<span>1.02x10^2 ml
Since molarity is defined as moles per liter, the product of the molarity and volume will remain constant as mole solvent is added. So let's set up an equality to express this
m0*v0 = m1*v1
where
m0, v0 = molarity and volume of original solution
m1, m1 = molarity and volume of final solution.
Solve for v0, then substitute the known values and calculate:
m0*v0 = m1*v1
v0 = (1.75 M * 500 ml)/8.61 M
v0 = (1.75 M * 500 ml)/8.61 M
V0 = 101.6260163
Rounding to 3 significant figures gives 102 ml.
So the original volume of the 8.61 M H2SO4 solution was 102 ml or 1.02x10^2 ml.</span>
1. 100.67
2. 168.55
3. 2.747
So I’m not sure if 2 or 3 are right sorry
An electron is found whizzing around the nucleus, so in subatomic particles.
Answer : The final temperature of the metal block is, 
Explanation :
As we know that,
![m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]](https://tex.z-dn.net/?f=m_1%5Ctimes%20c_1%5Ctimes%20%28T_%7Bfinal%7D-T_1%29%3D-%5Bm_2%5Ctimes%20c_2%5Ctimes%20%28T_%7Bfinal%7D-T_2%29%5D)
.................(1)
where,
q = heat absorbed or released
= mass of aluminum = 55 g
= mass of water = 0.48 g
= final temperature = ?
= temperature of aluminum =
= temperature of water =
= specific heat of aluminum = 
= specific heat of water= 
Now put all the given values in equation (1), we get
![55g\times 0.900J/g^oC\times (T_{final}-25)^oC=-[0.48g\times 4.184J/g^oC\times (T_{final}-25)^oC]](https://tex.z-dn.net/?f=55g%5Ctimes%200.900J%2Fg%5EoC%5Ctimes%20%28T_%7Bfinal%7D-25%29%5EoC%3D-%5B0.48g%5Ctimes%204.184J%2Fg%5EoC%5Ctimes%20%28T_%7Bfinal%7D-25%29%5EoC%5D)
Thus, the final temperature of the metal block is,
