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3 years ago

How much of a 15.0M stock solution do you need to prepare 250ml of a 2.35ml hf solution?

1 answer:

3 0

Using the relationship M1V1 = M2V2 where M1 and M2 are the molar concentrations (mol/L or mmol/ml) and V1 and V2 are the volumes of the solutions, we can arrive at the following answer for the given problem:

<span>15.0M (L of stock solution) = 2.35M (0.25L) *all volumes were converted to liters.

L of stock solution = (2.35*0.25)/15.0

Therefore, 0.0392L or 39.17 ml of stock solution is needed. </span>

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When an unknown substance is dissolved in water, hydronium ions form. What can you conclude about the substance?

Gala2k [10]

D.) The substance is an acid

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3 years ago

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PLEASE ANSWER...........WILL MARK BRAINLIEST!!!!!!!

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Explanation:

Alkali metal cations reacts vigorously with halogens to form ionic/electrovalent compounds.

The alkali metals are about the most reactive metals and they belong to group I .
In their outermost shell, they have one valence shell electron.
Alkali metals are largely electropositive willing to release their valence electron to attain stability.
The halogens have seven electrons in their outermost shell. To complete their configuration, they need just one electron.
All the halogens are strongly electronegative and they are readily reactive.
In the vicinity of group 1 elements, they react vigorously.
This is due to the electrostatic attraction between the metal and non-metal ion formed.
Therefore, they form strong ionic bonds between their compounds.

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4 years ago

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3 years ago

How many grams of testosterone, C19H28O2 (288.4 g/mol), must be dissolved in 299.0 grams of benzene to reduce the freezing point

iVinArrow [24]

Answer: 8.42 grams of testosterone

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=0.500^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

$K_f$ = freezing point constant of benzene= $5.12^0C/m$

m= molality

$\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}$

Weight of solvent (benzene)= 299.0 g = 0.299 kg

Molar mass of solute testosterone= 288.4 g/mol

Mass of solute testosterone added = ?

$0.500=1\times 5.12\times \frac{xg}{288.4 g/mol\times 0.299kg}$

$x=8.42g$

Thus 8.42 grams of testosterone must be dissolved in 299.0 grams of benzene to reduce the freezing point by 0.500°C.

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4 years ago

Why is the separation of mixtures into pure or relatively pure substances so important when performing a chemical analysis?

N76 [4]

Answer:

It is important to separate mixture into pure or relatively pure substances when performing a chemical analysis SO AS TO KNOW THE PROPERTIES COMING FROM EACH PART MIXTURE WHICH MAY INTERFERE WITH THE SEPARATION.

Explanation:

In chemistry, Mixture is the combination of two or more substances which are not combine chemically.

Mixture contain different substances with different physical and chemical properties.

It is important to purify the substances in a mixture so as to identify what properties are coming from each mixture and also some part of the mixture can interfere with the properties of other mixture present for skewing analysis.

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3 years ago