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3 years ago

Mass is conserved in a reaction if the of the initial reactant masses equals the of the masses of

Physics

1 answer:

KATRIN_1 [288]3 years ago

7 0

Answer:

mass of the products

Explanation:

The law of conservation of mass states that mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations. According to the law of conservation of mass, the <u>mass of the products </u>in a chemical reaction must equal the mass of the reactants.

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suppose a car of 1200kg is moving with a velocity of 40km/hr therefore its kinetic energy is not zero. 1. explain briefly what h

mezya [45]

Answer:

Explanation:

For starters begin with a warning not to touch the brake drums. All of the KE is transferred to the brake drums. The result is a large rise in temperature. Heat. If you press hard on the brakes, rubber is left on the road and there is heat involved in that too.

8 0

3 years ago

Read 2 more answers

A 0.500-mol sample of an ideal monatomic gas at 400 kPa and 300 K, expands quasi-statically until the pressure decreases to 160

Marianna [84]

Answer:

(a). The final volume is 2.5 times the initial volume.

The work done is 1143 J.

(b). The final temperature is 207.9 K

The work done is 574 J.

Explanation:

Given that,

Sample of an ideal gas = 0.500 mol

Initial pressure = 400 kPa

Final pressure = 160 kPa

Temperature = 300 K

(a) for isothermal,

Temperature will be same.

We need to calculate the volume of gas

$P_{f}V_{f}=P_{i}V_{i}$

$V_{f}=(\dfrac{P_{i}}{P_{f}})V_{i}$

Put the value into the fomrula

$V_{f}=(\dfrac{400}{160})V_{i}$

$V_{f}=2.5 V_{i}$

We need to calculate the work done

Using equation of energy

$dQ=dW$

$dQ=nRTln(\dfrac{P_{in}}{P_{f}})$

$dQ=0.500\times8.314\times300\times ln(\dfrac{400}{160})$

$dQ=1143\ J$

(b). For adiabatic,

No transfer of heat between system and surroundings

We need to calculate the final temperature

Using formula of gas

$P_{f}^{1-\gamma}T_{f}^{\gamma}=P_{i}^{1-\gamma}T^{\gamma}$

$T_{f}=(\dfrac{P_{i}}{P_{f}})^{\frac{1-\gamma}{\gamma}}T_{i}$

Put the value into the formula

$T_{f}=(\dfrac{400}{160})^{\frac{1-\frac{5}{3}}{\frac{5}{3}}}\times300$

$T_{f}=207.9\ K$

We need to calculate the wok done in adiabatic

Using formula of work done

$W=\dfrac{nR(T_{i}-T_{f})}{\gamma-1}$

$W=\dfrac{0.500\times8.314\times(300-207.9)}{\dfrac{5}{3}-1}$

$W=574\ J$

Hence, (a). The final volume is 2.5 times the initial volume.

The work done is 1143 J.

(b). The final temperature is 207.9 K

The work done is 574 J.

6 0

3 years ago

An athlet starting from stationary moves with an acceleration 2.2m/s^2 for 5 seconds, then for other 5 seconds

ycow [4]

Answer:

The graph is shown below.

The athlete's speed at the finish line is 11 m/s

Total distance covered is 82.5 m.

Explanation:

The graph of speed versus time is drawn below.

Acceleration in the first 5 seconds, $a=2.2$ m/s²

From the graph, the slope of line OA is the acceleration in the first 5 seconds.

Slope of line OA is given as:

$slope, a=\frac{AD}{OD}\\2.2=\frac{AD}{5}\\AD =2.2\times 5=11\textrm{ m/s}$

Now, the length AD is nothing but speed at point A or B as AB is a straight line.

Therefore, the speed when crossing the finish line is the speed at B which is equal to 11 m/s.

Distance covered is given by the total area under the graph.

The total area can be divided into two shapes; a triangle and a rectangle.

The area under the graph is the sum of areas of triangle OAD and rectangle ABCD.

Area of triangle OAD is, $A_{tri}=\frac{1}{2}\times OD\times AD=\frac{1}{2}\times 5\times 11=27.5$

Area of rectangle ABCD is, $A_{rec}=AB\times AD=5\times 11=55$

Therefore, the total distance covered till the finish line is given as:

$d_{total}=27.5+55=82.5\textrm{ m}$

4 0

3 years ago

What is the definition of Potential Energy?

Margarita [4]

Answer: its The ability to do work or cause change. Energy of an object due to its motion.

Explanation:In physics, potential energy is the energy held by an object because of its position relative to other objects, stresses within itself, its electric charge, or other factors.

5 0

4 years ago

When you push a 1.70 kg book resting on a tabletop it takes 2.60 N to start the book sliding. Once it is sliding, however, it ta

bezimeni [28]

Answer:

Coefficient of static friction ,μs = 0.15

Coefficient of kinetic friction ,μk = 0.089

Explanation:

Given that

m = 1.7 kg

Static friction :

The force required to start the motion F= 2.6 N

We know that when book is in rest condition then static friction force act on it.

We know that the maximum value of the static friction force

fr = μ m g

At limiting condition

F= fr

2.6 = μ m g

2.6 = μ x 1.7 x 9.81

$\mu=\dfrac{2.6}{1.7\times 9.81}$

μ = 0.15

Kinetic friction :

F= 1.5 N

When the book is in moving condition then kinetic friction force act on it.

We know that the maximum value of the kinetic friction force

fr = μ m g

F= fr

= μ m g

1.5 = μ x 1.7 x 9.81

$\mu=\dfrac{1.5}{1.7\times 9.81}$

μ = 0.089

Coefficient of static friction ,μs = 0.15

Coefficient of kinetic friction ,μk = 0.089

7 0

3 years ago

Read 2 more answers