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telo118 [61]
2 years ago
13

How many molecules are in a 6.82 gram sample of Hydrogen Sulfide (H2S)

Chemistry
1 answer:
Alinara [238K]2 years ago
5 0

Answer:

1.204×10^(23)

Explanation:

Molecular weight of H2S,  M = 1*2+32 = 34

weight of sample, m = 6.82 gram

now number of moles, n= m/M

n= 6.82/34 = 0.200 moles

1 mole has 6.023×10^(23) molecules of H2S.

therefore 0.2 moles of H2S will have  0.2×6.023×10^(23) = 1.204×10^(23) number of H2S molecules.

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ehidna [41]
<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2>----------------------------------------------------------------------</h2><h2>ELECTROPLATING </h2>

Electroplating the plating one metal on to the another metal, It is mostly used for preventing corrosion by using copper or chromium or decorate the object by using gold or silver plating.

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<h3>Principle:</h3>

When electricity is passed the thin layer of metal is deposited on another metal and water molecule given out as a By-Product, Thus this process works on the principle of Hydrolysis.

<h2>_____________________________________</h2><h2>QUESTION:</h2>

A) Electrolysis

B) Chromium prevents corrosion and gives the fine shining touch to the objects.

C) The metal which is deposited to the object i.e. spoon will be connected to the positive electrode of a battery, Thus it is anode. The spoon at which electroplating is need to be done is connected to the negative electrode, thus the Spoon is cathode.

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5 0
3 years ago
If you produced 76.10 grams of potassium chloride in the reaction how many grams of potassium were required?
11111nata11111 [884]

Answer:39.8375

Explanation:

The mole for the equation is 1:1

Then the molar mass of KCl is 74.5g

Molar mass of k is 39

74.5g of KCl gives 39g of k

76.10g of KCl gives xg of k

X= 76.10×39/74.5

X= 2967.9/7

X= 39.8375

6 0
3 years ago
What functional group is aspartic acid
Westkost [7]
The functional group of aspartic acid is -COOH.
8 0
3 years ago
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 help me!!!!!
kvasek [131]

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2 years ago
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