Answer:
The energy produced equals 140.760 kJ
Explanation:
The relation between power, current and voltage is
Applying the given values in the relation above we get
Now Since 
Again applying the calculated values we get
By scientific question, it really just means create a normal question. The example would be:
How will gas be affected if the temperature in an enclosed container is to be changed?
There are mannnyyy ways how to write a question for this, you just have to make sure you write what you are changing and how you are changing it, oh and make it sound smart haha.
I believe it will be A. Density, temperature, and viscosity, can be found in water through tests. Hope this helps! Mark as brainly please!
Water’s chemical formula is H2O
One atom of oxygen bonded to two attached atoms of hydrogen.
The hydrogen atoms are to one side of the oxygen atom, resulting in a water molecule having a positive charge on the side where the hydrogens reside and a negative charge on the other side, where the oxygen atom resides. This separation
of charge on opposite ends of the molecule is called polarity
I hope this right and can help you!
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>
