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lara31 [8.8K]
3 years ago
15

Help!! What is the concentration of an unknown with an absorbance of 0.4 using the plot?

Chemistry
1 answer:
sveta [45]3 years ago
6 0

Answer:3,4,1,2

Explanation:

hope this helped

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What particles are affecting by strong nuclear forces?
WINSTONCH [101]

Answer:

proton and neutron

I hope this helps :)

Explanation:

7 0
3 years ago
The mass of a proton is approximately equal to the mass of…
Misha Larkins [42]

Answer:

1.67262 × 10−27 kg,

Explanation:

I think its right 0_0

5 0
2 years ago
What mass of sodium hydroxide will completely neutralize 2.5 mol of sulfuric acid?
VMariaS [17]

Given :

2.5 mole of Sulfuric acid ( H_2SO_4 ) .

To Find :

Mass of sodium hydroxide will completely neutralize 2.5 mol of sulfuric acid

Solution :

Let us assume volume of water be 1 L .

Now , we know , to neutralize 1 mole of sulfuric acid we need 2 moles of NaOH .

So , for 2.5 mole sulfuric acid required 5 mole of NaOH .

Moles of NaOH ,

n=M\times Volume \\\\n=5\times 1=5\ moles  

Molecular mass of NaOH , M.M = 58.44 g/mol .

Mass of 5 moles of NaOH :

m=5\times 58.44\ g\\\\m=292.2\ g

Hence , this is the required solution .

7 0
3 years ago
Copper is commonly mined as an ore with a variable percent composition of copper (II) sulfide. This ore is also sometimes referr
pantera1 [17]

Answer:

m_{CuO}=93.6gCuO

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

CuS+\frac{3}{2} O_2\rightarrow CuO+SO_2

Thus, given the 1.00-kg of 12.5% ore, we can compute the theoretical yield of copper (II) oxide via stoichiometry:

m_{CuO}^{theoretical}=1.00kgCuS*\frac{1000gCuS}{1kgCuS} *\frac{12.5gCuS}{100gCuS} *\frac{1molCuS}{95.6gCuS} *\frac{1molCuO}{1molCuS} *\frac{79.5gCuO}{1molCuO} \\\\m_{CuO}^{theoretical}=103.95gCuO

Whereas the third factor accounts for the percent purity of the covellite. Then, given the percent yield, we can compute the actual yield by:

m_{CuO}=103.95gCuO*0.9\\\\m_{CuO}=93.6gCuO

Regards.

6 0
3 years ago
How many grams of oxygen are needed to react with 4.6 grams of titanium(IV) chloride
velikii [3]
The answer is: 0.78g
6 0
3 years ago
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