S is was the day I had to come home from work today so I’m going to go get food and get food for a while to go get some stuff to go to do get it
True
Explanation:
The position of sodium on the periodic table makes the element a solid at room temperature is very correct.
- Sodium is an elements with 11 electrons in its shell.
- It is located in group 1 on the periodic table because it contains an outer valence electron.
- In a free state Na metal, the atoms exhibit metallic bonding.
- Metallic bonding is responsible for the solid nature of this alkali metal at room temperature.
- It is highly reactive because it has just one valence electron.
- Metals are usually found left of the periodic table.
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Group 1 brainly.com/question/2154626
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Answer: it measures temperature
Explanation:
The amount of Calcium hydroxide : = 58.719 g
<h3>Further explanation</h3>
Given
100 g Nitric Acid-HNO₃
Required
The amount of Calcium hydroxide
Solution
Reaction(balanced) :
Ca(OH)₂ (s) + 2HNO₃ (aq) → Ca(NO₃)₂ (aq) + 2H₂O (l)
mol of Nitric acid (MW 63 g/mol) :
mol = mass : MW
mol = 100 : 63
mol = 1.587
From the equation, mol ratio of Ca(OH)₂ : HNO₃ = 1 : 2, so mol Ca(OH)₂ :
=1/2 x mol HNO₃
= 1/2 x 1.587
=0.7935
Mass of Ca(OH)₂ (MW=74 g/mol) :
= mol x MW
= 0.7935 x 74
= 58.719 g
Answer:
The pH of the solution is 11.48.
Explanation:
The reaction between NaOH and HCl is:
NaOH + HCl → H₂O + NaCl
From the reaction of 3.60x10⁻³ moles of NaOH and 5.95x10⁻⁴ moles of HCl we have that all the HCl will react and some of NaOH will be leftover:
Now, we need to find the concentration of the OH⁻ ions.
![[OH^{-}] = \frac{n_{NaOH}}{V}](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%20%5Cfrac%7Bn_%7BNaOH%7D%7D%7BV%7D%20)
Where V is the volume of the solution = 1.00 L
![[OH^{-}] = \frac{n_{NaOH}}{V} = \frac{3.01 \cdot 10^{-3} moles}{1.00 L} = 3.01 \cdot 10^{-3} mol/L](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%20%5Cfrac%7Bn_%7BNaOH%7D%7D%7BV%7D%20%3D%20%5Cfrac%7B3.01%20%5Ccdot%2010%5E%7B-3%7D%20moles%7D%7B1.00%20L%7D%20%3D%203.01%20%5Ccdot%2010%5E%7B-3%7D%20mol%2FL%20)
Finally, we can calculate the pH of the solution as follows:
![pOH = -log([OH^{-}]) = -log(3.01 \cdot 10^{-3}) = 2.52](https://tex.z-dn.net/?f=%20pOH%20%3D%20-log%28%5BOH%5E%7B-%7D%5D%29%20%3D%20-log%283.01%20%5Ccdot%2010%5E%7B-3%7D%29%20%3D%202.52%20)
Therefore, the pH of the solution is 11.48.
I hope it helps you!
