Why hf has higher boiling point greater than nh3

What ALWAYS occurs simultaneously with reduction?

Lena [83]

Answer:

Oxidation occurs simultaneously with reduction.

4 0

2 years ago

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hi, 100 points, , will also mark brainly-ist (what's the boiling point of water) ((you don't need to answer that))

BaLLatris [955]

Answer:

212 degrees F, and 100 degrees C.

Explanation:

If the temperature is held constant (which requires some heat input, since evaporation cools things) the liquid will all evaporate. If the temperature is much above 212 F, the water will boil. That means that it wont just evaporate from the surface but will form vapor bubbles, which then grow, inside the liquid itself. :)

6 0

3 years ago

Use the solubility rules to determine if:

xenn [34]

<u>Answer:</u>

A reaction is said to occur if there is a formation of an insoluble solid or a precipitate(s) or a liquid (l) or a gas(g).

If both the reactants and products are in aqueous state, No reaction takes place.

$2 \mathrm{NaCl}(a q)+M g B r_{2}(a q)>M g C l_{2}(a q)+2 \mathrm{NaBr}(a q)(\mathrm{NO} \text { REACTION })$

All chlorides and Bromides are soluble except that of Ag, Hg and Pb.

Hence, No reaction takes place since all the reactants and products are in aqueous states.

${KOH}(a q)+{NaCl}(a q)>K C l(a q)+{NaOH}(a q)(\mathrm{NO} \text { REACTION })$

Salts of Group IA are soluble. Hence No reaction takes place

$M g S(a q)+2 {NaOH}(a q)>M g(O H)_{2}(s)+N a_{2} S(a q)$

(REACTION TAKES PLACE)

All hydroxides are insoluble except that of Group IA, ammonium ion and Group IIA down from Calcium.

Hence Reaction takes place with the formation of $Mg(OH)_2$ precipitate

5 0

2 years ago

Give at least 3 real-world uses of paper chromatography used in the fields of chemistry and biology.

Alexxandr [17]

Folding , cutting , and to be written on

7 0

2 years ago

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Describe how you would prepare 250 mL of a 0.707 M<br> NaNO3 solution.

timurjin [86]

Answer:

To prepare 250 mL of a 0.707 M NaNO3 solution, take 15.045 g of NaNO₃ in 250 mL flask and fill with solvent up to mark.

Explanation:

Data given:

volume of solution = 250 mL

Convert mL to L

1 ml = 1000 L

250 ml = 250/1000 = 0.25

Molarity of solution = 0.707 M

How to prepare = ?

Solution:

Molarity:

This the term used for the concentration of the solution. It is the amount of solute in moles dissolve in 1 Liter of solution.

So we have to calculate amount of NaNO₃

So,

we have to know the number of moles of solution that will be required

Formula used for Molarity

Molarity = number of moles of solute / L of solution

Rearrange above equation

number of moles of solute = Molarity x L of solution . . . . . . . (1)

Put values in equation 1

number of moles of solute = 0.707 mol/L x 0.25 L

number of moles of solute = 0.177 mol

So we need 0.177 mol of NaNO₃ to prepare 250 mL of a 0.707 M solution.

Now convert moles to mass

Mass = no. of moles x Molar mass . . . . . . . . . . (2)

molar mass of NaNO₃ = 23 + 14 + 3(16)

molar mass of NaNO₃ = 85 g/mol

Put value in equation 2

Mass of NaNO₃= 0.177 mol x 85 g/mol

Mass of NaNO₃= 15.045 g

So now,

To prepare 250 mL of a 0.707 M NaNO3 solution, take 15.045 g of NaNO₃ in 250 mL flask and fill with solvent up to mark.

3 0

2 years ago