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3 years ago

Which of the following characteristics indicates that an atom is unstable?

Chemistry

2 answers:

elena55 [62]3 years ago

8 0

Answer:

Explanation:

An atom can be considered unstable in one or two ways. If it pickes up or loses an electron,it becomes electrically charged and highly reactive.Intabability can also occur in the nucleus when the number of protons and neutrons is unbalanced.

larisa86 [58]3 years ago

7 0

Answer:

the answer is A full valence shell

Explanation:

hope it helps

You might be interested in

Calculate the value of Key in the following reaction if the equilibrium concentrations are

MAXImum [283]

Answer:

The value of the equilibrium constant Kc is 5.45

Explanation:

The "Law of Mass Action" states:

"For a reversible reaction in chemical equilibrium at a given temperature, it is true that the product of the concentrations of the products raised to the stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients is a constant."

This constant was called the equilibrium constant. For a reaction:

aA + bB ⇄ cC + dD

the equilibrium constant Kc is:

$Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a} *[B]^{b} }$

In this case, the balanced reaction is:

2 H₂S → 2 H₂ + S₂

So, the equilibrium constant Kc is:

$Kc=\frac{[H_{2} ]^{2} *[S_{2} ] }{[H_{2} S]^{2} }$

The equilibrium concentrations are

[H₂S] =0.25 M
[H₂]= 0.88 M
[S₂]= 0.44M

Replacing in the definition of equilibrium constant:

$Kc=\frac{(0.88 M)^{2} *0.44 M }{(0.25 M)^{2} }$

Solving:

Kc= 5.45

The value of the equilibrium constant Kc is 5.45

7 0

3 years ago

The total pressure of a mixture of oxygen and hydrogen

Mandarinka [93]

Answer:

The composition of the original mixture in molepercent is 80% of H₂ and 20% of O₂.

Explanation:

We need to combine the ideal gas law (PV = nRT) and Dalton's law of partial pressure (Pt = Pa +Pb +Pc+...).

The total pressure of the mixture is Pt = P (H₂) + P (O₂)

The number of moles can be found by Pt = nt RT/V, in which nt = n (H₂) +n (O₂).

If Pt is 1 atm, nt is 1.0 mol.

Now we need to consider the chemical reaction below:

H₂ + 0.5O₂ → H₂O

This shows that for each mol of O₂ we need two mol of H₂.

We know that the remaining gas is pure hydrogen and that its pressure is 0.4atm. Since PV = nRT, by the end of the reaction, 0.4 mol of H₂ remains in the system.

This means that in the beginning we have n mol of H₂, and when x mol of H₂ reacts with 0,5x mol of O₂, 0.4 mol of H₂ reamains.

If we have 1 mol in the begining and 0.4 mol in the end, the total amount of gas that reacted (x + 0.5X) is equal to 0.6 mol

x + 0.5X = 0.6 mol ∴ x = 0.6 mol / 1.5 ∴ x = 0.4 mol

0.4 mol of H₂ reacted with 0.2 mol of O₂ and 0.4 mol of H₂ remained as excess.

Therefore, in the beginning we had 0.8 mol of H₂ and 0.2 mol of O₂. Thus the molepercent of the mixture is 80% of H₂ and 20% of O₂.

3 0

3 years ago

Can you help me on this since im struggling on this since day 1

emmasim [6.3K]

Of Ok here is what I am thinking.

Simple replacement means when one element or compound moves out of another element and is replaced by this compound.

For example if you have A+BC ----- AC + B

Most of the time the one which is replaced is found on the right and the left one is going be only one element or compound.

Double replacement means when the two compounds exchange the positive and negative ions. Ok let see this example:

AB+ CD------- AD+ BC
this means if A have + sign and D has - sign they will share to form new compound the same is true for DC. Don't forget that all compounds are formed from positive and negative ions.

QUESTIONS
CUSO4 + NA2S---- CUS + Naso4[ double replacement]
KBR +Pb[CLO3]2------K[clo3]2 +PbBr [ double replacement]
I do not think these questions can be single replacement.
I hope this may helps you.

3 0

3 years ago

A chemist titrates 190.0 mL of a 0.8125 M ammonia (NH) solution with 0.3733 M HCl solution at 25 °C. Calculate the pH at equival

stealth61 [152]

Answer:

Approximately $4.92$ .

