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Paha777 [63]
2 years ago
13

Most homes are built with some form of insulation separating the inside of the home from the outside. What does this insulation

do?
A. It decreases the temperature difference between the outside and inside.
B. It increases the rate of heat flow between the outside and inside.
C. It decreases the rate of heat flow between the outside and inside.
D. It generates heat and allows it to flow inside but not outside.
Chemistry
1 answer:
Ad libitum [116K]2 years ago
6 0

Answer:

A. It decreases the temperature difference between the outside and inside.

Explanation:

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If 74.5 m of oxygen are collected at a pressure of 98.0 kPa, what volume will the gas occupy if the
sergey [27]

Answer:

80.8 mL

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 74.5 mL

Initial pressure (P₁) = 98.0 kPa

Final pressure (P₂) = 90.4 kPa

Final volume (V₂) =?

The final volume of the gas can be by using the Boyle's laws equation as follow:

P₁V₁ = P₂V₂

98 × 74.5 = 90.4 × V₂

7301 = 90.4 × V₂

Divide both side by 90.4

V₂ = 7301 / 90.4

V₂ = 80.8 mL

Thus, the volume of the gas is 80.8 mL

4 0
3 years ago
1. Which statement is true about oxygen,
sergey [27]

Answer:

they belong to same grop

4 0
3 years ago
Helium is the lightest noble gas and the second most abundant element (after hydrogen) in the universe. The mass of a helium−4 a
dmitriy555 [2]

<u>Answer:</u> The fraction of atom's mass contributed by nucleus is 0.99

<u>Explanation:</u>

Nucleons are defined as the sub-atomic particles which are present in the nucleus of an atom. Nucleons are protons and neutrons.

The isotopic symbol of Helium-4 atom is _2^4\textrm{He}

Number of electrons = 2

Number of protons = 2

Number of neutrons = 4 - 2 = 2

We are given:

Mass of He-4 atom = 6.64648\times 10^{-24}g

Mass of 1 electron = 9.10939\times 10^{-28}g

Calculating the mass contributed by the nucleus = m_{He}-2(m_e)

Mass of the nucleus of He-4 atom = 6.64648\times 10^{-24}-(2\times (9.10939\times 10^{-28}))=(6.64648-0.0018219)\times 10^{-24}=6.64466\times 10^{-24}

To calculate the fraction of atom's mass contributed by the nucleus, we use the equation:

\text{Fraction of atom's mass contributed by nucleus}=\frac{\text{Mass of nucleus}}{\text{Mass of atom}}

Putting values in above equation, we get:

\text{Fraction of atom's mass contributed by nucleus}=\frac{6.64466\times 10^{-24}g}{6.64648\times 10^{-24}g}=0.99

Hence, the fraction of atom's mass contributed by nucleus is 0.99

8 0
3 years ago
How do you know which planet is farther away from the sun by years?
vladimir1956 [14]

I believe the answer is B

4 0
3 years ago
Which of the following is least likely to increase the rate of diffusion?A small concentration gradientB higher concentration of
Ksenya-84 [330]

Answer: Option (A) is the correct answer.

Explanation:

Rate of diffusion is defined as the total movement of molecules from a region of higher concentration to lower concentration.

The interaction between medium and the material is responsible for the rate of diffusion of a material or substance.

A small concentration gradient means small difference in the number of molecules taking part in a reaction. So, when there no large difference between the concentration then there won't be much difference in the rate of diffusion of a material.

Whereas a higher concentration of molecules will lead to more number of collisions due to which frequency of molecules increases. Therefore, rate of diffusion will also increase.

Small molecule size will also lead to increases in rate of diffusion. This is because according to Graham's law rate of diffusion is inversely proportional to molar mass of an element. Hence, smaller size molecule will have smaller mass. As a result, rate of diffusion will be more.

High temperature means more kinetic energy of molecules due to which more number of collisions will be there. Hence, rate of diffusion will also increase.

Thus, we can conclude that out of the given options a small concentration gradient  is least likely to increase the rate of diffusion.

4 0
3 years ago
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