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sashaice [31]
3 years ago
15

What is the name of this formula

Chemistry
1 answer:
WARRIOR [948]3 years ago
3 0

Lithium super oxide, Lithium super oxide is an inorganic compound which has only been isolated in matrix isolation experiments at 15-40 K. It is an unstable free radical that has been analyzed using infrared, Roman, electronic, electron spin resonance, soft X-ray spectroscopes, and a variety of theoretical methods

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Plz help with this it is the digestive system
Strike441 [17]

Answer:

see explanation for answer

Explanation:

salivary gland:  1

stomach: 2

small intestine: 6

liver: 4

gallbladder: 5

large intestine: 3

The answers correspond with the numbers on the text boxes, so you would drag number 1 to the salivary gland and so on.

5 0
3 years ago
Read 2 more answers
Write the complete balanced equation for the reaction between hydrochloric acid and magnesium hydroxide
Mashutka [201]
<span>M<span><span>g<span>(s)</span></span>+</span>2HC<span>l<span>(aq)</span></span>→MgC<span>l<span>2<span>(aq)</span></span></span>+<span>H<span>2<span>(g)</span></span></span></span> Explanation:

The reaction between magnesium and hydrochloric acid combine to form a salt of magnesium chloride and release hydrogen gas. This single replacement reaction is a classic example of a metal reacting in an acid to release hydrogen gas.


6 0
4 years ago
What are three elements that have similar chemical properties to oxygen
jok3333 [9.3K]
Sulfur, selenium, and tellurium
4 0
3 years ago
How many grams of silver bromide are produced when 505 grams of cobalt (III) bromide reacts completely in the following equation
DedPeter [7]
Mass = 473.2 g
Explanation:
Given data:
Mass of cobalt(III) nitrate = 206 g
Mass of silver bromide produced = ?
Solution:
Chemical equation:
CoBr₃ + 3AgNO₃ → 3AgBr + Co(NO₃)₃
Number of moles of cobalt(III) nitrate:
Number of moles = mass/ molar mass
Number of moles = 206 g/ 245 g/mol
Number of moles = 0.84 mol
Now we will compare the moles of cobalt(III) nitrate with silver bromide.
7 0
3 years ago
What is the molarity of 15.0 g of NaBR in 0.250 L of solution
Tom [10]

n = m/M = 15/102.894 = 0.1457811 moles

c = n/V = n / 0.25 = 0.58312438 moles/dm3

6 0
4 years ago
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