Explanation:

Initial volume of the solution: $V = 190.0\; \rm mL = 0.1900\; \rm L$ .

Initial quantity of $\rm NH_3$ :

$\begin{aligned} n({\rm NH_3}) &= c({\rm NH_3}) \cdot V({\rm NH_3}) \\ &= 0.3733\; \rm mol \cdot L^{-1} \times 0.1900\; \rm L \\ &\approx 0.154375\; \rm mol\end{aligned}$ .

Ammonia $\rm NH_3$ reacts with hydrochloric $\rm HCl$ acid at a one-to-one ratio:

$\rm NH_3 + HCl \to NH_4 Cl$ .

Hence, approximately $n({\rm HCl}) = 0.154375\; \rm mol$ of $\rm HCl\!$ molecules would be required to exactly react with the $\rm NH_3\!$ in the original solution and hence reach the equivalence point of this titration.

Calculate the volume of that $0.3733\; \rm mol \cdot L^{-1}$ $\rm HCl$ solution required for reaching the equivalence point of this titration:

$\begin{aligned}V({\rm HCl}) &= \frac{n({\rm HCl})}{c({\rm HCl})} \\ &\approx \frac{0.154375\; \rm mol}{0.3733\; \rm mol \cdot L^{-1}} \approx 0.413541\; \rm L\end{aligned}$ .

Hence, by the assumption stated in the question, the volume of the solution at the equivalence point would be approximately $0.413541\; \rm L + 0.1900\; \rm L \approx 0.6035\; \rm L$ .

If no hydrolysis took place, $0.154375\; \rm mol$ of $\rm NH_4 Cl$ would be produced. Because $\rm NH_4 Cl\!$ is a soluble salt, the solution would contain $0.154375\; \rm mol\!$ of $\rm {NH_4}^{+}$ ions. The concentration of $\rm {NH_4}^{+}\!$ would be approximately:

$\begin{aligned}c({\rm {NH_4}^{+}}) &= \frac{n({\rm {NH_4}^{+}})}{V({\rm {NH_4}^{+}})}\\ &\approx \frac{0.154375\; \rm mol}{0.6035\; \rm L} \approx 0.255782\; \rm mol \cdot L^{-1}\end{aligned}$ .

However, because $\rm NH_3 \cdot H_2O$ is a weak base, its conjugate $\rm {NH_4}^{+}$ would be a weak base.

$\begin{aligned}pK_{\rm a}({{\rm NH_4}}^{+}) &= pK_{\rm w} - pK_{\rm b}({\rm NH_3})\\ &\approx 13.99 - 4.75 = 9.25\end{aligned}$ .

Hence, the following reversible reaction would be take place in the solution at the equivalence point:

$\rm {NH_4}^{+} \rightleftharpoons NH_3 + H^{+}$ .

Let $x\; \rm mol \cdot L^{-1}$ be the increase in the concentration of $\rm H^{+}$ in this solution because of this reversible reaction. (Notice that $x \ge 0$ .) Construct the following $\text{RICE}$ table:

$\begin{array}{c|ccccc} \textbf{R}& \rm {\rm NH_4}^{+} & \rightleftharpoons & {\rm NH_3}& + & {\rm H}^{+}\\ \textbf{I} & 0.255782 \; \rm M \\ \textbf{C} & -x \;\rm M & & + x\;\rm M & & + x\; \rm M \\ \textbf{E} & (0.255782 - x)\; \rm M & & x\; \rm M & & x\; \rm M\end{array}$ .

Thus, at equilibrium:

Concentration of the weak acid: $[{\rm {NH_4}^{+}}] \approx (0.255782 - x) \; \rm M$ .
Concentration of the conjugate of the weak acid: $[{\rm NH_3}] = x\; \rm M$ .
Concentration of $\rm H^{+}$ : $[{\rm {H}^{+}}] \approx x\; \rm M$ .

$\displaystyle \frac{[{\rm NH_3}] \cdot [{\rm H^{+}}]}{[{ \rm {NH_4}^{+}}]} = 10^{pK_\text{a}({\rm {NH_4}^{+}})}$ .

$\displaystyle \frac{x^2}{0.255782 - x} \approx 10^{-9.25}$

Solve for $x$ . (Notice that the value of $x\!$ is likely to be much smaller than $0.255782$ . Hence, the denominator on the left-hand side $(0.255782 - x) \approx 0.255782$ .)