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RSB [31]
2 years ago
11

Which of the following has a mass of 10.0 g?

Chemistry
1 answer:
Serjik [45]2 years ago
5 0

Answer:

a.  0.119mol Kr

Explanation:

To solve this problem, we must understand that;

     Mass  = number of moles x molar mass

Molar mass of Kr  = 83.3g/mol

                        Ar = 40g/mol

                        He = 4g/mol

                        Ne = 20.18g/mol

a0.119 mol Kr               mass  = 0.119 x 83.3 = 9.9g

b 0.400 mol Ar             mass  = 0.4 x 40  = 16g

C 1.25 mol He               mass  = 1.25 x 4  = 5g

d 2.02 mol Ne              mass  = 2.02 x 20.18  = 40.8

Krypton is the answer

You might be interested in
4.2g of cerium reacted with oxygen to form 5.16g of an oxide of cerium. Find
olganol [36]

Answer:

CeO₂

Explanation:

Hello!

In this case, since we are given the mass of both cerium and the cerium oxide, we can first compute the moles of cerium and the moles of oxygen as shown below:

n_{Ce}=4.2gCe*\frac{1molCe}{140.12gCe}=0.03molCe\\

m_O=5.16g-4.2g=0.96gO\\\\n_O=0.96g*\frac{1molO}{16.0gO} =0.06molO

Now, we simply divide each moles by 0.03 as the fewest moles in the formula to obtain the simplest formula (empirical formula) of this oxide:

Ce=\frac{0.03}{0.03}=1\\\\O =\frac{0.06}{0.03}=2

Thus, the formula turns out:

CeO_2

Regards!

6 0
2 years ago
Sorry the last one wasn’t 99 points but this one is 14 points sorry! But 14 points is a lot right? so please be prepared! Hshsuu
vaieri [72.5K]

Answer:

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5 0
3 years ago
Read 2 more answers
Rank the organic compounds from most soluble to least soluble. To rank items as equivalent, overlap them.
denpristay [2]
The answer to your questions is as follows:

most soluble 
>CH3CH2OH 
>CH3OCH3 
>CH3CH3 
<span>least soluble
</span>
I hope my answer has come to your help. God bless and have a nice day ahead!
7 0
2 years ago
Equation 3x-5=-2x+ 10?<br> O<br> 0<br> x= 5<br> -5 = x<br> -15 = -5x<br> -5x = 15
Harman [31]

Answer:

are those the answer choice

7 0
3 years ago
Answer these please ASAP need help no idea how to do these
STALIN [3.7K]

Answer:

Explanation:

Cu:

Number of moles = Mass / molar masa

2 mol = mass / 64 g/mol

Mass = 128 g

Mg:

Number of moles = Mass / molar masa

0.5 mol = mass / 24 g/mol

Mass =  g

Cl₂:

Number of moles = Mass / molar masa

Number of moles  = 35.5 g / 24 g/mol

Number of moles = 852 mol

H₂:

Number of moles = Mass / molar mass

8 mol  = Mass / 2 g/mol

Mass =  16 g

P₄:

Number of moles = Mass / molar masa

2 mol  =  mass / 124 g/mol

Mass = 248 g

O₃:

Number of moles = Mass / molar masa

Number of moles  = 1.6 g /48  g/mol

Number of moles = 0.033 mol

H₂O

Number of moles = Mass / molar masa

Number of moles  = 54 g / 18 g/mol

Number of moles = 3 mol

CO₂

Number of moles = Mass / molar masa

2 mol  =  mass / 124 g/mol

Mass = 248 g

NH₃

Number of moles = Mass / molar masa

Number of moles  = 8.5 g / 17 g/mol

Number of moles = 0.5 mol

CaCO₃

Number of moles = Mass / molar masa

Number of moles  = 100 g / 100 g/mol

Number of moles = 1 mol

a)

Given data:

Mass of iron(III)oxide needed = ?

Mass of iron produced = 100 g

Solution:

Chemical equation:

F₂O₃ + 3CO    →    2Fe  + 3CO₂

Number of moles of iron:

Number of moles = mass/ molar mass

Number of moles = 100 g/ 56 g/mol

Number of moles = 1.78 mol

Now we compare the moles of iron with iron oxide.

                        Fe          :           F₂O₃                

                           2          :             1

                          1.78       :        1/2×1.78 = 0.89 mol

Mass of  F₂O₃:

Mass = number of moles × molar mass

Mass = 0.89 mol × 159.69 g/mol

Mass = 142.124 g

100 g of iron is 1.78 moles of Fe, so 0.89 moles of F₂O₃ are needed, or 142.124 g of iron(III) oxide.

b)

Given data:

Number of moles of Al = 0.05 mol

Mass of iodine = 26 g

Limiting reactant = ?

Solution:

Chemical equation:

2Al + 3I₂   →  2AlI₃

Number of moles of iodine = 26 g/ 254 g/mol

Number of moles of iodine = 0.1 mol

Now we will compare the moles of Al and I₂ with AlI₃.

                          Al            :         AlI₃    

                          2             :           2

                         0.05         :        0.05

                           I₂            :         AlI₃

                           3            :          2

                         0.1           :           2/3×0.1 = 0.067

Number of moles of AlI₃ produced by Al are less so it will limiting reactant.

Mass of AlI₃:                            

Mass = number of moles × molar mass

Mass = 0.05 mol × 408 g/mol

Mass = 20.4 g

26 g of iodine is 0.1 moles. From the equation, this will react with 2 moles of Al. So the limiting reactant is Al.

c)

Given data:

Mass of lead = 6.21 g

Mass of lead oxide = 6.85 g

Equation of reaction = ?

Solution:

Chemical equation:

2Pb + O₂   → 2PbO

Number of moles of lead = mass / molar mass

Number of moles = 6.21 g/ 207 g/mol

Number of moles = 0.03 mol

Number of moles of lead oxide = mass / molar mass

Number of moles = 6.85 g/ 223 g/mol

Number of moles = 0.031 mol

Now we will compare the moles of oxygen with lead and lead oxide.

               Pb         :        O₂

                2          :         1

               0.03     :      1/2×0.03 = 0.015 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 0.015 mol × 32 g/mol

Mass =  0.48 g

The mass of oxygen that took part in equation was 0.48 g. which is 0.015 moles of oxygen. The number of moles of Pb in 6.21 g of lead is 0.03 moles. So the balance equation is

2Pb + O₂   → 2PbO

   

6 0
2 years ago
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