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sergij07 [2.7K]
3 years ago
7

How do you write the formula for potassium hydroxide

Chemistry
2 answers:
makvit [3.9K]3 years ago
7 0

KOH that or H3C-CH2-OH

velikii [3]3 years ago
3 0

You can use KOH in my opinion

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How many kilograms of solvent (water) must 0.71 moles of KI be dissolved in to produce a 1.93 m solution?
GalinKa [24]

Answer: kg= 0.37

Explanation:

Use the molality formula.

M= m/kg

6 0
3 years ago
For the following electrochemical reaction: Al3+(aq) + 3e -> Al(s) Eº = -1.66 V E° = 2.87 F2(g) + 2e -> 2F (aq) Calculate
makvit [3.9K]

<u>Answer:</u> The standard electrode potential of the cell is 4.53 V.

<u>Explanation:</u>

We are given:

E^o_{(F_2/F^-)}=2.87V\\E^o_{(Al^{3+}/Al)}=-1.66V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Aluminium will undergo oxidation reaction and will get oxidized.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=2.87-(-1.66)=4.53V

Hence, the standard electrode potential of the cell is 4.53 V.

6 0
3 years ago
Is one mole of zinc the same as one atom of zinc?
Papessa [141]

Answer: No.

Explanation: One mole of zinc is not the same as one atom of zinc. In one mole of zinc, there are approximately 6.022*10^23 atoms of zinc.

3 0
3 years ago
1.Which form of energy is due to the motion of an object's particles .
tamaranim1 [39]

Answer:

thermal

Explanation:

4 0
3 years ago
A piece of an unknown substance weighing 124.0 grams is heated in boiling water to 100.0oc. when the substance is placed in a ca
MakcuM [25]

The heat cause 300g water temperature increase from 20 to 26 celcius. The heat transferred would be: 300g * (26 °C -20 °C) *4.2 joule/gram °C= 7560J

The unknown substance is added to the water, so its final temperature should be the same as the water. The calculation would be:

7560J= 124g * (100-26)* specific heat

specific heat= 7560J / 124g / 74 °C= 0.8238 J/gram °C

3 0
3 years ago
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