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3 years ago

Cc28417 avatar

Chemistry

1 answer:

Tanya [424]3 years ago

8 0

Answer:

Explanation:

i don't know the answer so if you don't mind can you help me ?

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As liquid water boils and changes into water vapor (steam), what happens

Lunna [17]

Answer:

D.The space between the water particles becomes much larger.

Explanation:

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3 years ago

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What group represents the alkaline, alkali, halogens, noble gases, and transition metals elements?

motikmotik

Alkaline metal-group 1
Alkaline earth metals-group 2
halogens=group 7
noble gases-group 8 or o
transition metals are the actinides and the lactanides below the periodic table

4 0

3 years ago

Cân bằng phương trình Mg+HCl->MgCl2+H2 bằng phương pháp thăng bằng

nignag [31]

quá trình oxi hóa: Mg0 -> Mg2+ + 2e-

quá trình khử: 2H+ + 2e- -> H2 (cần 2 H+ để tạo 1 H2)

----> chỉ cần thêm 2 vào trước HCl (vì H+ là do HCl cung cấp)

----> Mg + 2HCl -> MgCl2 + H2

yay :)

8 0

2 years ago

The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,

Masja [62]

Answer: The concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

Explanation:

The given cell is:

$Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)$

Half reactions for the given cell follows:

Oxidation half reaction: $H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V$ ( × 3)

Reduction half reaction: $Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V$ ( × 2)

Net reaction: $3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=1.50-0=1.50V$

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = 1.23 V

$E^o_{cell}$ = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

$[Au^{3+}]=?M$

$[H^{+}]=1.0M$

Putting values in above equation, we get:

$1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})$

$[Au^{3+}]=1.87\times 10^{-14}M$

Hence, the concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

7 0

3 years ago

What mass is in 5 moles of helium?

ArbitrLikvidat [17]

Answer:

Mass = 20 g

Explanation:

Given data:

Number of moles of He = 5 mol

Mass of He = ?

Solution:

Formula:

Number of moles = mass/ molar mass

Molar mass = 4 g/mol

by putting values,

5 mol = Mass / 4 g/mol

Mass = 5 mol × 4 g/mol

Mass = 20 g

3 0

3 years ago