Given that Ag-107 (51.839% abundance) and the mass ratio: Ag-109/Ag-107=1.0187.
Calculate the mass of Ag-107. (Atomic mass of Ag = 107.87 amu)
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Answer:

Sn1 mechanism reaction

Explanation:

In this case, we have to start with the protonation of the "OH" group by the attack of the lone pairs in the alcohol group to the "H" in the HBr producing a positive charge in the oxygen. Then, water is produced and a carbocation is generated. This carbocation can be stabilized by a hydride shift. We can move a hydrogen atom to the positive charge and we will obtain a tertiary carbocation. Finally, the [tex]Br^- [/tex] will attack to produce the final halide.

See figure 1.

I hope it helps!

Answer:

B

Explanation:

The Ca+2 ion means that 2 electrons have been given away. So when you try and find the answer, you have to count backwards from Calcium. When you do, you get K+ first and then Argon which is either column 8 orc column 18 depending on your periodic table.

The element you hit is Argon.

The answer is B

Answer:

B ar

Explanation:

pen foster answer

Answer:

b. Binary compounds of carbon and hydrogen

Explanation:

Before proceeding, Hydrocarbons refers to organic chemical compounds composed exclusively of hydrogen and carbon atoms. This means the only elements present in an hydrocarbon are;

- Carbon

- Hydrogen

Looking through the options;

- Option A: This is wrong because the hydroxyl group contains oxygen and hydrocarbons contain only hydrogen and carbon.

- option B: This is correct. Binary compounds refers to compounds with just two elements.

- option C: This is wrong because water contains oxygen and hydrocarbons contain only hydrogen and carbon.

- option D: Carbon atoms can contain other elements so this option is wrong.

- option E: This also wrong because we had already gotten the correct option.

Answer:

Pb(NO3)2(aq) + 2KCl(aq) ------> 2KNO3(aq) + PbCl2(s)

3.52 g of PbCl2

3.76 g of KCl

Explanation:

The equation of the reaction is;

Pb(NO3)2(aq) + 2KCl(aq) ------> 2KNO3(aq) + PbCl2(s)

Number of moles of Pb(NO3)2 =mass/molar mass 5.06g/331.2 g/mol = 0.0153 moles

Number of moles of KCl= mass/ molar mass= 6.03g/74.5513 g/mol= 0.081 moles

Next we obtain the limiting reactant; the limiting reactant yields the least number of moles of products.

For Pb(NO3)2;

1 mole of Pb(NO3)2 yields 1 mole of PbCl2

Therefore 0.0153 moles of Pb(NO3)2 yields 0.0153 moles of PbCl2

For KCl;

2 moles of KCl yields 1 mole of PbCl2

0.081 moles of KCl yields 0.081 moles ×1/2 = 0.041 moles of PbCl2

Therefore Pb(NO3)2 is the limiting reactant.

Theoretical Mass of precipitate obtained = 0.0153 moles of PbCl2 × 278.1 g/mol = 4.25 g of PbCl2

% yield = actual yield/theoretical yield ×100

Actual yield = % yield × theoretical yield /100

Actual yield= 82.9 ×4.25/100

Actual yield = 3.52 g of PbCl2

If 1 mole of Pb(NO3) reacts with 2 moles of KCl

0.0153 moles of Pb(NO3)2 reacts with 0.0153 moles × 2 = 0.0306 moles of KCl

Amount of excess KCl= 0.081 moles - 0.0306 moles = 0.0504 moles of KCl

Mass of excess KCl = 0.0504 moles of KCl × 74.5513 g/mol = 3.76 g of KCl

Answer: The oxidation number of oxygen in this compound is -2.

Explanation:

For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.  

In [tex]Na_2O[/tex], sodium is having an oxidation state of +1 called as [tex]Na^+[/tex] cation and oxygen is an anion with oxidation state of -2. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral [tex]Na_2O[/tex]

The cations and anions being oppositely charged attract each other through strong coloumbic forces and form an ionic bond.

Answer:

Sodium Oxide: Na2O, Barium Phosphide: Ba3P2

Explanation:

Just did the question

Answer:

If the temperature increases the molecular movement as well, and if it increases the same it will happen with the molecular movement.

Pressure, volume and temperature are three factors that are closely related since they increase the temperature, the pressure usually decreases due to the dispersion of the molecules that can be generated, so the volume also increases.

If the temperature drops, the material becomes denser, its molecules do not collide with each other, their volume and pressure increases.

Explanation:

The pressure is related to the molecular density and the movement that these molecules have.

