Answer:
1.53 L
Explanation:
Step 1: Given data
- Mass of oxygen (m): 11.2 g
- Ideal gas constant (R): 0.0821 atm.L/mol.K
Step 2: Calculate the moles (n) corresponding to 11.2 g of oxygen
The molar mass of oxygen is 32.00 g/mol.
11.2 g × (1 mol/32.00 g) = 0.350 mol
Step 3: Calculate the volume of oxygen
We will use the ideal gas equation.
P × V = n × R × T
V = n × R × T / P
V = 0.350 mol × (0.0821 atm.L/mol.K) × 415 K / 7.78 atm
V = 1.53 L
Write as a proportion, showing the relationship of both given information:
68.0g 0.3g
---------- = -----------
1L x ( your answer)
Cross multiply: 68.0g× X = 0.3g × 1L
68.0g (X)= 0.3g/L
Solve for X by dividing both sides by 68.0 g
68.0g (X) = 0.3g/L
------------- ------------------
68.0g 68.0g
Then enter into calculator 0.3/68 and that will be your solution. Make sure you round up.
<span>Blood pH has an ideal level of about 7.3 to 7.4. It is important for the pH ofblood to remain constant because if your blood pH varies, itcan be deadly.<span>hope this helps </span></span>
Answer:
27.6mL of LiOH 0.250M
Explanation:
The reaction of lithium hydroxide (LiOH) with chlorous acid (HClO₂) is:
LiOH + HClO₂ → LiClO₂ + H₂O
<em>That means, 1 mole of hydroxide reacts per mole of acid</em>
Moles of 20.0 mL = 0.0200L of 0.345M chlorous acid are:
0.0200L ₓ (0.345mol / L) = <em>6.90x10⁻³ moles of HClO₂</em>
To neutralize this acid, you need to add the same number of moles of LiOH, that is 6.90x10⁻³ moles. As the LiOH contains 0.250 moles / L:
6.90x10⁻³ moles ₓ (1L / 0.250mol) = 0.0276L of LiOH =
<h3>27.6mL of LiOH 0.250M</h3>
