When the water level of a stream or river exceeds its natural banks, it has reached its

1. According to the equation, what mass of hydrogen fluoride is necessary to produce 2.3 g of sodium fluoride?

Vanyuwa [196]

Answer:

1.096g

Explanation:

You must know the atomic mass of Hydrogen, Fluorine, and Sodium before you can start:

Hydrogen: 1.008g/mol

Fluorine: 18.99g/mol

Sodium: 22.98g/mol

Next, find the composition percentage of NaF

22.98 + 18.99 = 41.97

Fluorine is 18.99/41.97 =45.25%

Sodium is 100-45.25 = 54.75%

Ultimately we want to know about HF so find how much F is in 2.3g: 2.3 * 0.4525 = 1.041g

Find comp. percentage of HF

18.99+1.008 = 19.998; H/total F/total

Hydrogen 5.041%

Fluorine 94.959%

Laws of conservation of say we have 1.041g of fluorine in our HF. We know 1.041 is 94.959% of the mass of HF so do some simple math to find the remaining: 1.041/0.94959 = 1.096g

4 0

3 years ago

What can you say about an atom with 4 protons, 2 neutrons, and 5 electrons?

iren2701 [21]

Explanation:

This is a neutral beryllium atom

5 0

3 years ago

You want to determine the protein content in milk with the Kjeldahl method. You take 100 g whole milk and use 100 mL of 0.5 M hy

stepladder [879]

Answer:

3.38%

Explanation:

Given that;

the mass of the whole milk sample = 100 g

volume of HCl = 100 mL = 0.1 L

molarity of HCl = 0.5 M

volume of NaOH = 34.50 mL = 0.0345 L

molarity of NaOH = 0.3512 M

Since we knew the molarity and volume of both HCl and NaOH; we can calculate their corresponding number of moles present.

So, number of moles of HCl = molarity of HCl × volume of HCl

number of moles of HCl = 0.5 M × 0.100 mL

= 0.05 mole

number of moles of NaOH = Molarity of NaOH × Volume of NaOH

number of moles of NaOH = 0.3512 M × 0.0345 L

= 0.012 mole

From the question, we can deduce that the number of HCl that is consumed by NH₃ is equal to the number of moles of HCl that is consumed by NaOH.

SO, number of moles of HCl consumed by NH₃ = Total moles of HCl - moles of HCl consumed by NaOH

= 0.05 mole - 0.012 mole

= 0.038 mole

However, to determine the mole of NH₃ present , we have:

number of moles of NH₃ present = number of moles of HCl consumed by NH₃ = 0.038

∴ the mass of Nitrogen with the molecular weight (14.0 g/mol) = 0.038 moles × 14.0 g/mol

= 0.530 g

Now, the percentage of Nitrogen can be calculated as;

$percentage of nitrogen =\frac{mass of nitrogen}{mass of the whole milk sample} *100$

$percentage of nitrogen =\frac{0.530g}{100g} *100$

$percentage of nitrogen =0.530%$ %

the percentage of protein in the sample = CP × %age of N

where CP is given as 6.38

∴ the percentage of protein in the sample = 6.38 × 0.530%

the percentage of protein in the sample = 3.3814%

the percentage of protein in the sample = 3.38%

6 0

3 years ago

An electrochemical cell at 25°C is composed of pure copper and pure lead solutions immersed in their respective ionis. For a 0.6

ExtremeBDS [4]

Answer :

(a) The concentration of $Pb^{2+}$ is, 0.0337 M

(b) The concentration of $Pb^{2+}$ is, $6.093\times 10^{32}M$

Solution :

<u>(a) As per question, lead is oxidized and copper is reduced.</u>

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Pb\rightarrow Pb^{2+}+2e^-$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

The balanced cell reaction will be,

$Pb(s)+Cu^{2+}(aq)\rightarrow Pb^{2+}(aq)+Cu(s)$

Here lead (Pb) undergoes oxidation by loss of electrons, thus act as anode. Copper (Cu) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^o_{[Pb^{2+}/Pb]}=-0.13V$

$E^o_{[Cu^{2+}/Cu]}=+0.34V$

$E^o=E^o_{[Cu^{2+}/Cu]}-E^o_{[Pb^{2+}/Pb]}$

$E^o=0.34V-(-0.13V)=0.47V$

Now we have to calculate the concentration of $Pb^{2+}$ .

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Pb^{2+}]}{[Cu^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = 0.507 V

Now put all the given values in the above equation, we get:

$0.507=0.47-\frac{0.0592}{2}\log \frac{[Pb^{2+}]}{(0.6)}$

$[Pb^{2+}]=0.0337M$

Therefore, the concentration of $Pb^{2+}$ is, 0.0337 M

<u>(b) As per question, lead is reduced and copper is oxidized.</u>

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Cu\rightarrow Cu^{2+}+2e^-$

Reduction half reaction: $Pb^{2+}+2e^-\rightarrow Pb$

The balanced cell reaction will be,

$Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)$

Here Copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^o_{[Pb^{2+}/Pb]}=-0.13V$

$E^o_{[Cu^{2+}/Cu]}=+0.34V$

$E^o=E^o_{[Pb^{2+}/Pb]}-E^o_{[Cu^{2+}/Cu]}$

$E^o=-0.13V-(0.34V)=-0.47V$

Now we have to calculate the concentration of $Pb^{2+}$ .

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}]}{[Pb^{2+}]}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = 0.507 V

Now put all the given values in the above equation, we get:

$0.507=-0.47-\frac{0.0592}{2}\log \frac{(0.6)}{[Pb^{2+}]}$

$[Pb^{2+}]=6.093\times 10^{32}M$

Therefore, the concentration of $Pb^{2+}$ is, $6.093\times 10^{32}M$

6 0

3 years ago

Iron(II) chloride is formed from the reaction between iron and copper(II) chloride. Fe + CuCl2 FeCl2 + Cu If the reactants have

elena55 [62]

Answer: see figure attached and explanation below.

Explanation:

1) Chemical equation (given):

Fe + CuCl₂ → Cu + FeCl₂

2) ΔHf reactants: -256 kJ/mol (given)

3) ΔHf products: - 321 kJ/mol (given)

4) ΔH reaction = ΔHf products - ΔHf reactants = - 321 kJ/mol - (- 256 kJ/mol) = - 65 kJ/mol

5) Conclusion:

i) Since ΔHf of products is less (more negative) than ΔHf of reactants, the reaction is exhotermic: the reaction released energy, which is the reason why the products content less potential energy than the reactants.

ii) Then, the energy diagram is the typical one of an exothermic reaction: the products start a certain potential energy level, the energy incrases until reaching the activation energy (the energy barrier to form the activated complex) and then energy decreases until a level below the energy of the reactants.

iii) See the attached figure with such kind of diagram showing the products at a lower level than the reactans

3 0

3 years ago

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