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AlladinOne [14]
3 years ago
5

A drop of water placed on a smooth, dry surface will form a dome-shaped droplet instead of flowing outward in different directio

ns. Which of these best explains this observation?
A. The bonds between hydrogen and oxygen atoms are too strong.
B. The electrons in the atom attract the electrons in other atoms.
C. Water molecules at the surface experience fewer hydrogen bonds than water molecules within the liquid
D. Water molecules near the surface produce more buoyant force than water molecules within the liquid.
Chemistry
2 answers:
Dafna1 [17]3 years ago
7 0
Your answer is C Water molecules at the surface experience fewer hydrogen bonds than water molecules within the liquid.
KIM [24]3 years ago
4 0
<h2>Answer:</h2>

The correct answer is option C which is,<u> "Water molecules at the surface experience fewer hydrogen bonds than water molecules within the liquid".</u>

<h3>Explanation:</h3>
  • In some conditions, a water droplet remains in the droplet form on the dry surface rather than spreading on the whole surface. It covers only remains on the mall part of the surface and making a dome shape.
  • It is due to the fact that the hydrogen bond between the water molecules is much strong than the attraction between the water molecules and surface.
  • Hence the option C is correct.

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On the reaction below, label the BSA, BSB, CA, and CB. CH3COOH + H2O → CH3COO– + H3O+
Ostrovityanka [42]

Answer:

Acid(BSA) = CH₃COOH

Base (BSB) = H₂O

Conjugate base (CB) = CH₃COO⁻

Conjugate acid (CA) = H₃O⁺

Explanation:

Equation of reaction;

CH₃COOH + H₂O → CH₃COO⁻ + H₃O⁺

Hello,

From my understanding of the question, we are required to identify the

1) Acid

2) Base

3) conjugate acid

4) conjugate base in the reaction

Acid (BSA) = CH₃COOH

Base (BSB) = H₂O

CA = conjugate acid = H₃O⁺

CB = conjugate base = CH₃COO⁻

7 0
2 years ago
The molar mass is determined by measuring the freezing point depression of an aqueous solution. A freezing point of -5.20°C is r
Dima020 [189]

Answer:

The empirical formula is C2H4O3

The molecular formula is C4H8O6

The molar mass is 152 g/mol

Explanation:

The complete question is: An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of 31.57% C and 5.30% H. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of -5.20°C is recorded for a solution made by dissolving 10.56 g of the compound in 25.0 g water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.

Step 1: Data given

Mass % of Carbon = 31.57 %

Mass % of H = 5.30 %

Freezing point = -5.20 °C

10.56 grams of the compound dissolved in 25.0 grams of water

Kf water = 1.86 °C kg/mol

Step 2: Calculate moles of Carbon

Suppose 31.57% = 31.57 grams

moles C = mass C / Molar mass C

moles C = 31.57 grams / 12.0 g/mol = 2.63 moles

Step 3: Calculate moles of Hydrogen:

Moles H = 5.30 grams / 1.01 g/mol

moles H = 5.25 moles

Step 4: Calculate moles of Oxygen

Moles O = ( 100 - 31.57 - 5.30) / 16 g/mol

Moles O = 3.95 moles

Step 5: We divide by the smallest number of moles

C: 2.63 / 2.63 = 1 → 2

H: 5.25/2.63 = 2 → 4

O: 3.95/ 2.63 = 1.5 → 3

The empirical formula is C2H4O3

The molar mass of the empirical formula = 76 g/mol

Step 6: Calculate moles solute

Freezing point depression = 5.20 °C = m * 1.86

m = 5.20 / 1.86

m = 2.80 molal = 2.80 moles / kg

2.80 molal * 0.025 kg = 0.07 moles

Step 7: Calculate molar mass

Molar mass = mass / moles

Molar mass = 10.56 grams / 0.07 moles

Molar mass = 151 g/mol

Step 8: Calculate molecular formula

151 / 76 ≈  2

We have to multiply the empirical formula by 2

2*(C2H4O3) = C4H8O6

The molecular formula is C4H8O6

The molar mass is 152 g/mol

6 0
3 years ago
You have two sealed jars of water at the same temperature. In the first jar there is a large amount of water. In the second jar
melisa1 [442]

since both the jars are kept at the same temperature the vapor pressure will be same in both the cases.


3 0
3 years ago
What is the relationship between atmospheric pressure and the density of gas particles in an area of increasing pressure? As air
Marina86 [1]

We need to know the relationship between atmospheric pressure and the density of gas particles in an area of increasing pressure.

The relationship is: As air pressure in an area increases, the density of the gas particles in that area increases.

For any gaseous substance, density of gas is directly proportional to pressure of gas.

This can be explained from idial gas edquation:

PV=nRT

PV=\frac{w}{M}RT [where, w= mass of substance, M=molar mass of substance]

PM=\frac{w}{V}RT

PM=dRT [where, d=density of thesubstance]

So, for a particular gaseous substance (whose molar mass is known), at particular temperature, pressure is directly related to density of gaseous substance.

Therefore, as air pressure in an area increases, the density of the gas particles in that area increases.

3 0
3 years ago
Los hermanos Ana Victoria y José Leonardo están patinando en el hielo, con 25 y 20 kg respectivamente de masa, si Victoria empuj
andreyandreev [35.5K]

Utiliando las leyes de Newton encontraremos que la aceleración de Ana es -1.6 m/s^2.

La segunda ley de Newton dice que:

F = m*a

Fuerza es igual a masa por aceleración.

La tercera ley de Newton dice que cuando dos objetos interactuan, cadan objeto ejerce una fuerza de <u>igual magnitud pero opuesta direccion</u> en el otro.

Ahora veamos como aplicar esto.

Sabemos que la niña empuja al niño, asumamos que con una fuerza F.

Tendremos entonces la ecuación:

F = 20kg*(2m/s^2) = 40N

Y por la tercer ley de Newton, esta misma fuerza (pero en opuesta dirección) se aplica a la niña, entonces tendremos:

-40N = 25kg*a

-40N/25kg = a = -1.6 m/s²

La aceleración con la que retrosede la niña es -1.6 m/s²

Sí quieres aprender más, puedes leer:

brainly.com/question/17123407

3 0
2 years ago
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