Answer:
q = - 2067.2 J of Heat is giving out when 85.0g of lead cools from 200.0 c to 10.0 c.
Explanation:
The Specific Heat capacity of Lead is 0.128 
This means, increase in temperature of 1 gm of lead by 
will require 0.128 J of heat.
Formula Used :
q = amount of heat added / removed
m = mass of substance in grams = 85.0 g
c = specific heat of the substance = 0.128
= Change in temperature
= final temperature - Initial temperature
= 10 - 200
= -
put value in formula
q = - 
On calculation,
q = - 2067.2 J
- sign indicates that the heat is released in the process
Volume. Gases and liquids are typically measured in milliliters (mL) or cubic centimeters (cm^3) - both of which are equivalent (1 mL = 1 cm^3).
Slavery played a big role in this. the confederacy allowed slavery while the union was oppose to it.
Answer: B) 2 (as indicated by electron distribution shown), but taking into account the real properties of this element, 4,7,8 also occur (see below).
Explanation:
This is the electron complement/atomic number of ruthenium, which actually has the structure [Kr] 5s1 4d7
Nevertheless, Ru does not form Ru(I) compounds and few Ru(II) compounds (RuCl2, RuBr2, RuI2). It also forms Ru(III)Cl3 and a larger number of Ru(IV) compounds, e.g. RuO2, RuS2. It also forms RuO4
Answer:
18.76atm
Explanation:
Using the formula V1P1/T1 = V2P2/T2, from combined gas law. Volume is constant since we have not been given. Therefore the formula comes to be; P1/T1 = P2/T1
To get P2 = T2(P1/T1)
Where P2 is final pressure
P2 = 239K ( 23atm/293K)
=18.76atm
