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2 years ago

____________ electricity is a quick burst of energy which is produced by charged particles.

Chemistry

2 answers:

Brut [27]2 years ago

7 0

Static is the answer

sukhopar [10]2 years ago

6 0

It’s static electricity!

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What is the molarity of 122.5 g of AlCl3 in 1.0 L of solution? (MM = 133 g/mol)

yawa3891 [41]

Answer: Molarity is defined as moles of solute per liter of solution. So, find the moles of solute and divide by the liters of solution.
molar mass AlCl3 = 133g/mole
moles AlCl3 = 127 g x 1 mole/133 g = 0.955 moles
liters of solution = 400 ml x 1 liter/1000 ml = 0.400 liters
Molarity = 0.955 moles/0.400 liters = 2.39 M
Explain: I looked it up on wyzant.com

4 0

3 years ago

Name the compound that makes up the white ring

Brut [27]

Answer:

ammonium chloride

Explanation:

if it's the white ring I think your talking abt

3 0

2 years ago

Read 2 more answers

1. Magnesium chloride solution reacts with silver nitrate solution to form magnesium nitrate

Whitepunk [10]

a. 1,4332 g

b. 7.54~g

<h3>Further explanation</h3>

Given

Reaction

MgCl2 (s) + 2 AgNO3 (aq) → Mg(NO3)2 (aq) + 2 AgCl (s)

20 cm of 2.5 mol/dm^3 of MgCl2

20 cm of 2.5 g/dm^3 of MgCl2

Required

the mass of silver chloride - AgCl

Solution

a. mol MgCl2 :

$\tt 20~cm^3=20\times 10^{-3}~dm^3\\\\mol=M\times V\\\\mol=2.5~mol/dm^3\times 20\times 10^{-3}DM^3=0.05$

From equation, mol AgCl = 2 x mol MgCl2=2 x 0.05=0.1

mass AgCl(MW=143,32 g/mol)= 0.1 x 143,32=1,4332 g

b. mol MgCl2 (MW=95.211 /mol):

$\tt mol=M\times V\\\\mol=\dfrac{2.5~g/dm^3}{95,211 g/mol}=0.0263~mol/dm^3$

From equation, mol AgCl = 2 x mol MgCl2=2 x 0.0263=0.0526

mass AgCl(MW=143,32 g/mol)= 0.0526 x 143,32=7.54~g

6 0

2 years ago

For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia

Fittoniya [83]

Answer : The value of $E^o_{(Cu^{2+}/Cu)}$ is, 0.34 V

Explanation :

Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

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

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

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

$Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$