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ValentinkaMS [17]
3 years ago
10

Name the following bases -Ba(OH)2 - Ca(OH)2 -RbOH

Chemistry
2 answers:
Nimfa-mama [501]3 years ago
5 0
~Ba(OH)2 is Barium hydroxide

~Ca(OH)2 is Calcium hydroxide

~RbOH is Rubidium hydroxide

i believe are the correct answers,
Harman [31]3 years ago
3 0

Ba(OH)_2\rightarrow \bold{\green{Barium\: hydroxide}}

Ca(OH)_2\rightarrow \bold{\red{Calcium\: hydroxide}}

RbOH \rightarrow \bold{\blue{Rubidium \:hydroxide}}

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How many grams of mercury would be contained in 15 compact fluorescent light bulbs?
zhenek [66]

Answer:

Grams of mercury= 0.06 g of Hg

Note: The question is incomplete. The complete question is as follows:

A compact fluorescent light bulb contains 4 mg of mercury. How many grams of mercury would be contained in 15 compact fluorescent light bulbs?

Explanation:

Since one fluorescent light bulb contains 4 mg of mercury,

15 such bulbs will contain 15 * 4 mg of mercury = 60 mg

1 mg = 0.001 g

Therefore, 60 mg = 0.001 g * 60 = 0.06 g of mercury.

Compact fluorescent lightbulbs (CFLs) are tubes containing mercury and noble gases. When electricity is passed through the bulb, electron-streams flow from a tungsten-coated coil. They collide with mercury atoms, exciting their electrons and creating flashes of ultraviolet light. A phosphor coating on the inside of the tube absorbs this UV light flashes and re-emits it as visible light. The amount of mercury in a fluorescent lamp varies from 3 to 46 mg, depending on lamp size and age.

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3 years ago
Validate how valence electrons determine the chemical reactivity of an element: give 2 examples of elements with high reactivity
Ostrovityanka [42]
Nitrogen is more reactive than oxygen and oxygen than chlorine
8 0
3 years ago
Which of the following pairs of reactants will react together to produce water (H2O) as one of the products?
Rainbow [258]
1) <span>NaNO3 and H2O - no reaction , it is dissolution
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so Answer number 3)
</span> 2Fe(OH)3 +3 H2SO4 ------>   Fe2(SO4)3 + 6H2O<span>

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7 0
3 years ago
Read 2 more answers
Assuming that the experiments performed in the absence of inhibitors were conducted by adding 5 μl of a 2 mg/ml enzyme stock sol
prohojiy [21]

Hey there!:

From the given data ;

Reaction  volume = 1 mL , enzyme content = 10 ug ( 5 ug in 2 mg/mL )

Enzyme mol Wt = 45,000 , therefore [E]t is 10 ug/mL , this need to be express as "M" So:

[E]t in molar  = g/L * mol/g

[E]t  = 0.01 g/L * 1 / 45,000

[E]t = 2.22*10⁻⁷

Vmax = 0.758 umole/min/ per mL

= 758 mmole/L/min

=758000 mole/L/min => 758000 M

Therefore :

Kcat = Vmax/ [E]t

Kcat = 758000 / 2.2*10⁻⁷ M

Kcat = 3.41441 *10¹² / min

Kcat = 3.41441*10¹² / 60 per sec

Kcat = 5.7*10¹⁰ s⁻¹

Hence   kcat of   xyzase is  5.7*10¹⁰ s⁻¹


Hope that helps!



4 0
3 years ago
Please please please no guessing I had to type this all out. How do oceanographers define saltanily?
Gre4nikov [31]
The correct answer is D. Number of grams of salt per kilometer of water.
5 0
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