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pshichka [43]
3 years ago
6

Which option is an example of a physical property?

Chemistry
1 answer:
KatRina [158]3 years ago
3 0

Answer:

The answer is freezing point . Freezing point is a physical property of an atom

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Calculate how much heat is absorbed by a sample that weighs 12 kilograms, has a specific heat of 0.231 kg/CJ, and is heated from
Andrei [34K]

Answer:

97 J

Explanation:

Step 1: Given data

  • Mass of the sample (m): 12 kg
  • Specific heat capacity (c): 0.231 J/kg.°C (this can also be expressed as 0.231 J/kg.K)
  • Initial temperature: 45 K
  • Final temperature: 80 K

Step 2: Calculate the temperature change

ΔT = 80 K - 45 K = 35 K

Step 3: Calculate the heat required (Q)

We will use the following expression.

Q = c × m × ΔT

Q = 0.231 J/kg.K × 12 kg × 35 K = 97 J

4 0
3 years ago
You used a calorimeter in the heat transfer lab. Explain how the calorimeter works, and how to calculate the heat given off or a
Butoxors [25]
A calorimeter contains reactants and a substance to absorb the heat absorbed. The initial temperature (before the reaction) of the heat absorbent is measured and then the final temperature (after the reaction) is also measured. The absorbent's specific heat capacity and mass are also known. Given all of this data, the equation:
Q = mcΔT 
To find the heat released.
3 0
3 years ago
describe the placement of the crucible lid on the crucible when heating the magnesium. Why is it important that this be done cor
Andreas93 [3]
What
:(((((((((((((((((((((
7 0
3 years ago
A sample of gas is observed to effuse through a pourous barrier in 4.98 minutes. Under the same conditions, the same number of m
kogti [31]

Answer:

The molar mass of the unknown gas is \mathbf{ 51.865 \  g/mol}

Explanation:

Let assume that  the gas is  O2 gas

O2 gas is to effuse through a porous barrier in time t₁ = 4.98 minutes.

Under the same conditions;

the same number of moles of an unknown gas requires  time t₂  =  6.34 minutes to effuse through the same barrier.

From Graham's Law of Diffusion;

Graham's Law of Diffusion states that, at a constant temperature and pressure; the rate of diffusion of a gas is inversely proportional to the square root of its density.

i.e

R \  \alpha  \ \dfrac{1}{\sqrt{d}}

R = \dfrac{k}{d}  where K = constant

If we compare the rate o diffusion of two gases;

\dfrac{R_1}{R_2}= {\sqrt{\dfrac{d_2}{d_1}}

Since the density of a gas d is proportional to its relative molecular mass M. Then;

\dfrac{R_1}{R_2}= {\sqrt{\dfrac{M_2}{M_1}}

Rate is the reciprocal of time ; i.e

R = \dfrac{1}{t}

Thus; replacing the value of R into the above previous equation;we have:

\dfrac{R_1}{R_2}={\dfrac{t_2}{t_1}}

We can equally say:

{\dfrac{t_2}{t_1}}=  {\sqrt{\dfrac{M_2}{M_1}}

{\dfrac{6.34}{4.98}}=  {\sqrt{\dfrac{M_2}{32}}

M_2 = 32 \times ( \dfrac{6.34}{4.98})^2

M_2 = 32 \times ( 1.273092369)^2

M_2 = 32 \times 1.62076418

\mathbf{M_2 = 51.865 \  g/mol}

7 0
3 years ago
Please answer ASAP!!
Alenkasestr [34]

Answer:

C.0.28 V  

Explanation:

Using the standard cell potential we can find the standard cell potential for a voltaic cell as follows:

The most positive potential is the potential that will be more easily reduced. The other reaction will be the oxidized one. That means for the reactions:

Cu²⁺ + 2e⁻ → Cu E° = 0.52V

Ag⁺ + 1e⁻ → Ag E° = 0.80V

As the Cu will be oxidized:

Cu → Cu²⁺ + 2e⁻

The cell potential is:

E°Cell = E°cathode(reduced) - E°cathode(oxidized)

E°cell = 0.80V - (0.52V)

E°cell = 1.32V

Right answer is:

<h3>C.0.28 V </h3>

<h3 />

4 0
3 years ago
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