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krek1111 [17]
3 years ago
9

Enter the appropriate symbol for an isotope of phosphorus-32 corresponding to the isotope notation AZX. Express your answer as a

chemical symbol using isotope notation.
Chemistry
1 answer:
Lelechka [254]3 years ago
8 0

Answer:

^{32}_{15}P

Explanation:

Atomic number : It is defined as the number of electrons or number of protons present in a neutral atom.

However, when we talk about the atomic number of the ion, it is not equal to the number of electrons as electron can be gained or loosed.

This is why, more appropriately, the number of the protons which are present in the nucleus of the atom is called the atomic number.

Thus, atomic number of phosphorus = 15

Mass number is the number of the entities present in the nucleus which is the equal to the sum of the number of protons and electrons.

Given, Mass number = 32

Thus, the symbol of the isotope is:-

^{32}_{15}P

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Answer:

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Explanation:

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6 0
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Anyone think they can help me with this one?
BARSIC [14]
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Then identify the groups that are connected off the parent chain, here we have a methyl group on carbon 2.

Thus the name would be 2 - methyl - 3 - heptyne. I believe.
4 0
3 years ago
The solubility product constant of pbcl2 is 1.7 × 10−5 . what is the maximum concentration of pb2+ that can be in ocean water th
Sunny_sXe [5.5K]
Answer is: the maximum concentration of Pb²⁺ is 6.8·10⁻³ M.
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c(NaCl) = c(Cl⁻) = 0.0500 M.
Ksp(PbCl₂) = c(Pb²⁺) · c(Cl⁻)².
c(Pb²⁺) = Ksp(PbCl₂) ÷ c(Cl⁻)².
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3 0
3 years ago
In a 0.730 M solution, a weak acid is 12.5% dissociated. Calculate Ka of the acid.
Mamont248 [21]

Answer:

Approximately 1.30 \times 10^{-2}, assuming that this acid is monoprotic.

Explanation:

Assume that this acid is monoprotic. Let \rm HA denote this acid.

\rm HA \rightleftharpoons H^{+} + A^{-}.

Initial concentration of \rm HA without any dissociation:

[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}.

After 12.5\% of that was dissociated, the concentration of both \rm H^{+} and \rm A^{-} (conjugate base of this acid) would become:

12.5\% \times 0.730\; \rm mol \cdot L^{-1} = 0.09125\; \rm mol \cdot L^{-1}.

Concentration of \rm HA in the solution after dissociation:

(1 - 12.5\%) \times 0.730\; \rm mol \cdot L^{-1} = 0.63875\; \rm mol\cdot L^{-1}.

Let [{\rm HA}], [{\rm H}^{+}], and [{\rm A}^{-}] denote the concentration (in \rm mol \cdot L^{-1} or \rm M) of the corresponding species at equilibrium. Calculate the acid dissociation constant K_{\rm a} for \rm HA, under the assumption that this acid is monoprotic:

\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}.

5 0
3 years ago
If the CaCO3 weighed 983 g and the CaO weighed 551 g, how many grams of CO2 were formed in the reaction?
stira [4]

Answer:

The answer to your question is 432 g of CO₂

Explanation:

Data

CaCO₃  = 983 g

CaO = 551 g

CO₂ = ?

Balanced reaction

                               CaCO₃ (s)   ⇒   CaO (s)   +  CO₂ (g)

This reaction is balanced, to solve this problem just remember the Lavoisier Law of conservation of mass that states that the mass of the reactants is equal to the mass of the products.

                    Mass of reactants = Mass of products

                    Mass of CaCO₃   = Mass of CaO + Mass of CO₂

Solve for CO₂

                    Mass of CO₂  = Mass of CaCO₃ - Mass of CaO                    

                     Mass of CO₂ = 983 g - 551 g

Simplification

                     Mass of CO₂ = 432 g                        

         

5 0
3 years ago
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