Answer:
ΔH°comb=-5899.5 kJ/mol
Explanation:
First, consider the energy balance:
Where
is the calorimeter mass and
is the number of moles of the samples;
is the combustion enthalpy. The energy balance says that the energy that the reaction release is employed in rise the temperature of the calorimeter, which is designed to be adiabatic, so it is suppose that the total energy is employed rising the calorimeter temperature.
The product
is the heat capacity, so the balance equation is:

So, the enthalpy of combustion can be calculated:

I will be happy to solve any doubt you have.
Concept: The magnification of spherical mirror can be defined by two ways.
(i) In terms of the height of the object and image.
The magnification of the spherical mirror is defined as the ratio of the height of the image'
' to the height of the object '
'. It is denoted by letter 'm'.
Mathematically, it can be written as

(ii) In terms of the object's and image's distances.
The magnification of the spherical mirror is defined as the negative ratio of the image distance'
' to the object distance '
'.
Mathematically, it can be written as

Now, from equation (1) and (2) we have,

Given: Spherical Concave Mirror,
We will consider positive sign for object's and image's distance because both are in front of the mirror.
Object distance
.
Image distance 
Object's height 
Image's height 
Now, apply equation (3)


Or, hi = - 8 mm
Here; negative sign means, the image will be inverted.
The image's height will be 8 mm.
Answer:

Explanation:
<u>Friction Force</u>
When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.
There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.
Please find the free body diagrams in the figure provided below.
The equilibrium condition for the mass 1 is

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa
![\displaystyle F_a=F_{r1}+F_{r2}.....[1]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F_a%3DF_%7Br1%7D%2BF_%7Br2%7D.....%5B1%5D)
The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

The friction forces are computed by


Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.
Replacing in [1]

Simplifying

Plugging in the values
![\displaystyle F_{a}=0.25(9.8)[400+2(100)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F_%7Ba%7D%3D0.25%289.8%29%5B400%2B2%28100%29%5D)

Answer: alpha particle. i think
Explanation: