Answer:
58.61 grams
Explanation:
Taking The molecular weight of NaCl = 58.44 grams/mole
<u>Determine how many grams of NaCl to prepare the bath solution </u>
first we will calculate the moles of NaCl that is contained in 6L of 170 mM of NaCI solution
= ( 6 * 170 ) / 1000
= 1020 / 1000 = 1.020 moles
next
determine how many grams of NaCl
= moles of NaCl * molar mass of NaCl
= 1.020 * 58.44
= 58.61 grams
Answer:
The molecular formula of cacodyl is C₄H₁₂As₂.
Explanation:
<u>Let's assume we have 1 mol of cacodyl</u>, in that case we'd have 209.96 g of cacodyl and the<u> following masses of its components</u>:
- 209.96 g * 22.88/100 = 48.04 g C
- 209.96 g * 5.76/100 = 12.09 g H
- 209.96 g * 71.36/100 = 149.83 g As
Now we convert those masses into moles:
- 48.04 g C ÷ 12 g/mol = 4.00 mol C
- 12.09 g H ÷ 1 g/mol = 12.09 mol H
- 149.83 g As ÷ 74.92 g/mol = 2.00 mol As
Those amounts of moles represent the amount of each component in 1 mol of cacodyl, thus, the molecular formula of cacodyl is C₄H₁₂As₂.
Answer:
The final pressure is approximately 0.78 atm
Explanation:
The original temperature of the gas, T₁ = 263.0 K
The final temperature of the gas, T₂ = 298.0 K
The original volume of the gas, V₁ = 24.0 liters
The final volume of the gas, V₂ = 35.0 liters
The original pressure of the gas, P₁ = 1.00 atm
Let P₂ represent the final pressure, we get;



∴ The final pressure P₂ ≈ 0.78 atm.
Answer:
0.129g MgCl2
Explanation:
For this we need to understand the concept of molarity.
Molarity is number of moles of solute/litres of solution
M=n/L
Here we are given molarity of 0.054M and volume of 25ml. we just plug this in formula to find moles of MgCl2
0.054=x/(25/1000) (we divided 25 by 1000 to convert it to litres of solution)
x=0.00135 moles of MgCl2 (we are not done yet the question asks for grams so to convert to grams we multiply by molar mass of MgCl2.)
0.00135molesMgCl2 x 95.211g MgCl2/1molMgCl2
= 0.129g MgCl2
1) Balanced chemical reaction:
2H2 + O2 -> 2H20
Sotoichiometry: 2 moles H2: 1 mol O2 : 2 moles H2O
2) Reactant quantities converted to moles
H2: 5.00 g / 2 g/mol = 2.5 mol
O2: 50.0 g / 32 g/mol = 1.5625 mol
Limitant reactant: H2 (because as per the stoichiometry it will be consumed with 1.25 mol of O2).
3) Products
H2 totally consumed -> 0 mol at the end
O2 = 1.25 mol consumed -> 1.5625 mol - 1.25 mol = 0.3125 mol at the end
H2O: 2.5 mol H2 produces 2.5 mol H2O -> 2.5 mol at the end.
Total number of moles: 0.3125mol + 2.5 mol = 2.8125 mol
4) Pressure
Use pV = nRT
n = 2.8125
V= 9 liters
R = 0.082 atm*lit/K*mol
T = 35 C + 273.15 = 308.15K
p = nRT/V = 7.9 atm