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Amiraneli [1.4K]
3 years ago
10

Faraday's law calculates how many amperes are generated. True or false?

Chemistry
2 answers:
Bezzdna [24]3 years ago
7 0

The answer would be False

Assoli18 [71]3 years ago
5 0
Hmm well thinking of how amperes are generated it should be true
You might be interested in
List 5 spheres and give an example of something you would find in each sphere . Ill give 50 points
Brilliant_brown [7]
Hydrosphere, atmosphere, geosphere, biosphere, anthrosphere.

Hydrosphere: Liquid, vapour or ice.

Atmosphere: Oxygen.

Geosphere: Minerals.

Biosphere: Ecosystem.

Anthrosphere: Human habitats
3 0
2 years ago
Read 2 more answers
The ratio of sizes between the ionic radii of anions and cations in a cell has no influence on the manner of packing for that ce
Morgarella [4.7K]

Answer:

The answer is B. False

Explanation:

The ratio of sizes between the ionic radii of cations and anions in a cell influences the manner of packing for that cell thereby predicting the possible cation/anion coordination number in any compound and establishing the structure of ionic solids.

5 0
3 years ago
How many milliliters of 1.50 M magnesium sulfate soulution is required to supply 2.50 mole of this salt?
arsen [322]

Answer:

1670 ml

Explanation:

molarity x Volume (Liters) = moles => Volume (Liters) = moles/Molarity

Volume needed = 2.50mol/1.50M = 1.67 Liters = 1670 ml.

3 0
3 years ago
A sodium ion has a radius of 1.16 x 10^-10 m and a nearby fluoride ion has a radius of 1.9 x 10^-10 m. Determine the distance be
Ann [662]

The answer is: the distance between two nuclei is 2.35×10⁻¹⁰ m.

r(Na⁺) = 1.16×10⁻¹⁰ m; radius of sodium cation.

r(F⁻) = 1.9×10⁻¹⁰ m; radius of fluoride anion.

d(NaF) = r(Na⁺) + r(F⁻).

d(NaF) = 1.16×10⁻¹⁰ m + 1.9×10⁻¹⁰ m.

d(NaF) = 2.35×10⁻¹⁰ m; distance between two nuclei.

The sum of ionic radii of the cation and anion gives the distance between the ions in a crystal lattice.

4 0
3 years ago
You are given a sulfuric acid solution of unknown concentration. You dispense 10.00 mL of the unknown solution into an Erlenmeye
dmitriy555 [2]

Answer:

"0.053457 M" of sulfuric acid.

Explanation:

The given values are:

V = 10 mL solution

V_{added} = 12.20 mL

V_{total} = 22.20 mL

then,

M 0.103 M of NaOH,

V_{rinsed} = experiment  will not be affected

V_{total \ base} = 10.38 mL

Now,

⇒  mol of NAOH = MV

                            = 0.103\times 10.38

                            =  1.06914  \ m

Whether Sulfuric acid, then

⇒  H_{2}SO_{4} + 2NaOH = Na_{2}SO_{4} + 2H_{2}O

⇒  mol \ of \ acid =\frac{1}{2}\times \ mol \   of  \ base

⇒  1.06914 \ m \ mol \ of \ base = \frac{1}{2}\times 1.06914 = 0.53457 \ m \ mol \ of \ acid

Before any dilution:

V_{sample} = 10  \ mL

⇒  M \ acid = \frac{m \ mol}{V}

                 =\frac{ 0.53457 }{10}

                 =0.053457 \ M (Sulfuric acid)

6 0
3 years ago
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