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anygoal [31]
3 years ago
15

Se ha quemado magnesio (reacción con el oxígeno) y se obtuvieron 12g de óxido de magnesio (II). ¿Cuánto magnesio se quemó? ¿Qué

volumen de oxígeno medido en condiciones normales se quemó?
Physics
1 answer:
Sedaia [141]3 years ago
7 0

Answer:

The first answer is 7,24g Mg

The second answer is 3,36L O2

Explanation:

Atomic masses: Mg= 24,3 ; O=16

Solution: MMgO = 40,3 g/mol

2 Mg + O2 => 2 MgO

1) 12 g MgO x (2x24,3 g Mg / 2x 40,3 MgO) = 7,24 g Mg

2)

At  first calculate moles from Oxygen:

12 g MgO x ( 1 mol from Oxygen/ 2 x 40,3 g MgO ) = 0,15 moles from Oxygen

then ....

0,15 moles from Oxygen x (22,4 lts / 1 mol from Oxygen ) = 3,36 lts

22.4 liters is the constant that we use in this equation, it is a value that is repeated since we are in the presence of a gas with ideal behavior, so it is considered that the normal molar volume of ideal gaseous substances is always 22.4 liters, estimated with a temperature of 0º C and a pressure of 1 atmosphere. (constant temperature and pressure)

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3 years ago
water in a cup and a kettle can have the same temperature even though the quantities are different . give reasons​
jekas [21]

Answer:

The reason is because both are exposed to a virtually infinite heat sink, due to the virtually infinite mass  and of the surrounding environment, compared to the sizes of either the cup or the kettle such that the equilibrium temperature, T_{(equilibrium)} reached is the same for both the cup and the kettle as given by the relation;

\infty M_{(environ)} \times  c_{(environ)} \times (T_2 - T_1) = m_{1} \times  c_{(water)} \times (T_3 - T_2) + m_{2} \times  c_{(water)} \times (T_4 - T_2)

Due to the large heat sink, T₂ - T₁ ≈ 0 such that the temperature of the kettle and that of the cup will both cool to the temperature of the environment

Explanation:

4 0
3 years ago
Four solid plastic cylinders all have radius 2.41 cm and length 5.94 cm. Find the charge of each cylinder given the following ad
Paladinen [302]

Answer:

Check explanation

Explanation:

QUICK NOTE: THE QUESTION IS NOT COMPLETE. Although it is not, we can make assumptions, since we only need values for the UNIFORM CHARGE DENSITY.

SO, LET US BEGIN;

To solve this question we are to use the equation (1) below;

Charge,Q = uniform charge density,p × Total area of the cylinder,A ------------------------------------------------------------------------(1).

From the question, we are given radius, R to be 2.41 cm and length, L to be 5.94 cm.

Step one: calculate for the total area of the cylinder, A.

Total area of the cylinder, A= area of the top surface + area of the buttom + area of the curved surface of the cylinder.

Hence, total area of the cylinder,A is;

==> πR^2 + πR^2 + 2πRL. -------------------------------------------------------------------------(2).

Then, total area of the cylinder,A is;

==> (L + R)2πR.

Step two: find the charge of each cylinder.

===> For the first cylinder; we have the uniform charge density to be 35 nC/m^2.

Therefore, the combination of equation (1) and (3) gives;

Charge Q= p × (L + R)2πR...----------------------------(4)

Hence, Q= 35 × [(5.94 + 2.41) 2× 3.143 × 5.94].= 10912.615 coulumb.

====> For the second cylinder, we have a uniform charge density of 50 nC/m^2.

Using equation (4), charge,Q= 15,589.45 Coulumb

=====> For THE third cylinder, the uniform charge density is 600, we make use of equation (4);

Charge,Q= 600×311.789.

Charge,Q= 187,073.4 coulumb.

====> For THE fourth cylinder, the uniform charge density is 750 nC/m^2.., we make use of equation (4);

Charge,Q= 233,841.75 coulumb.

7 0
3 years ago
The magnetic field in a solenoid is . A circular wire of radius 8 cm is concentric with a solenoid of radius 2 cm and length d =
defon

Answer:

6.03 mV

Explanation:

length of solenoid, L = 2 m, N = 12000, di/dt = 40 A/s,

Magnetic field due to solenoid

B = μ0 n i = μ0 N i / L

dB/dt = μ0 N / L x di / dt

dB /dt = (4 x 3.14 x 10^-7 x 12000 x 40) / 2 = 0.3 T/s

Induced emf, e = rate of change of magnetic flux

e = dΦ / dt = A x dB / dt

e = 3.14 x 0.08 x 0.08 x 0.3 = 6.03 x 10^-3 V = 6.03 mV

7 0
3 years ago
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