1 is D 2 is c 3 is A and 4 the nucleus is so big because it is one of our largest cells in our body... hope this helps!!!!!!!!!!!
<span>The student should
follow following steps to make 1 L of </span>2.0 M CaCl₂.<span>
<span>
1. First he should
calculate the number of moles of 2.0 M CaCl</span></span>₂ in 1 L solution.<span>
</span>Molarity of the solution = 2.0 M<span>
Volume of solution which should be prepared = 1 L
Molarity =
number of moles / volume of the solution
Hence, number of moles in 1 L = 2 mol
2. Find
out the mass of dry CaCl</span>₂ in 2 moles.<span>
moles =
mass / molar mass
Moles of CaCl₂ =
2 mol</span><span>
Molar mass of CaCl₂ = </span><span>110.98 g/mol
Hence, mass of CaCl</span>₂ = 2 mol x <span>110.98 g/mol
= 221.96
g
3. Weigh the mass
accurately
4. Then take a cleaned and dry1 L volumetric flask and place a funnel top of it. Then carefully add the salt into the volumetric flask and
finally wash the funnel and watch glass
with de-ionized water. That water also should be added into the volumetric
flask.
5. Then add some
de-ionized water into
the volumetric flask and swirl well until all salt are
dissolved.
<span>6. Then top up to
mark of the volumetric flask carefully.
</span></span>
7. As the final step prepared solution should be labelled.
Answer:
pH = 12.22
Explanation:
<em>... To make up 170mL of solution... The temperature is 25°C...</em>
<em />
The dissolution of Barium Hydroxide, Ba(OH)₂ occurs as follows:
Ba(OH)₂ ⇄ Ba²⁺(aq) + 2OH⁻(aq)
<em>Where 1 mole of barium hydroxide produce 2 moles of hydroxide ion.</em>
<em />
To solve this question we need to convert mass of the hydroxide to moles with its molar mass. Twice these moles are moles of hydroxide ion (Based on the chemical equation). With moles of OH⁻ and the volume we can find [OH⁻] and [H⁺] using Kw. As pH = -log[H⁺], we can solve this problem:
<em>Moles Ba(OH)₂ molar mass: 171.34g/mol</em>
0.240g * (1mol / 171.34g) = 1.4x10⁻³ moles * 2 =
2.80x10⁻³ moles of OH⁻
<em>Molarity [OH⁻] and [H⁺]</em>
2.80x10⁻³ moles of OH⁻ / 0.170L = 0.01648M
As Kw at 25°C is 1x10⁻¹⁴:
Kw = 1x10⁻¹⁴ = [OH⁻] [H⁺]
[H⁺] = Kw / [OH⁻] = 1x10⁻¹⁴/0.01648M = 6.068x10⁻¹³M
<em>pH:</em>
pH = -log [H⁺]
pH = -log [6.068x10⁻¹³M]
<h3>pH = 12.22</h3>
Based on the assumption that the reaction involves N and O to produce NO, if 25.0 g of NO are produced, the amount of N gas used would be 11.66 grams
<h3>Stoichiometric calculation</h3>
From the equation of the reaction:
N + O ---------> NO
Mole ratio of N to NO is 1:1
Mole of 25.0 g of NO = 25/30.01 = 0.833 moles
Equivalent mole of N = 0.833 moles
Mass of 0.833 moles N = 0.833 x 14 = 11.66 grams
More on stoichiometric calculations can be found here: brainly.com/question/8062886
Answer : The fraction of carbonic acid present in the blood is 5.95%
Explanation :
The mixture consists of carbonic acid ( H₂CO₃) and bicarbonate ion ( HCO₃⁻). This represents a mixture of weak acid and its conjugate which is a buffer.
The pH of a buffer is calculated using Henderson equation which is given below.
![pH = pKa + log \frac{[Base]}{[Acid]}](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2B%20log%20%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D)
We have been given,
pH = 7.5
pKa of carbonic acid = 6.3
Let us plug in the values in Henderson equation to find the ratio Base/Acid.
![7.5 = 6.3 + log \frac{[base]}{[acid]}](https://tex.z-dn.net/?f=7.5%20%3D%206.3%20%2B%20log%20%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D)
![1.2 = log \frac{[base]}{[acid]}](https://tex.z-dn.net/?f=1.2%20%3D%20log%20%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D)
![\frac{[Base]}{[Acid]} = 10^{1.2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%20%3D%2010%5E%7B1.2%7D)
![\frac{[Base]}{[Acid]} = 15.8](https://tex.z-dn.net/?f=%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%20%3D%2015.8)
![[Base] = 15.8 \times [Acid]](https://tex.z-dn.net/?f=%5BBase%5D%20%3D%2015.8%20%5Ctimes%20%5BAcid%5D)
The total of mole fraction of acid and base is 1. Therefore we have,
![[Acid] + [Base] = 1](https://tex.z-dn.net/?f=%5BAcid%5D%20%2B%20%5BBase%5D%20%3D%201)
But Base = 15.8 x [Acid]. Let us plug in this value in above equation.
![[Acid] + 15.8 \times [Acid] = 1](https://tex.z-dn.net/?f=%5BAcid%5D%20%2B%2015.8%20%5Ctimes%20%5BAcid%5D%20%3D%201)
![16.8 [Acid] = 1](https://tex.z-dn.net/?f=16.8%20%5BAcid%5D%20%3D%201)
![[Acid] = \frac{1}{16.8}](https://tex.z-dn.net/?f=%5BAcid%5D%20%3D%20%5Cfrac%7B1%7D%7B16.8%7D)
![[Acid] = 0.0595](https://tex.z-dn.net/?f=%5BAcid%5D%20%3D%200.0595)
[Acid] = 0.0595 x 100 = 5.95 %
The fraction of carbonic acid present in the blood is 5.95%