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Alenkinab [10]
3 years ago
13

What is the mass of 18.8 moles of helium atoms?

Chemistry
1 answer:
Firlakuza [10]3 years ago
5 0
18.8 mol= mass of He X atomic mass
18.8 mol= mass of He X 4 g mol^-1
mass of He= (18.8 X 4)g = 75.2 g
You might be interested in
What is the molar mass of a molecule that contains one sodium atom, one oxygen atom, and one hydrogen atom?
GrogVix [38]

wherears one hydrogen atom has a mass of approximately 1 u ,1 mol of h atoms has a mass of approximately 1 gram and wherears one sodium atom has an approximately 1 gram and one sodium atom have approximate mass of 23 u

7 0
3 years ago
A 25.0 ml sample of a 0.2900 m solution of aqueous trimethylamine is titrated with a 0.3625 m solution of hcl. calculate the ph
Flauer [41]
a) After adding 10 mL of HCl

first, we need to get moles of (CH3)3N = molarity * volume

                                                                 = 0.29 m * 0.025 L
                                                                 = 0.00725M moles
then, we need to get moles of HCl = molarity * volume

                                                          = 0.3625 m * 0.01L
                                                          = 0.003625 moles

so moles of (CH3)3N remaining    = moles of (CH3)3N - moles of HCl
                                                         = 0.00725 - 0.003625

                                                      = 0.003625 moles

and when the total volume = 0.01 L + 0.025L = 0.035 L

∴ [(CH3)3N] = moles remaining / total volume

                    = 0.003625 moles / 0.035L
                    = 0.104 M

when we have Pkb so we can get Kb :

pKb = - ㏒Kb
4.19 = -㏒Kb

∴Kb = 6.5 x 10^-5

when Kb = [(CH3)3NH+] [OH-] / [(CH3)3N]

and by using ICE table we assume we have:

[(CH3)3NH+] = X & [OH] = X 

and [(CH3)3N] = 0.104 -X

by substitution:

∴ 6.5 x 10^-5 = X^2 / (0.104-X)  by solving for X

∴X = 0.00257 M
∴[OH-] = X = 0.00257 M

∴POH = -㏒[OH]

           = -㏒0.00257
           = 2.5

∴ PH = 14 - POH
         = 14 - 2.5
         = 11.5
b) after adding 20ML of HCL:

moles of HCl = molarity * volume
                       = 0.3625 m * 0.02 L

                       = 0.00725 moles

  

the complete neutralizes of (CH3)3N we make 0.003625 moles of (CH3)3NH+ So, now we need the Ka value of (CH3)3NH+:

and when the total volume = 0.02L + 0.025 = 0.045L

∴ [ (CH3)3NH+] = moles / total volume

                          = 0.003625 / 0.045L
                          = 0.08 M
 
when Ka = Kw / Kb 

and we have Kb = 6.5 x 10^-5 & we know that Kw = 1 x 10^-14 

so, by substitution:

Ka = (1 x 10^-14) / (6.5 x 10^-5)

    = 1.5 x 10^-10


when Ka expression = [(CH3)3N][H+] / [(CH3)3NH+]

by substitution:

∴ 1.5 x 10^-10 = X^2 / (0.08 - X)  by solving for X

∴X = 3.5 x 10^-6  M

∴ [H+]= X = 3.5 x 10^-6 M

∴PH = -㏒[H+]

        = -㏒(3.5 x 10^-6)

       = 5.5


C) after adding 30ML of HCl:

moles of HCl = molarity * volume 

                       = 0.3625m * 0.03L

                       = 0.011 moles

and when moles of (CH3)3N neutralized = 0.003625 moles 

∴ moles of HCl remaining    = moles HCl - moles (CH3)3N neutralized

                                                = 0.011moles - 0.003625moles

                                                = 0.007375 moles
when total volume = 0.025L + 0.03L

                                = 0.055L

∴[H+] = moles / total volume

           = 0.007375 mol / 0.055L

            = 0.134 M

∴PH = -㏒[H+]

        = -㏒ 0.134

        = 0.87
8 0
3 years ago
A chunk of pure silver weighs 20.0 g. What is its volume? Silver has a density of 10.49 g/cm cubed. Use unit analysis, NOT the D
julia-pushkina [17]

Answer: 1.91cm3

Explanation:

10.49g of silver dissolves is 1cm3 of solution.

Therefore 20g of silver will dissolve in = 20/10.49 = 1.91cm3 of the solution.

8 0
3 years ago
Consider the balanced equation below.
wolverine [178]

Answer:

a) H2S:SO2 = 2:2 and O2:H2O = 3:2

Explanation:

  • 2H2S + 3O2 → 2SO2 + 2H2O

⇒ H2S:SO2 = 2 mol H2S / 2 mol SO2 ≡ 2 : 2

⇒ O2:H2O = 3 mol O2 / 2 mol H2O ≡ 3 : 2

8 0
3 years ago
A certain element X has four isotopes. 5.845% of X has a mass of 53.93961 amu. 91.75% of X has a mass of 55.93494 amu. 2.123% of
Akimi4 [234]

Answer:

Average atomic mass of element X is 56.19 amu

Explanation:

Given:

Four isotopes of X:

5.845% of X has a mass of 53.93961 amu.

91.75% of X has a mass of 55.93494 amu.

2.123% of X has a mass of 56.93539 amu.

0.2820% of X has a mass of 57.93328 amu.

To find: average atomic mass of element X

Solution:

Isotopes are variants of a particular chemical element such that they have the same number of protons but different number of neutrons.

Average atomic mass of element X = \frac{53.93961+55.93494+56.93539+57.93328}{4}=\frac{224.74322}{4}=56.185805\approx 56.19

So, the average atomic mass of element X is 56.19 amu.

5 0
3 years ago
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