Answer:
m H2O = 56 g
Explanation:
∴ The heat ceded (-) by the Aluminum part is equal to the heat received (+) by the water:
⇒ - (mCΔT)Al = (mCΔT)H2O
∴ m Al = 25.0 g
∴ Mw Al = 26.981 g/mol
⇒ n Al = (25.0g)×(mol/26.981gAl) = 0.927 mol Al
⇒ Q Al = - (0.927 mol)(24.03 J/mol°C)(26.8 - 86.4)°C
⇒ Q Al = 1327.64 J
∴ mH2O = Q Al / ( C×ΔT) = 1327.64 J / (4.18 J/g.°C)(26.8 - 21.1)°C
⇒ mH2O = 55.722 g ≅ 56 g
Hope this helps :) remember your conversions and just practice it's fairly easy:)
Answer:
1. 31.25 mL
2. 1.98 g/L
3. 0.45 g/mL
Explanation:
For each of the problems, you need to perform unit conversions. You need to use the information given to you to convert to a specific unit.
1. You need volume (mL). You have density (g/mL) and mass (g). Divide mass by density. You will cancel out mL and be left with g.
(50.0 g)/(1.60 g/mL) = 31.25 mL
2. You are given grams and liters. You need to find density with units g/L. This means that you have to divide grams by liters.
(0.891 g)/(0.450 L) = 1.98 g/L
3. You have to find density again but this time with units g/mL. Divide the given mass by the volume.
(10.0 g)/(22.0 mL) = 0.45 g/mL
Answer:
The answer is A hope this helps
Explanation:
<u>Answer:</u> The concentration of solute is 0.503 mol/L
<u>Explanation:</u>
To calculate the concentration of solute, we use the equation for osmotic pressure, which is:
where,
= osmotic pressure of the solution = 24 atm
i = Van't hoff factor = 2 (for NaCl)
c = concentration of solute = ?
R = Gas constant = 
T = temperature of the solution = ![25^oC=[273+25]=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B273%2B25%5D%3D298K)
Putting values in above equation, we get:
Hence, the concentration of solute is 0.503 mol/L
