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IRINA_888 [86]
3 years ago
8

n the laboratory, two forms of sodium phosphate will be available (the monobasic monohydrate NaH2PO4·H2O, F.W. = 137.99 g/mol, a

nd the dibasic Na2HPO4, F.W. = 141.96 g/mol). Explain the most efficient way of making this phosphate buffer by using only one of the compounds, and specify which compound you will use.
Chemistry
1 answer:
const2013 [10]3 years ago
6 0

Answer:

The compound you will use is the Dibasic phosphate

Explanation:

Simple phosphate buffer is used ubiquitously in biological experiments, as it can be adapted to a variety of pH levels, including isotonic. This wide range is due to phosphoric acid having 3 dissociation constants, (known in chemistry as a triprotic acid) allowing for formulation of buffers near each of the pH levels of 2.15, 6.86, or 12.32. Phosphate buffer is highly water soluble and has a high buffering capacity,

In this case the most efficient way is to disolve the dibasic compound which in the reaction with the water will form the monobasic phosphate.

To make the buffer you have to prepare the amount of distillate water needed, disolve the dibasic phospate, and then adjust with HCl or NaOH depending on the pH needed.

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Relative formula mass of glucose? (C6H12O6)
Rom4ik [11]

To find the mass of glucose, you must multiply the atomic weight of each of the elements in the molecule by the subscripts in the formula:

C_{6}=6*(12.01g)=72.06g

H_{12}=12*(1.008g)=12.096g

O_{6}=6*(15.999g)=95.994g

Then you add all of them together:

72.06g+12.096g+95.994g=180.15g

Therefore, the molar weight of glucose is 180.15 grams.

3 0
3 years ago
Write the empirical formula of at least four binary ionic compounds that could be formed from the following ions: Zn2+, Ni4+, F-
hram777 [196]

Ionic compounds are formed between oppositely charged ions.

A binary ionic compound is composed of ions of two different elements - one of which is a positive ion(metal), and the other is negative ion (nonmetal).

To write the empirical formula of binary ionic compound we must remember that one ion should be positive and other ion should be negative, then only the correct formula should be written. To write the empirical formula the charges of opposite ions should be criss-crossed.

First empirical formula of binary ionic compound is written betweenZn^{2+} (Positive ion)and F^{-} (Negative ion)

First Formula would be ZnF_{2}

Second empirical formula is between Zn^{2+}(Positive ion) and O^{2-}(Negative ion)

Second Formula would be Zn_{2}O_{2}

Note : When the subscript are same they get cancel out, so Zn_{2}O_{2} would be written as ZnO

Third empirical formula is between Ni^{4+}(Positive ion) and F^{-}(Negative ion)

Third Formula would be :NiF_{4}

Forth empirical formula is between Ni^{4+}(Positive ion)and O^{2-}(negative ion)

Forth Formula would be : Ni_{2}O_{4} or NiO_{2}

Note- The subscript will be simplified and the formula will be written as NiO_{2}.

The empirical formula of four binary ionic compounds are : ZnF_{2}, ZnO, NiF_{4},NiO_{2}


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Answer:
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Explanation:
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