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,
Answer:
see explanation for answer
Explanation:
salivary gland: 1
stomach: 2
small intestine: 6
liver: 4
gallbladder: 5
large intestine: 3
The answers correspond with the numbers on the text boxes, so you would drag number 1 to the salivary gland and so on.
Answer:
34.7mL
Explanation:
First we have to convert our grams of Zinc to moles of zinc so we can relate that number to our chemical equation.
So: 6.25g Zn x (1 mol / 65.39 g) = 0.0956 mol Zn
All that was done above was multiplying the grams of zinc by the reciprocal of zincs molar mass so our units would cancel and leave us with moles of zinc.
So now we need to go to HCl!
To do that we multiply by the molar coefficients in the chemical equation:
This leaves us with 2(0.0956) = 0.1912 mol HCl
Now we use the relationship M= moles / volume , to calculate our volume
Rearranging we get that V = moles / M
Now we plug in: V = 0.1912 mol HCl / 5.50 M HCl
V= 0.0347 L
To change this to milliliters we multiply by 1000 so:
34.7 mL
The chloroplasts i believe is the answer
Answer:
% composition O = 19.9%
% composition Cu = 80.1%
Explanation:
Given data:
Total mass of compound = 3.12 g
Mass of copper = 2.50 g
Mass of oxygen = 3.12 - 2.50 = 0.62 g
% composition = ?
Solution:
Formula:
<em>% composition = ( mass of element/ total mass)×100</em>
% composition Cu = (2.50 g / 3.12 g)×100
% composition Cu = 0.80 ×100
% composition Cu = 80.1%
For oxygen:
<em>% composition = ( mass of element/ total mass)×100</em>
% composition O = (0.62 g / 3.12 g)×100
% composition O = 0.199 ×100
% composition O = 19.9%
