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Aliun [14]
3 years ago
15

Why is a standard system of measurement important?

Chemistry
2 answers:
GrogVix [38]3 years ago
8 0
I believe the correct response is A.
Angelina_Jolie [31]3 years ago
7 0
So the question ask to determine which of the following is the reason why is a standard system of measurement is important and the best answer would be letter A. Scientist want to share measurement data that they can understand. I hope you are satisfied with my answer and feel free to ask for more .

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Explain the difference between mechanical weathering and chemical weathering​
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Discribe how to prepare 50.00mL of a 3.00M hydrochloric acid solution using 12.0 M HCl.
Leokris [45]

Answer:

The answer to your question is given below.

Explanation:

To prepare 50mL of 3M HCl, we must calculate the volume of the stock solution needed. This can obtained as follow:

Molarity of stock solution (M1) = 12M

Volume of stock solution needed (V1) =?

Molarity of diluted solution (M2) = 3M

Volume of diluted solution (V2) = 50mL

The volume of the stock solution needed can be obtained by using the dilution formula as shown below:

M1V1 = M2V2

12 x V1 = 3 x 50

Divide both side by 12

V1 = (3 x 50)/12

V1 = 12.5mL

The volume of the stock solution needed is 12.5mL

Therefore, to prepare 50mL of 3M HCl, we must measure 12.5mL of the stock solution i.e 12M HCl and then, add water to the mark in a 1L volumetric flask. Now we can measure out 50mL of the solution.

6 0
3 years ago
2KMnO4= K2MnO4+ MnO2+O2 how many grams of KMnO4 are required to produce 1.60 grams of O2
Sergeu [11.5K]

Answer: 15.8 g of KMnO_4 will be required to produce 1.60 grams of O_2

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} O_2=\frac{1.60g}{32g/mol}=0.05moles

2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2

According to stoichiometry :

As 1 mole of O_2 is given by = 2 moles of KMnO_4

Thus 0.05 moles of O_2 is given by =\frac{2}{1}\times 0.05=0.10moles  of KMnO_4

Mass of KMnO_4=moles\times {\text {Molar mass}}=0.10moles\times 158g/mol=15.8g

Thus 15.8 g of KMnO_4 will be required to produce 1.60 grams of O_2

5 0
3 years ago
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