What is the mass in grams of 1.204x10^24 atoms of carbon

Please help me I need these answers

Cerrena [4.2K]

1 is true

2 is d 7

3 is

1 e

2 b

3 d

4 a

5 c

8 0

3 years ago

How did knowing the number of valence electrons in one

Orlov [11]

Explanation:

Knowing the number of valence electrons in one of the alien elements helps in identifying it because the number of valence electrons can help categorize the alien element. Similar elements have the same valence electrons and knowing the category of the element can help further analyze the element.

5 0

3 years ago

Read 2 more answers

If 21.00 mL of a 0.68 M solution of C6H5NH2 required 6.60 mL of the strong acid to completely neutralize the solution, what was

Andru [333]

Answer:

pH = 2.46

Explanation:

Hello there!

In this case, since this neutralization reaction may be assumed to occur in a 1:1 mole ratio between the base and the strong acid, it is possible to write the following moles and volume-concentrations relationship for the equivalence point:

$n_{acid}=n_{base}=n_{salt}$

Whereas the moles of the salt are computed as shown below:

$n_{salt}=0.021L*0.68mol/L=0.01428mol$

So we can divide those moles by the total volume (0.021L+0.0066L=0.0276L) to obtain the concentration of the final salt:

$[salt]=0.01428mol/0.0276L=0.517M$

Now, we need to keep in mind that this is an acidic salt since the base is weak and the acid strong, so the determinant ionization is:

$C_6H_5NH_3^++H_2O\rightleftharpoons C_6H_5NH_2+H_3O^+$

Whose equilibrium expression is:

$Ka=\frac{[C_6H_5NH_2][H_3O^+]}{C_6H_5NH_3^+}$

Now, since the Kb of C6H5NH2 is 4.3 x 10^-10, its Ka is 2.326x10^-5 (Kw/Kb), we can also write:

$2.326x10^{-5}=\frac{x^2}{0.517M}$

Whereas x is:

$x=\sqrt{0.517*2.326x10^{-5}}\\\\x=3.47x10^-3$

Which also equals the concentration of hydrogen ions; therefore, the pH at the equivalence point is:

$pH=-log(3.47x10^{-3})\\\\pH=2.46$

Regards!

6 0

3 years ago

Attempt 2

Assoli18 [71]

Lets let our mass equal 3 on alletals and solve using d=m/v equation

Aluminum
V=3/2.70=1.11
Silver
V=3/10.5=.286
Rhenium
V=3/20.8=.144
Nickel
V=3/8.90=.337
This gives us the following list from largest to smallest Aluminum, Nickel, Silver, and Rhenium

4 0

3 years ago

What mass of cu(s) is electroplated by running 26.5 a of current through a cu2+(aq) solution for 4.00 h?

kramer

T = 14400 s
26.5 x 14400=381600 C
381600/96500=3.95 Faradays
Cu2+ + 2e- = Cu
3.95 faradays ( 1 mol/ 2 Faradays) = 1.97
mass = 1.97 x 63.55 g/mol=125 g

moles Au = 33.1 / 196.967 g/mol=0.168
Au+ + 1e- = Au
0.168 ( 1 Faraday/ 1mol)= 0.168 Faraday
0.168 x 96500=16217 Coulombs
16217 / 5.00=3243 s => 54 min

7 0

3 years ago