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Schach [20]
3 years ago
6

Oxygen and tin 4 chloride react form chlorine and tin 3 oxide

Chemistry
1 answer:
Deffense [45]3 years ago
4 0
Your question wasn’t very clear, but that isn’t right.
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Which of the following is the correct sequence for the discovery of subatomic particles? a Electrons> Protons > Neutrons b
laila [671]

Answer:

C

Explanation;

Maybe is C

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What is the oxidizing agent in the following equation? Al (s) + 3 Ag+ (aq) produces Al^+3 (aq) + 3 Ag (s)
Keith_Richards [23]

Answer:

Ag is the oxidizing agent

Explanation:

oxidizing agent in the following equation?

Al (s) + 3 Ag+ (aq) = Al+3 (aq) + 3 Ag (s)

Left side

Al = 1

Ag = 3

Right Side

Al = 1

Ag = 3

So it's balanced already good.

Define

oxidizing agent = An oxidizing agent is the substance that gains electrons and is reduced in a chemical reaction.

Al is the reducing agent.

Ag is the oxidizing agent

7 0
3 years ago
Recently, the United States government decided to construct a huge crypt in the middle of Yucca Mountain in Nevada to bury high-
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Structural formula for 4-nonene and fluorine gas
goldenfox [79]
I THINK it's <span>1,1-Difluorononane, or </span>C_9H_{18}F_2.<span>
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6 0
3 years ago
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Question 3 0
Tanya [424]

Answer:

Option D. KBr < KCl < NaCl

Explanation:

We'll begin by calculating the number of mole of each sample.

This can be obtained as follow:

For NaCl:

Mass = 1 g

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mole of NaCl =?

Mole = mass /Molar mass

Mole of NaCl = 1/58.5

Mole of NaCl = 0.0171 mole

For Kbr:

Mass = 1 g

Molar mass of KBr = 39 + 80 = 119 g/mol

Mole of KBr =?

Mole = mass /Molar mass

Mole of KBr = 1/119

Mole of KBr = 0.0084 mole

For KCl:

Mass = 1 g

Molar mass of KCl = 39 + 35.5 = 74.5 g/mol

Mole of KCl =?

Mole = mass /Molar mass

Mole of KCl = 1/74.5

Mole of KCl = 0.0134 mole

Summary

Sample >>>>>>>> Number of mole

NaCl >>>>>>>>>> 0.0171

KBr >>>>>>>>>>> 0.0084

KCl >>>>>>>>>>> 0.0134

Arranging the number of mole of the sampl in increasing order, we have:

KBr < KCl < NaCl

5 0
3 years ago
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