The movement is regulated by temperature, since if it increases, the friction and collision of the molecules also.

On the other hand, the higher the volume, the less pressure there will be on the molecules, since they are more dispersed among themselves.

(in the opposite case that the volume decreases, the pressure increases)

Answer:

Approximately [tex]10.88[/tex].

Explanation:

[tex]\rm OBr^{-}[/tex] can act as a weak Bronsted-Lowry base:

[tex]\rm OBr^{-}\; (aq) + H_2O\; (l) \rightleftharpoons HOBr\; (l) + OH^{-}\; (aq)[/tex].

(Side note: the state symbol of [tex]\rm HOBr[/tex] in this equation is [tex]\rm (l)[/tex] (meaning liquid) because [tex]\rm HOBr[/tex] is a weak acid.)

However, the equilibrium constant of this reaction, [tex]K_\text{eq}[/tex], isn't directly given. The idea is to find [tex]K_\text{eq}[/tex] using the [tex]\rm pH[/tex] value at the half-equivalence point. Keep in mind that this system is at equilibrium all the time during the titration. If temperature stays the same, then the same [tex]K_\text{eq}[/tex] value could also be used to find the [tex]\rm pH[/tex] of the solution before the acid was added.

At equilibrium:

[tex]\displaystyle K_\text{eq} = \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]}[/tex].

At the half-equivalence point of this titration, exactly half of the base, [tex]\rm OBr^{-}[/tex], has been converted to its conjugate acid, [tex]\rm HOBr[/tex]. Therefore, the half-equivalence concentration of [tex]\rm OBr^{-}[/tex] and [tex]\rm HOBr[/tex] should both be equal to one-half the initial concentration of [tex]\rm OBr^{-}[/tex].

As a result, the half-equivalence concentration of [tex]\rm OBr^{-}[/tex] and [tex]\rm HOBr[/tex] should be the same. The expression for [tex]K_\text{eq}[/tex] can thus be simplified:

[tex]\begin{aligned}& K_\text{eq} \\&= \frac{\left(\text{half-equivalence $[\rm HOBr\; (l)]$}\right)\cdot \left(\text{half-equivalence $[\rm OH^{-}\; (aq)]$}\right)}{\text{half-equivalence $[\rm OBr^{-}\; (l)]$}}\\ &=\text{half-equivalence $[\rm OH^{-}\; (aq)]$}\end{aligned}[/tex].

In other words, the [tex]K_\text{eq}[/tex] of this system is equal to the [tex]\rm OH^{-}[/tex] concentration at the half-equivalence point. Assume that [tex]\rm p\mathnormal{K}_\text{w}[/tex] the self-ionization constant of water, is [tex]14[/tex]. The concentration of [tex]\rm OH^{-}[/tex] can be found from the [tex]\rm pH[/tex] value:

[tex]\begin{aligned}& \text{half-equivalence $[\rm OH^{-}\; (aq)]$} \\ &= 10^{\rm pH - p\mathnormal{K}_\text{w}}\;\rm mol \cdot L^{-1} \\ &= 10^{7.75 - 14}\; \rm mol \cdot L^{-1}\\ &= 10^{-6.25}\; \rm mol \cdot L^{-1}\end{aligned}[/tex].

Therefore, [tex]\begin{aligned} K_\text{eq} &= 10^{-6.25}\end{aligned}[/tex].

Again, since [tex]\rm KOBr[/tex] is a soluble salt, all that [tex]0.200\; \rm M[/tex] of [tex]\rm KOBr[/tex] in this solution will be in the form of [tex]\rm K^{+}[/tex] and [tex]\rm OBr^{-}[/tex] ions. Before any hydrolysis takes place, the concentration of [tex]\rm OBr^{-}[/tex] should be equal to that of [tex]\rm KOBr[/tex]. Therefore:

[tex]\text{$[\rm OBr^{-}\; (aq)]$ before hydrolysis} = 0.200\; \rm M[/tex].

Let the equilibrium concentration of [tex][\rm OH^{-}\; (aq)][/tex] be [tex]x\; \rm M[/tex]. Create a RICE table for this reversible reaction:

[tex]\begin{array}{c|ccccccc} & \rm OBr^{-}\; (aq) &+&\rm H_2O\; (l)& \rightleftharpoons & \rm HOBr\; (l)& + & \rm OH^{-}\; (aq) \\ \textbf{I}& 0.200\; \rm M & & & & 0 \; \rm M & & 0\; \rm M \\ \textbf{C} & -x\; \rm M & & & & +x \; \rm M & & +x\; \rm M \\ \textbf{E}& (0.200 + x)\; \rm M & & & & x \; \rm M & & x\; \rm M \end{array}[/tex].

