Answer:
0.26 ml
Explanation:
d = m/V
=> V = m/d = 15.2/58 = 0.26 ml
Answer:
V₂ = 16.5 L
Explanation:
To solve this problem we use <em>Avogadro's law, </em>which applies when temperature and pressure remain constant:
V₁/n₁ = V₂/n₂
In this case, V₁ is 22.0 L, n₁ is [mol CO + mol NO], V₂ is our unknown, and n₂ is [mol CO₂ + mol N₂].
- n₁ = mol CO + mol NO = 0.1900 + 0.1900 = 0.3800 mol
<em>We use the reaction to calculate n₂</em>:
2CO(g) + 2NO(g) → 2CO₂(g) + N₂(g)
0.1900 mol CO * 
0.1900 mol CO₂
0.1900 mol NO * 
0.095 mol N₂
- n₂ = mol CO₂ + mol N₂ = 0.1900 + 0.095 = 0.2850 mol
Calculating V₂:
22.0 L / 0.3800 mol = V₂ / 0.2850 mol
V₂ = 16.5 L
I'm pretty sure the answer is Temperature
Answer:
(a) adding 0.050 mol of HCl
Explanation:
A buffer is defined as the mixture of a weak acid and its conjugate base -or vice versa-.
In the buffer:
1.0L × (0.10 mol / L) = 0.10 moles of HF -<em>Weak acid-</em>
1.0L × (0.050 mol / L) = 0.050 moles of NaF -<em>Conjugate base-</em>
-The weak acid reacts with bases as NaOH and the conjugate base reacts with acids as HCl-
Thus:
<em>(a) adding 0.050 mol of HCl:</em> The addition of 0.050moles of HCl produce the reaction of 0.050 moles of NaF producing HF. That means after the reaction, all NaF is consumed and you will have in solution just the weak acid <em>destroying the buffer</em>.
(b) adding 0.050 mol of NaOH: The NaOH reacts with HF producing more NaF. Would be consumed just 0.050 moles of HF -remaining 0.050 moles of HF-. Thus, the buffer <em>wouldn't be destroyed</em>.
(c) adding 0.050 mol of NaF: The addition of conjugate base <em>doesn't destroy the buffer</em>
Answer:
i know 3, reproduction, organisms, and variations in traits
Explanation:
