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Luba_88 [7]
3 years ago
6

What kind of bond is created by a weak electrical attraction between polar molecules

Chemistry
2 answers:
Lyrx [107]3 years ago
8 0

your answer is hydrogen bond.

mars1129 [50]3 years ago
5 0
Hydrogen bonds is the correct answer
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A breathing mixture used by deep-sea divers contains helium, oxygen, and carbon dioxide. What is the partial pressure of oxygen
Kay [80]
Answer is: pressure of oxygen is 31,3 kPa.
The total pressure<span> of an ideal gas mixture is the sum of the </span>partial pressures<span> of the gases in the mixture.
p(mixture) = p(helium) + p(oxygen) + p(carbon dioxide).
p(oxygen) = p(mixture) - (p(helium) + p(carbon dioxide)).
p(oxygen) = 101,4 kPa - (68,7 kPa + 1,4 kPa).
p(oxygen) = 101,4 kPa - 70,1 kPa.
p(oxygen) = 31,3 kPa.

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3 0
3 years ago
What are formed when sodium ions and chlorine ions combine to produce nacl?
GREYUIT [131]
<span>bright yellow light and lots of heat-energy.</span>
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3 years ago
The combustion of 0.570 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera
torisob [31]

Answer:

The temperature change from the combustion of the glucose is 6.097°C.

Explanation:

Benzoic acid;

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.570 g

Moles of benzoic acid = \frac{0.570 g}{122.12 g/mol}=0.004667 mol

Energy released by 0.004667 moles of benzoic acid on combustion:

Q=3,228 kJ/mol \times 0.004667 mol=15.0668 kJ=15,066.8 J

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.053°C

Q=C\times \Delta T

15,066.8 J=C\times 2.053^oC

C=7,338.92 J/^oC

Glucose:

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=2.900 g

Moles of glucose = \frac{2.900 g}{180.16 g/mol}=0.016097 mol

Energy released by the 0.016097 moles of calorimeter  combustion:

Q'=2,780 kJ/mol \times 0.016097 mol=44.7491 kJ=44,749.1 J

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

Q'=C\times \Delta T'

44,749.1 J=7,338.92 J/^oC\times \Delta T'

\Delta T'=6.097^oC

The temperature change from the combustion of the glucose is 6.097°C.

6 0
3 years ago
What is the acknowledgment to ''Do all liquids evaporate at the same rate?'
natita [175]
<span>'Do all liquids evaporate at the same rate
 that would be false

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4 0
3 years ago
Khjgjhgfmhgxfhkugfdnhmvngchnvnhvhgh
igor_vitrenko [27]

Answer:

I don’t know

Explanation:

5 0
3 years ago
Read 2 more answers
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