Answer: 0.0508mL
Explanation: Using the basic formula that states: C acid * V acid = C base * V base. we have:0.568 * 17.88 = 20 * C base.
therefore concentration of the base is 1.0156/20 = 0.0508 mL
Answer:
<em>yes</em>
Explanation:
the cuttle fish tell the difference between blue and yellow
Answer:

Explanation:
In this case, we have to start with the <u>chemical reaction</u>:

So, if we start with <u>10 mol of cyclohexanol</u> (
) we will obtain 10 mol of cyclohexanol (
). So, we can calculate the grams of cyclohexanol if we<u> calculate the molar mass:</u>

With this value we can calculate the grams:

Now, we have as a product 500 mL of
. If we use the <u>density value</u> (0.811 g/mL). We can calculate the grams of product:

Finally, with these values we can calculate the <u>yield</u>:
%= (405.5/820)*100 = 49.45 %
See figure 1
I hope it helps!
Potassium oxide is an ionic compound. The potassium has a charge of <span>K+</span> and oxygen has a charge of <span>O<span>2−</span></span>. We need 2 potassium ions to balance one oxide ion making the formula <span><span>K2</span>O</span>.
Potassium hydroxide is an ionic compound. The potassium has a charge of <span>K+</span> and hydroxide has a charge of <span>OH−</span>. We need 1 potassium ion to balance one hydroxide ion making the formula KOH.
<span><span>K2</span>O+<span> H2</span>O→KOH</span>
To balance the equation we place a coefficient of 2 in front of the potassium hydroxide.
<span><span>K2</span>O+<span>H2</span>O→2KOH</span>
I hope this was helpful.
Answer:
(a) 7.11x10⁻⁴ M/s
(b) 2.56 mol.L⁻¹.h⁻¹
Explanation:
(a) The reaction is:
O₃(g) + NO(g) → O₂(g) + NO₂(g) (1)
The reaction rate of equation (1) is given by:
(2)
<u>We have:</u>
k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹
[O₃]₀ = 2.35x10⁻⁶ M
[NO]₀ = 7.74x10⁻⁵ M
Hence, to find the inital reacion rate we will use equation (2):
Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s
(b) The number of moles of NO₂(g) produced per hour per liter of air is:
t = 1 h
V = 1 L
![\frac{\Delta[NO_{2}]}{\Delta t} = rate](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%5BNO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%20%3D%20rate)
![\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%5BNO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%20%3D%207.11%20%5Ccdot%2010%5E%7B-4%7D%20M%2Fs%2A%5Cfrac%7B3600%20s%7D%7B1%20h%7D%20%3D%202.56%20mol.L%5E%7B-1%7D.h%7B-1%7D)
Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹
I hope it helps you!