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olasank [31]
2 years ago
13

Write anode and cathode in Zn-Ag galvanic cell​

Chemistry
1 answer:
melomori [17]2 years ago
8 0

Explanation:

Zinc is the anode (solid zinc is oxidised). Silver is the cathode (silver ions are reduced).

By convention in standard cell notation, the anode is written on the left and the cathode is written on the right. So, in this cell: Zinc is the anode (solid zinc is oxidised). Silver is the cathode (silver ions are reduced).

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If 23.5 g of ammonia (NH3) is dissolved in 1.0 L of solution, what is the molarity?
r-ruslan [8.4K]

Answer:

1.38 M

Explanation:

Need to use the Molarity equation M=n/L

23.5g/ 17.031g/mol NH3 = 1.38 moles

1.38 moles/ 1.0 L = 1.38 M

4 0
2 years ago
Darin skateboarded to the north at 4 m/s. In 5 seconds, he sped up to 7 m/s. What was his acceleration?
Julli [10]
To solve this problem we can use following equation.

v =u + at

Where v is the final velocity (m/s), u is the initial velocity (m/s), a is the acceleration (m/s²) and t is the time taken (s).

v = 7 m/s
u = 4 m/s
a = ?
t = 5 s

By applying the equation, we can get
    7 m/s = 4 m/s + a x 5 s
    3 m/s = a x 5 s
          a = 0.6 m/s²

Hence, the acceleration is 0.6 m/s² towards north.

Answer is "C".
4 0
3 years ago
What does RNA do?
MariettaO [177]

Answer:

b, its a polymer that translates genetic information.

7 0
3 years ago
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How was the periodic table arranged
Andrews [41]
The elements on the periodic table are arranged in order, by increasing atomic number.<span />
5 0
3 years ago
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In an experiment, 3.25 g of NH3 are allowed to react with 3.50 g of O2. 4NH3 + 5O2 &gt; 4NO + 6H2O a. Which reactant is the limi
cluponka [151]

Answer:

2.61 g of NO will be formed

The limiting reagent is the O₂

Explanation:

The reaction is:

4NH₃  +  5O₂  →  4NO  +  6H₂O

We convert the mass of the reactants to moles:

3.25g / 17 g/mol = 0.191 moles of NH₃

3.50g / 32 g/mol =0.109 moles of O₂

Let's determine the limiting reactant by stoichiometry:

4 moles of ammonia react with 5 moles of oxygen

Then, 0.191 moles of ammonia will react with (0.191 . 5) / 4 = 0.238 moles of oxygen. We only have 0.109 moles of O₂ and we need 0.238, so as the oxygen is not enough, this is the limiting reagent

Ratio with NO is 5:4

5 moles of oxygen produce 4 moles of NO

0.109 moles will produce (0.109 . 4)/ 5 = 0.0872 moles of NO

We convert the moles to mass, to get the answer

0.0872 mol . 30g / 1 mol = 2.61 g

7 0
2 years ago
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