Answer: The answer is 167
Explanation: This is because that was right on edg. so yea heart this tho plsss
Answer:
The mass of tin is 164 grams
Explanation:
Step 1: Data given
Specific heat heat of tin = 0.222 J/g°C
The initial temeprature of tin = 80.0 °C
Mass of water = 100.0 grams
The specific heat of water = 4.184 J/g°C
Initial temperature = 30.0 °C
The final temperature = 34.0 °C
Step 2: Calculate the mass of tin
Heat lost = heat gained
Qlost = -Qgained
Qtin = -Qwater
Q = m*c*ΔT
m(tin)*c(tin)*ΔT(tin) = -m(water)*c(water)*ΔT(water)
⇒with m(tin) = the mass of tin = TO BE DETERMINED
⇒with c(tin) = the specific heat of tin = 0.222J/g°C
⇒with ΔT(tin) = the change of temperature of tin = T2 - T1 = 34.0°C - 80.0°C = -46.0°C
⇒with m(water) = the mass of water = 100.0 grams
⇒with c(water) = the specific heat of water = 4.184 J/g°C
⇒with ΔT(water) = the change of temperature of water = T2 - T1 = 34.0° C - 30.0 °C = 4.0 °C
m(tin) * 0.222 J/g°C * -46.0 °C = -100.0g* 4.184 J/g°C * 4.0 °C
m(tin) = 163.9 grams ≈ 164 grams
The mass of tin is 164 grams
What an electron and a neutron have in common is that <u>each particle exists inside an atom,</u>
Atoms consist of three particles: protons (which are positively charged), electrons (which are negatively charged), and neutrons (which have no charge).
Answer : The equilibrium constant 
for the reaction is, 0.1133
Explanation :
First we have to calculate the concentration of 
.
Now we have to calculate the dissociated concentration of .
The balanced equilibrium reaction is,
Initial conc. 1.731 M 0
At eqm. conc. (1.731-x) (2x) M
As we are given,
The percent of dissociation of = 
= 1.2 %
So, the dissociate concentration of = 
The value of x = 0.2077 M
Now we have to calculate the concentration of 
at equilibrium.
Concentration of = 1.731 - x = 1.731 - 0.2077 = 1.5233 M
Concentration of 
= 2x = 2 × 0.2077 = 0.4154 M
Now we have to calculate the equilibrium constant for the reaction.
The expression of equilibrium constant for the reaction will be :
![K_c=\frac{[Br]^2}{[Br_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BBr%5D%5E2%7D%7B%5BBr_2%5D%7D)
Now put all the values in this expression, we get :
Therefore, the equilibrium constant for the reaction is, 0.1133
