Answer:
2 moles
Explanation:
The following were obtained from the question:
Molarity = 0.25 M
Volume = 8L
Mole =?
Molarity is simply defined as the mole of solute per unit litre of solution. It is represented mathematically as:
Molarity = mole of solute/Volume of solution.
With the above equation, we can easily find the number of mole of MgCl2 present in 8 L of 0.25 M MgCl2 solution as follow:
Molarity = mole of solute/Volume of solution.
0.25 = mole of MgCl2 /8
Cross multiply to express in linear form
Mole of MgCl2 = 0.25 x 8
Mole of MgCl2 = 2 moles
Therefore, 2 moles of MgCl2 are present in 8 L of 0.25 M MgCl2 solution
Hey there!:
1) The additional stability that accompanies the formation of the network<span>Crystalline is measured as network enthalpy.
</span>2) The reticular energy is the energy released when the solid Crystal isform from separate ions in the gaseous state. Always exothermic.<span>
3) </span>The enthalpy of the network depends directly on the size of the loads and conversely in the distance between the ions .
hope this helps!
V1 = 2.0 L
T1 = 25.0 oC = 298 K V2 = V1T2 = (2.0 L)(244 K) = 1.6 L
V2 = ? t1(298 K)
T2 = –28.9 oC = 244 K
Answer:
the mass of the air in the room is 4.96512 kg ( in 0°C)
Answer:
The new force will be \frac{1}{100} of the original force.
Explanation:
In the context of this problem, we're dealing with the law of gravitational attraction. The law states that the gravitational force between two object is directly proportional to the product of their masses and inversely proportional to the square of a distance between them.
That said, let's say that our equation for the initial force is:
![F = G\frac{m_1m_2}{R^2}The problem states that the distance decrease to 1/10 of the original distance, this means:[tex]R_2 = \frac{1}{10}R](https://tex.z-dn.net/?f=F%20%3D%20G%5Cfrac%7Bm_1m_2%7D%7BR%5E2%7D%3C%2Fp%3E%3Cp%3EThe%20problem%20states%20%20that%20%20the%20distance%20decrease%20to%201%2F10%20of%20the%20original%20distance%2C%20this%20means%3A%3C%2Fp%3E%3Cp%3E%5Btex%5DR_2%20%3D%20%5Cfrac%7B1%7D%7B10%7DR)
And the force at this distance would be written in terms of the same equation:

Find the ratio between the final and the initial force:

Substitute the value for the final distance in terms of the initial distance:

Simplify:

This means the new force will be \frac{1}{100} of the original force.