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RoseWind [281]
3 years ago
9

Which pair of elements would most likely bond to form a covalently bonded compound?

Chemistry
1 answer:
anastassius [24]3 years ago
3 0

Answer:

two nonmetal elements join together to form covalent compounds

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When a 2.5 mol of sugar (C12H22O11) are added to a certain amount of water the boiling point is raised by 1 Celsius degree. If 2
Mademuasel [1]

Answer is: boiling point will be changed by 4°C.

Chemical dissociation of aluminium nitrate in water: Al(NO₃)₃ → Al³⁺(aq) + 3NO⁻(aq).

Change in boiling point: ΔT =i · Kb · b.

Kb - molal boiling point elevation constant of water is 0.512°C/m, this the same for both solution.

b -  molality, moles of solute per kilogram of solvent., this is also same for both solution, because ther is same amount of substance.

i - Van't Hoff factor.

Van't Hoff factor for sugar solution is 1, because sugar do not dissociate on ions.

Van't Hoff factor for aluminium nitrate solution is approximately 4, because it dissociates on four ions (one aluminium cation and three nitrate anions). So ΔT is four times bigger.

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3 years ago
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Which one allows you to obtain the most precise volume? Circle one.the first one on the right because it's too ee and also helps
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Answer:50 ml ☺️

Explanation:

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3 years ago
HELP PLEASE! There is a water-filled continental rift in Iceland. What type of plate boundary would cause this rift? *
ad-work [718]
1.A- convergent
2.A- lower mantle

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WHAT IS THE PH OF LEMON JUICE IF IT HAS A HYDROGEN ION CONCENTRATION [AT]ON[H^ + ]=5.0*10^ -2 M.
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8 0
3 years ago
Express the concentration of a 0.0390 M aqueous solution of fluoride, F − , in mass percentage and in parts per million (ppm). A
Vadim26 [7]

Answer:

Mass percentage → 0.074 %

[F⁻] = 741 ppm

Explanation:

Aqueous solution of flouride → [F⁻] = 0.0390 M

It means that in 1L of solution, we have 0.0390 moles of F⁻

We need the mass of solution and the mass of 0.0390 moles of F⁻

Mass of solution can be determined by density:

1g/mL = Mass of solution / 1000 mL

Note: 1L = 1000mL

Mass of solution: 1000 g

Moles of F⁻ → 0.0390 moles . 19g /1 mol = 0.741 g

Mass percentage → (Mass of solute / Mass of solution) . 100

(0.741 g / 1000 g) . 100 = 0.074 %

Ppm = mass of solute . 10⁶ / mass of solution (mg/kg)

0.741 g . 1000 mg/1g = 741 mg

1000 g . 1 kg/1000 g = 1kg

741 mg/1kg = 741 ppm

5 0
3 years ago
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