_{2} gives 1 mol of Zn^{2+} and 2 moles of CH_{3}COO^{-} upon dissociation.

so,   [Zn^{2+}] = 0.234 M and [CH_{3}COO^{-}] = (2\times 0.234)M=0.468M

so, K_{sp}=(0.234)\times (0.468)^{2}=0.0513          

8 0
3 years ago
How do producers make their own food
creativ13 [48]
Photosynthesis is how producers make their own food
8 0
3 years ago
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