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3 years ago

Where is the magnetic field of a bar magnet the strongest?

1 answer:

4 0

The correct answer is A. The magnetic field is strongest at the north pole and the south pole of a bar magnet. At the poles, the magnetic field is equally strong while the force is weaker at the middle of the bar. The field lines are closely packed at each pole and it gets wider as the lines get further from the pole.

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Newton's Laws of Motion

Ira Lisetskai [31]

Answer:

a: 1st paragraph from the left side indicates Newton's first law.

b:, 2nd paragraph from the left side indicates Newton's third law.

c: 1st paragraph from right side indicates Newton's first law.

d: 2nd paragraph from right side indicates Newton's second law.

Explanation:

5 0

3 years ago

If 335 g water at 35.5C loses 5750 J of heat, what is the final temperature of the water?

White raven [17]

Answer:

The final temperature is 39.58 degree Celsius

Explanation:

As we know

Q = m * c * change in temperature

Specific heat of water (c) = 4.2 joules per gram per Celsius degree

Substituting the given values we get -

5750 = 335 * 4.2 * (X - 35.5)

X = 39.58 degree Celsius

3 0

3 years ago

The vapor pressure of water is 1.00 atm at 373 K, and the enthalpy of vaporization is 40.68 kJ mol!. Estimate the vapor pressure

Yuki888 [10]

Answer:

The vapor pressure at temperature 363 K is 0.6970 atm

The vapor pressure at 383 K is 1.410 atm

Explanation:

To calculate $\Delta H_{vap}$ of the reaction, we use clausius claypron equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = vapor pressure at temperature $T_1$

$P_2$ = vapor pressure at temperature $T_2$

$\Delta H_{vap}$ = Enthalpy of vaporization

R = Gas constant = 8.314 J/mol K

1) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =363 K

$T_2$ = final temperature =373 K

$P_2=1 atm, P_1=?$

Putting values in above equation, we get:

$\ln(\frac{1 atm}{P_1})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{363}-\frac{1}{373}]$

$P_1=0.69671 atm \approx 0.6970 atm$

The vapor pressure at temperature 363 K is 0.6970 atm

2) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =373 K

$T_2$ = final temperature =383 K

$P_1=1 atm, P_2?$

Putting values in above equation, we get:

$\ln(\frac{P_2}{1 atm})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{383}]$

$P_2=1.4084 atm \approx 1.410 atm$

The vapor pressure at 383 K is 1.410 atm

8 0

3 years ago

The solid product from reaction of sulfuric acid with sucrose is?

Mama L [17]

$\huge{\mathbb{\tt {QUESTION:}}}$

The solid product from reaction of sulfuric acid with sucrose is?

$\huge{\mathbb{\tt {ANSWER:}}}$

Concentrated <u>sulfuric acid</u> is added to sucrose forming carbon, steam and <u>sulfur</u> dioxide.

<h3>RELATED TO THE QUESTION </h3>

$\huge{\mathbb{\tt {What \: is \: solid?:}}}$

<u>Solid is one of the four fundamental states of matter</u>. The molecules in a solid are closely packed together and contain the least amount of kinetic energy.
<u>A solid is characterized by structural rigidity and resistance to a force applied to the surface</u>.

$\huge{\mathbb{\tt {What \: is \: sulfur \: acid?}}}$

<u>Sulfuric acid or sulphuric acid, also known as oil of vitriol, is a mineral acid composed of the elements sulfur, oxygen and hydrogen</u>, with molecular formula H₂SO₄. It is a colorless, odorless and viscous liquid that is miscible with water at all concentrations.

$\huge{\mathbb{\tt{What \: is \: sucrose?}}}$

<u>Sucrose is common sugar. It is a disaccharide</u>, a molecule composed of two monosaccharides: glucose and fructose. Sucrose is produced naturally in plants, from which table sugar is refined. It has the molecular formula C₁₂H₂₂O₁₁.

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6 0

2 years ago

A sample of Francium-212 will decay to one sixteenth of its original amount after 80 minutes. What is the half-life of Francium-

lbvjy [14]

20 minutes.

The sample would lose one half the quantity of francium in each half-life.

$1/16 = 1/2^{4} = 1/2 \times1/2 \times1/2 \times1/2$

Thus a mass decrease by a factor of 16 would correspond to a period of four half-lives. It took 80 minutes for the sample to lose all these francium, therefore

$t_{1/2} = 80/4 = 20$ minutes.

6 0

3 years ago