Answer:
7.32 g of F₂
Solution:
The equation is as follow,
2 LiI + F₂ → 2 LiF + I₂
According to equation,
51.88 g (2 mole) of LiF is produced from = 37.99 g (1 mole) F₂
So,
10 g of LiF will be produced by = X g of F₂
Solving for X,
X = (10 g × 37.99 g) ÷ 51.88 g
X = 7.32 g of F₂
Answer:
<h2>Density = 0.00026 g/mL</h2>
Explanation:
The density of a substance can be found by using the formula
<h3>
</h3>
From the question
mass of air = 1.2 g
volume = 4,555 mL
Substitute the values into the above formula and solve for the density
That's
<h3>
</h3>
= 0.0002634
We have the final answer as
<h3>Density = 0.00026 g/mL</h3>
Hope this helps you
Isn't it nitrogen AND chlorine? Both are nonmetals, and they need to gain electrons in order to bond.
Answer:
In 1897, the British physicist J. J. Thomson (1856–1940) proved that atoms were not the most basic form of matter. He demonstrated that cathode rays could be deflected, or bent, by magnetic or electric fields, which indicated that cathode rays consist of charged particles (Figure 2.2.2 ). More important, by measuring the extent of the deflection of the cathode rays in magnetic or electric fields of various strengths, Thomson was able to calculate the mass-to-charge ratio of the particles. These particles were emitted by the negatively charged cathode and repelled by the negative terminal of an electric field. Because like charges repel each other and opposite charges attract, Thomson concluded that the particles had a net negative charge; these particles are now called electrons. Most relevant to the field of chemistry, Thomson found that the mass-to-charge ratio of cathode rays is independent of the nature of the metal electrodes or the gas, which suggested that electrons were fundamental components of all atoms.
Explanation:
Answer : The energy removed must be, 29.4 kJ
Explanation :
The process involved in this problem are :
The expression used will be:
![Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{fusion}]+[m\times c_{p,s}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=Q%3D%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2B%5Bm%5Ctimes%20%5CDelta%20H_%7Bfusion%7D%5D%2B%5Bm%5Ctimes%20c_%7Bp%2Cs%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
= heat released for the reaction = ?
m = mass of benzene = 94.4 g
= specific heat of solid benzene = 
= specific heat of liquid benzene = 
= enthalpy change for fusion = 
Now put all the given values in the above expression, we get:
![Q=[94.4g\times 1.73J/g.K\times (279-322)K]+[94.4g\times -125.6J/g]+[94.4g\times 1.51J/g.K\times (205-279)K]](https://tex.z-dn.net/?f=Q%3D%5B94.4g%5Ctimes%201.73J%2Fg.K%5Ctimes%20%28279-322%29K%5D%2B%5B94.4g%5Ctimes%20-125.6J%2Fg%5D%2B%5B94.4g%5Ctimes%201.51J%2Fg.K%5Ctimes%20%28205-279%29K%5D)
Negative sign indicates that the heat is removed from the system.
Therefore, the energy removed must be, 29.4 kJ