Assume that external factors (such as temperature) stays the same. The [tex]K_\text{eq}[/tex] found at the half-equivalence point should apply here, as well.

[tex]\displaystyle K_\text{eq} = \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]}[/tex].

At equilibrium:

[tex]\displaystyle \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]} = \frac{x^2}{0.200 + x}[/tex].

Assume that [tex]x[/tex] is much smaller than [tex]0.200[/tex], such that the denominator is approximately the same as [tex]0.200[/tex]:

[tex]\displaystyle \frac{[\rm HOBr\; (l)]\cdot [\rm OH^{-}\; (aq)]}{[\rm OBr^{-}\; (aq)]} = \frac{x^2}{0.200 + x} \approx \frac{x^2}{0.200}[/tex].

That should be equal to the equilibrium constant, [tex]K_\text{eq}[/tex]. In other words:

[tex]\displaystyle \frac{x^2}{0.200} \approx K_\text{eq} = 10^{-6.25}[/tex].

Solve for [tex]x[/tex]:

[tex]x \approx 3.35\times 10^{-4}[/tex].

In other words, the [tex]\rm OH^{-}[/tex] before acid was added was approximately [tex]3.35\times 10^{-4}\; \rm M[/tex], which is the same as [tex]3.35\times 10^{-4}\; \rm mol \cdot L^{-1}[/tex]. Again, assume that [tex]\rm p\mathnormal{K}_\text{w} = 14[/tex]. Calculate the [tex]\rm pH[/tex] of that solution:

[tex]\begin{aligned}\rm pH &= \rm p\mathnormal{K}_\text{w} + \log [\mathrm{OH^{-}}] \approx 10.88\end{aligned}[/tex].

(Rounded to two decimal places.)

Answer:

The Henry's law constant for argon is [tex]k=2.11*10^{-3}\frac{ M}{atm}[/tex]

Explanation:

Henry's Law indicates that the solubility of a gas in a liquid at a certain temperature is proportional to the partial pressure of the gas on the liquid.

C = k*P

where C is the solubility, P the partial pressure and k is the Henry constant.

So, being the concentration [tex]C=\frac{ngas}{V}[/tex]  

where ngas is the number of moles of gas and V is the volume of the solution, you must calculate the number of moles ngas. This is determined by the Ideal Gas Law: P*V=n*R*T where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. So [tex]n=\frac{P*V}{R*T}[/tex]

In this case:

Then:

[tex]n=\frac{1 atm*5.16*10^{-2} L}{0.082 \frac{atm*L}{mol*K} *298K}[/tex]

Solving:

n= 2.11 *10⁻³ moles

So: [tex]C=\frac{ngas}{V}=\frac{2.11*10^{-3} moles}{1 L} =2.11*10^{-3} \frac{moles}{L}= 2.11*10^{-3} M[/tex]

Using Henry's Law and being C=CAr and P =PAr:

2.11*10⁻³ M= k* 1 atm

Solving:

[tex]k=\frac{2.11*10^{-3} M}{1 atm}[/tex]

You get:

[tex]k=2.11*10^{-3}\frac{ M}{atm}[/tex]

The Henry's law constant for argon is [tex]k=2.11*10^{-3}\frac{ M}{atm}[/tex]

The Henry's law constant for argon gas in 1 litre of water is 2.1 × 10⁻³M/atm.

Henry's law of gas states that solubility of a gas in any liquid at particular temperature is directly proportional to the partial pressure of the gas.

C∝P

C = kP, where

k = Henry's constant

P = partial pressure of gas

C is the solubility and it is present in the form of concentration and will be calculated as:
C = n/V

n = no. of moles

V = volume

And moles of the gas will be calculated by using the ideal gas equation as:

PV = nRT

n = (1)(5.16×10⁻²) / (0.082)(298) = 2.1 × 10⁻³ moles

And Concentration in liquid will be:

C = 2.1 × 10⁻³mol / 1L = 2.1 × 10⁻³ M

Now we put all these values in the first equation to calculate the value of k as:

k = (2.1 × 10⁻³M) / (1atm) = 2.1 × 10⁻³M/atm

Hence required value of k is 2.1 × 10⁻³M/atm.

To know more about Henry's law, viist the below link:

https://brainly.com/question/12823901