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Bas_tet [7]
4 years ago
14

One of these notations represents the definition of one mole. That is

Chemistry
1 answer:
just olya [345]4 years ago
3 0

Answer:

6.02*10^23

Explanation:

This is the number for one mole. Just like one dozen = 12, one mole = 6.02*10^23.

Fun fact, if you had a mole of pennies you could spend 1 million dollars every second of your life and not have even spent 1% of it by the time you die at 100 years old.

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ollegr [7]

Answer:

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Explanation:

3 0
3 years ago
Read 2 more answers
How many atoms of hydrogen are in 0.500 mol of ch3oh molecules?
cestrela7 [59]
In 1 mol of CH3OH, you have 4 H-atoms (because 3 H-atoms are attached to the C-atom, and one H-atom in the OH group). That means in 0.500 mol of CH3OH, you have 2 H-atoms since it is halved. And then we have Avogadro's constant: 6.02 * 1023.

The question asks for how many hydrogen atoms there are in 0.500 mol CH3OH. Using the numbers that we have (Avogadro's constant and no. of H-atoms), the answer of the question will be something like:

<span>H-atoms in CH3OH = 2 * 6.02 * </span>1023<span> = ~1.2 * 10</span>24

 


8 0
3 years ago
What is the volume of the sphere? Round your answer<br> to the nearest tenth.<br> cm
myrzilka [38]

Answer:

volume

v = 4/3π r^3

Explanation:

it isn't specific enough but that is the equation of how to get any volume

volume equals four thirds times pi times radios to the power of three

4 0
4 years ago
What happens when sodium and sulfur combine
german

Answer:

It emits hydrogen sulfide...smells like rotten eggs..

ty:)pls let me know whether this is ryt:D

6 0
3 years ago
Read 2 more answers
Hydrogen iodide, HI, is formed in an equilibrium reaction when gaseous hydrogen and iodine gas are heated together. If 20.0 g of
Kaylis [27]

Answer: D. 19.9 g hydrogen remains.

Explanation:

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

a) moles of H_2

\text{Number of moles}=\frac{20.0g}{2g/mol}=10.0moles

b) moles of I_2

\text{Number of moles}=\frac{20.0g}{254g/mol}=0.0787moles

H_2(g)+I_2(g)\rightarrow 2HI(g)

According to stoichiometry :

1 mole of I_2 require 1 mole of H_2

Thus 0.0787 moles of l_2 require=\frac{1}{1}\times 0.0787=0.0787moles of H_2

Thus l_2 is the limiting reagent as it limits the formation of product and H_2 acts as the excess reagent. (10.0-0.0787)= 9.92 moles of H_2are left unreacted.

Mass of H_2=moles\times {\text {Molar mass}}=9.92moles\times 2.01g/mol=19.9g

Thus 19.9 g of H_2 remains unreacted.

5 0
3 years ago
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