Metamorphic rocks are formed at high temperature and medium to low pressure

Percent Yield Lab Report

Vlada [557]

Based on the data obtained from the reaction, the following conclusions can be made;

according to the law of conservation of mass, the mass of the product was higher than the reactant because of the mass of oxygen added during the combustion.
the percent yield is less than 100% because of loss in mass of either reactants or products.
the errors could have occurred during the weighing and transfer of reactants and products
repeated measurements are required in order to improve accuracy

<h3>What is the percent yield of the reaction?</h3>

Equation of the reaction is given below:

2 Mg + O₂ ----> 2 MgO

Trial 1

Mass of empty crucible with lid = 26.698 (g)

Mass of Mg metal, crucible, and lid = 27.040 (g)

Mass of MgO, crucible, and lid = 27.198 (g)

Mass of metal = 27.040 - 26.698 = 0.342

Mass of MgO = 27.198 - 26.698 = 0.500

<h3>Moles of Mg used</h3>

moles of Mg = mass/molar mass

molar mass of Mg = 24 g

moles of Mg = 0.342/24 = 0.01425

<h3>Moles of MgO expected</h3>

Based on the equation of reaction;

moles of MgO expected = 0.01425

moles of MgO produced = mass/molar mass

molar mass of MgO = 40 g/mol

moles of MgO produced = 0.500/40 = 0.0125

<h3>Percent yield</h3>

Percent yield = (moles of MgO produced/moles of MgO expected) * 100%

Percentage yield = 0.0125/0.01425 * 100%

Percent yield of MgO = 87.7%

Trial 2

Mass of empty crucible with lid = 26.691 (g)

Mass of Mg metal, crucible, and lid = 27.099 (g)

Mass of MgO, crucible, and lid = 27.361 (g)

Mass of metal = 27.099 - 26.691 = 0.408

Mass of MgO = 27.361 - 26.691 = 0.670

<h3>Moles of Mg used</h3>

moles of Mg = mass/molar mass

molar mass of Mg = 24 g

moles of Mg = 0.408/24 = 0.0170

<h3>Moles of MgO expected</h3>

Based on the equation of reaction;

moles of MgO expected = 0.0170

moles of MgO produced = mass/molar mass

molar mass of MgO = 40 g/mol

moles of MgO produced = 0.670/40 = 0.01675

<h3>Percent yield</h3>

Percent yield = (moles of MgO produced/moles of MgO expected) * 100%

Percent yield = 0.01675/0.0170 * 100%

Percent yield of MgO = 98.5%

Average percent yield = (87.7 + 98.5)% / 2

Average percent yield = 89.0%

Based on the data obtained from the reaction, the following conclusion can be made;

according to the law of conservation of mass, the mass of the product was higher than the reactant because of the mass of oxygen added during the combustion.
the percent yield is less than 100% because of loss in mass of either reactants or products.
the errors could have occurred during the weighing and transfer of reactants and products
repeated measurements are required in order to improve accuracy

Learn more about law of conservation of mass at: brainly.com/question/1824546

6 0

3 years ago

In an open circuit like the picture

Dmitry_Shevchenko [17]

C. The light bulb will be off
This is due to the reason that the wire doesn’t go back to the battery :)

5 0

3 years ago

Read 2 more answers

Que aceleración adquiere un cuerpo de 0.5 kg cuando actúa una fuerza n2

san4es73 [151]

La aceleración que adquiere el cuerpo de 0.5Kg es:

5 0

3 years ago

If an element has 5 electrons how many of them will be in the 2nd energy level

egoroff_w [7]

Oi mate 3 electrons will be in tha 2nd energy level you silly willy.

6 0

3 years ago

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A car of 1000 kg with good tires on a dry road can decelerate (slow down) at a steady rate of about 5.0 m/s2 when braking. If a

vredina [299]

(a) 4.0 s

The acceleration of the car is given by

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time interval

For this car, we have

v = 0 (the final speed is zero since the car comes to a stop)

u = 20 m/s is the initial velocity

$a=-5.0 m/s^2$ is the deceleration of the car

Solving the equation for t, we find the time needed to stop the car:

$t=\frac{v-u}{a}=\frac{0-(20 m/s)}{-5.0 m/s^2}=4 s$

(b) 40 m

The stopping distance of the car can be calculated by using the equation

$v^2 - u^2 = 2ad$

where

v = 0 is the final velocity

u = 20 m/s is the initial velocity

a = -5.0 m/s^2 is the acceleration of the car

d is the stopping distance

Solving the equation for d, we find

$d=\frac{v^2-u^2}{2a}=\frac{0^2-(20 m/s)^2}{2(-5.0 m/s^2)}=40 m$

(c) $-5.0 m/s^2$

The deceleration is given by the problem, and its value is $-5.0 m/s^2$ .

(d) 5000 N

The net force applied on the car is given by

$F=ma$

where

m is the mass of the car

a is the magnitude of the acceleration

For this car, we have

m = 1000 kg is the mass

$a=5.0 m/s^2$ is the magnitude of the acceleration

Solving the formula, we find

$F=(1000 kg)(5.0 m/s^2)=5000 N$

(e) $2.0\cdot 10^5 J$

The work done by the force applied by the car is

$W=Fd$

where

F is the force applied

d is the total distance covered

Here we have

F = 5000 N

d = 40 m (stopping distance)

So, the work done is

$W=(5000 N)(40 m)=2.0\cdot 10^5 J$

(f) The kinetic energy is converted into thermal energy

Explanation:

when the breaks are applied, the wheels stop rotating. The car slows down, as a result of the frictional forces between the brakes and the tires and between the tires and the road. Due to the presence of these frictional forces, the kinetic energy is converted into thermal energy/heat, until the kinetic energy of the car becomes zero (this occurs when the car comes to a stop, when v = 0).

8 0

4 years ago