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nadya68 [22]
4 years ago
4

During the rGFP purification experiment, the instructor will have to make breaking buffer for the students to use. This buffer c

ontains 150mM NaCl. Given a bottle of crystalline NaCl (M.W. = 40g/mole), describe how you would make 500ml of 150mM NaCl. Include the type and sizes of any measuring vessels that you use to make the solution.
Chemistry
1 answer:
Leya [2.2K]4 years ago
6 0

Answer:

dealing with a buffer

use the formula C = n/V...........1

therefore say n = m/Mr.........2

then substitute equation 2 into equation 1

∴ CV = m/Mr

then m = CVMr

Since we are looking for mass of  Nacl to make a buffer solution.

m = 150×10^{-3} ml^{-1} × 500×10^{-3} l × 40 g/mol

mass of Nacl = 3g

Explanation: You first weight out 3g of 150mM of NaCl and then dissolve it into a 1000ml beaker dilute it using a  volume of 500ml of water to make volume.

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4 years ago
Which of the following equations does not demonstrate the law of conservation of mass?
enot [183]

The third option does not obey the law of conservation of mass.

Option 3.

Explanation:

The law of conservation of mass states that the sum of the masses of reactants should be equal to the sum of the masses of the products.

For example, if we consider the first option to verify if it obeys law of conservation of mass or not, 2 Na + Cl₂ → 2 NaCl

So one way to verify it is to find the mass of Na, then multiply it with 2, and then add this with 2 times of mass of chlorine. So this sum should be equal to the 2 times mass of NaCl. But it is somewhat lengthy.

Another way to easily determine this is to check if the elements are present equally in both sides. Such as, in reactant side and product side 2 atoms of Na is present . Similarly, the Cl atoms are also present in equal number in both reactant and product side. Thus this obeyed the law of conservation of mass.

Like this, if we see the second option, there also 1 atom of Na is present in reactant and product side and 2 molecules of H is present in reactant and product side, 1 oxygen is present in reactant and product side and 1 Cl is present in reactant and product side. So it also obeys the law of conservation of mass.

But in the third option, P₄ + 5 O₂→ 2 P₄O₁₀, here, there is 4 atoms of P in reactant side but in product side there is (4*2) = 8 atoms of P. Similarly, the number of atoms of oxygen in reactants and product side is also not same. So the third option does not obey the law of conservation of mass.

The fourth option also obeys the law of conservation of mass as the number of atoms of each element is same in both the product and reactant side.

Thus, the third option does not obey the law of conservation of mass.

5 0
4 years ago
Pls help me with this question
Drupady [299]
The answer is c hopefully I helped you
4 0
3 years ago
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