The Lewis structure of the carbonate ion is an example of a resonance structure. True False

What is the average mass of a single silicon atom in grams?

torisob [31]

1 mol of Silicon = 28.0855 g (in average)

then

1 mol = 6.022*10^23 atoms

then

28.0855/(6.022*10^23) g/atom

4.66381*10^-23 g per atom

5 0

3 years ago

What does the suffix ite stand for

insens350 [35]

Answer: The -ite suffix is used on the oxyanion with one oxygen atom fewer (like sulfite SO32- or nitrite NO2-).

8 0

3 years ago

Assuming you have 6.24 x 1014 electrons and the surface area of the pail is 0.2 m2, what is the charge density (C/m2)?

pentagon [3]

Answer:

σ = 4.998 E-4 C/m²

Explanation:

1 Coulomb (C) ≡ 6.241509 E18 electrons (e)

∴ # elect = 6.24 E14 elect

charge (Q):

⇒ Q = (6.24 E14 elect)/( 1 C /6.241509 E18 elect) = 9.998 E-5 C

charge density (σ):

σ = Q/S

∴ surface area (S) = 0.2 m²

⇒ σ = ( 9.998 E-5 C ) / ( 0.2 m²)

⇒ σ = 4.998 E-4 C/m²

4 0

3 years ago

What is love and why are people scared to show love even if they like boys or girls

myrzilka [38]

Because society has shaped our mind on what we shall think about things

6 0

4 years ago

Read 2 more answers

In a titration of 47.41 mL of 0.3764 M ammonia with 0.3838 M aqueous nitric acid, what is the pH of the solution when 47.41 mL +

Volgvan

Answer: The pH of the solution is 1.136

Explanation:

To calculate the moles from molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

For ammonia:

Molarity of ammonia = 0.3764 M

Volume of ammonia = 47.41 mL = 0.04741 L (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

$0.3764mol/L=\frac{\text{Moles of ammonia}}{0.04741L}\\\\\text{Moles of ammonia}=0.01784mol$

For nitric acid:

Molarity of nitric acid = 0.3838 M

Volume of ammonia = (47.41 + 10.00) mL = 57.41 mL= 0.05741 L

Putting values in above equation, we get:

$0.3838mol/L=\frac{\text{Moles of nitric acid}}{0.05741L}\\\\\text{Moles of nitric acid}=0.02203mol$

After the completion of reaction, amount of nitric acid remained = 0.022 - 0.0178 = 0.0042 mol

For the reaction of ammonia with nitric acid, the equation follows:

$NH_3+HNO_3\rightarrow NH_4NO_3$

At $t=0$ 0.0178 0.022

Completion 0 0.0042 0.0178

As, the solution of the reaction is made from strong acid which is nitric acid and the conjugate acid of weak base which is ammonia. So, the pH of the reaction will be based totally on the concentration of nitric acid.

To calculate the pH of the reaction, we use the equation:

$pH=-\log[H^+]$

where,

$[H^+]=\frac{0.0042mol}{0.05741L}=0.0731M$

Putting values in above equation, we get:

$pH=-\log(0.0731)\\\\pH=1.136$

Hence, the pH of the solution is 1.136

8 0

3 years ago