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madreJ [45]
3 years ago
13

A contains 2.60g of impure h2so4 in 5oocm³ of a solution 5.30g of na2co3 in 1.00dm³ of solution

Chemistry
1 answer:
Tatiana [17]3 years ago
3 0

Answer: a. 0.05mol/dm3

bi. 0.045mol/dm3

bii. 4.41g/dm3

c. 84.8%

Explanation:Please see attachment for explanation

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You need to list the elements but remember that fluorine has the highest electronegativity out of the entire periodic table
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A sample of hydrated magnesium sulfate (MgSO4)
yaroslaw [1]

Answer:

MgSO4.7H2O

Explanation:

Let the formula for the hydrated magnesium sulphate be MgSO4.xH2O

Mass of the hydrated salt (MgSO4.xH2O) = 12.845g

Mass of anhydrous salt (MgSO4) = 6.273g

Mass of water molecule(xH2O) = Mass of the hydrated salt — Mass of anhydrous salt = 12.845 — 6.273 = 6.572g

Now,we can obtain the number of mole of water molecule present in the hydrated salt as follows:

Molar Mass of hydrated salt (MgSO4.xH2O) = 24 + 32 + (16x4) + x(2 + 16) = 24 + 32 + 64 + x(18) = 120 + 18x

Mass of xH2O/ Molar Mass of MgSO4.xH2O = Mass of water / mass of hydrated salt

18x/120 + 18x = 6.572/12.845

Cross multiply to express in linear form

18x x 12.845 = 6.572(120 + 18x)

231.21x = 788.64 + 118.296x

Collect like terms

231.21x — 118.296x = 788.64

112.914x = 788.64

Divide both side by 112.914

x = 788.64 /112.914

x = 7

Therefore the formula for the hydrated salt (MgSO4.xH2O) is MgSO4.7H2O

6 0
3 years ago
Which is an example of a solution?
Rudiy27
The answer is C. Salt and water is a solution 
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Answer:

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The Henderson-Hasselbalch is used to calculate the pH of a buffer solution. It depends on the weak acid approximation.

Since the weak acid ionizes only to a small extent, then we can say that [HA] ≈ [HA]i

Where [HA] = final concentration of the acid and [HA]i = initial concentration of the acid.

It also follows that [A^-] ≈ [A^-]i where [A^-] and[A^-]i refer to final and initial concentrations of the conjugate base hence the answer above.

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If a liquid has a Ph of 3 what is it​
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Lower than 7 is acid greater than 7 is a base
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