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Keith_Richards [23]
2 years ago
12

Write a balanced half-reaction describing the reduction of gaseous dichlorine to aqueous chloride anions.

Chemistry
1 answer:
Annette [7]2 years ago
4 0

Answer:- Cl_2(g)+2e^-\rightarrow 2Cl^-(aq)

Explanations:- In reduction the electrons are accepted and so they are written on the reactant side. When an atom accepts electrons then it forms anion. Chlorine has 7 valence electrons and it needs one more electron to complete it's octet. Since, dichlorine has two Cl atoms and each Cl atom needs one more electron to complete it's octet, two electrons are accepted by dichorine to make aqueous chloride ion. For balancing the equation, there would be two chloride ions as the reactant side has two chlorine atoms.

Cl_2(g)+2e^-\rightarrow 2Cl^-(aq)

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If you perform the experiment described in investigation #15 by mixing 10 g of glue with 13 g of water and 8 g of sodium borate
Phantasy [73]

10 g of glue with 13 g of water ,

Mass ratio of the material can be calculated as:    

\frac{Mass of glue}{Mass of glue + Mass of water}  

\frac{10 g}{10 g + 13 g }  

\frac{10 g}{23 g }

8 g of sodium borate suspended in 11 g of water, mass ratio can be calculated as:

\frac{Mass of sodium borate}{Mass of sodium borate + Mass of water}  

\frac{ 8 g}{ 8 g + 11 g }  

\frac{8 g}{19 g }

3 0
2 years ago
The average concentration of bromde ion in seawater is 65 mg of bromide ion per kg of seawater. what is the molarity of the brom
allsm [11]
Usually concentrations are expressed as molarity, or moles of solute per liter solution. First, convert the mass of bromide ion to moles. The molar mass of bromine is 79.904 g/mol.

Moles of bromine = 65 mg * 1 g/1000 mg * 1 mol/79.904 g = 8.135×10⁻⁴ moles

Next, convert the mass of seawater to volume using the density.

Volume of seawater =  1 kg * 1 m³/ 1,025 kg * 1000 L/1 m³ = 0.976 L

Thus,
Molarity = 8.135×10⁻⁴ moles/0.976 L = 8.335×10⁻⁴ M
5 0
2 years ago
BRAINLIESTTT ASAP!!!
Stels [109]

While I am not the brainliest I can certainly answer.  

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8 0
2 years ago
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How many moles of sodium bicarbonate is needed to neutralize 0.8 ml of sulphuric acid?
svet-max [94.6K]

Answer:

n NaHCO3 = 9.6 E-3 mol

Explanation:

balanced reaction:

  • 2 NaHCO3(s) + H2SO4(ac) ↔ Na2SO4(ac) + 2 CO2(g) + 2 H2O(l)
  • assuming a concentration of H2SO4 6M....normally worked in the lab

⇒ n H2SO4 = 8 E-4 L * 6 mol/L = 4.8 E-3 mol H2SO4

according to balanced reaction, we have that for every mol of H2SO4 there are two mol of NaHCO3 ( sodium bicarbonate)

⇒ mol NaHCO3 = 4.8 E-3 mol H2SO4 * ( 2 mol NaHCO3 / mol H2SO4 )

⇒ ,mol NaHCO3 = 9.6 E-3 mol

So 9.6 E-3 mol NaHCO3,  are the minimun moles necessary to neutralize the acid.

6 0
2 years ago
Find the volume of a balloon of a gas at 842 mm Hg and -23 celcius if it’s volume is 915 milliliters at a pressure of 1170 mm Hg
Ronch [10]

The volume of a balloon f a gas at 842 mm Hg and -23 celsius if it’s volume is 915 milliliters at a pressure of 1170 mm Hg And a temperature of 24 celsius is 0.22 litres

Explanation:

Data given:

Initial volume of the balloon having gas V1= 915ml OR 0.195 L

initial pressure of the gas P1= 1170 mm Hg OR 1.53 atm

initial temperature of the gas T1 = 24 celsius or 273.15 + 24 = 297.15 K

Final pressure of the gas P2 = 842 mm Hg or 1.10 atm

final temperature of the gas T2 = -23 degrees or 273.15 - 23 = 250.15 K

Final volume at final temperature and pressure V2=?

The formula used is of Gas Law:

\frac{P1V1}{T1} = \frac{P2V2}{T2}

V2 = \frac{P1V1T2}{T1P2}

putting the values in the equation:

V2 =  \frac{1.53 X 0.195 X 250.15}{297.15 X 1.10}

V2 = 0.22 litres is the volume

The volume is 0.22 litres at a pressure of 1170 mmHg and temperature of -23 degrees.

5 0
2 years ago
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