The noble gases are grouped together in the periodic table. where are the noble gases in the table?

KengaRu [80]

To neutralize acids, a weak base is used. Bases have a bitter or astringent taste and a pH greater than 7. Common bases are sodium hydroxide, potassium hydroxide and ammonium hydroxide. Bases are neutralized by using a weak acid.

7 0

2 years ago

A zinc rod is placed in 0.1 MZnSO4 solution at 298 K. Write the electrode reaction and calculate the potential of the electrode.

yanalaym [24]

Answer: The potential of electrode is -0.79 V

Explanation:

When zinc is dipped in zinc sulfate solution, the electrode formed is $Zn^{2+}(aq.)/Zn(s)$

Reduction reaction follows: $Zn^{2+}(0.1M)+2e^-\rightarrow Zn(s);(E^o_{Zn^{2+}/Zn}=-0.76V)$

To calculate the potential of electrode, we use the equation given by Nernst equation:

$E_{(Zn^{2+}/Zn)}=E^o_{(Zn^{2+}/Zn)}-\frac{0.059}{n}\log \frac{[Zn]}{[Zn^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = ?V

$E^o_{cell}$ = standard electrode potential of the cell = -0.76 V

n = number of electrons exchanged = 2

$[Zn]=1M$ (concentration of pure solids are taken as 1)

$[Zn^{2+}]=0.1M$

Putting values in above equation, we get:

$E_{cell}=-0.76-\frac{0.059}{2}\times \log(\frac{1}{0.1})\\\\E_{cell}=-0.79V$

Hence, the potential of electrode is -0.79 V

4 0

3 years ago

A 400 mL sample of hydrogen gas is collected over water at 20°C and 760 torr the vapor pressure of water at 20°C is 17.5 torr. w

o-na [289]

Answer:

V2 = 17371.43ml

Explanation:

We use Boyles laws

since temperature is constant

P1V1=P2V2

760 x 400 = 17.5 x V2

304000 = 17.5 x V2

V2 = 304000/17.5

V2 = 17371.43ml

4 0

3 years ago

A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefully weighs out 385.mg of oxalic acid H2

Vlada [557]

Answer:

The molarity of the sodium hydroxide solution is 0.0692 M

Explanation:

Step 1: Data given

Mass of H2C2O4 = 385 mg = 0.385 grams

volume = 250 mL = 0.250 L

Volume of NaOH = 123.7 mL = 0.1237 L

Molar mass H2C2O4 = 90.03 g/mol

Step 2: The balanced equation

2NaOH + H2C2O4 → Na2C2O4 + 2H2O

Step 3: Calculate moles H2C2O4

Moles H2C2O4 = mass H2C2O4 / molar mass H2C2O4

Moles H2C2O4 = 0.385 grams / 90.03 g/mol

Moles H2C2O4 = 0.00428 moles

Step 4: Calculate molarity of H2C2O4

Molarity H2C2O4 = moles / volume

Molarity H2C2O4 = 0.00428 moles / 0.250 L

Molarity H2C2O4 = 0.01712 M

Step 5: Calculate molarity of NaOH

2*Ca*Va = n*Cb*Vb

⇒ with Ca = Molarity of H2C2O4 = 0.01712 M

⇒ Va = volume of H2C2O4 = 0.250 L

⇒Cb = molarity of NaOH = TO BE DETERMINED

⇒ Vb = volume of NaOH = 0.1237 L

Cb = (2*0.01712*0.250)/0.1237

Cb = 0.0691 M

The molarity of the sodium hydroxide solution is 0.0692 M

4 0

3 years ago

What other signs provide evidence of chemical reactions and show that a chemical change has occurred?

Dmitry_Shevchenko [17]

Color change
new substance formed
heat or light change
new smell
irreversible

4 0

3 years ago