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musickatia [10]
3 years ago
9

Find value of x(6xx) + (7x100-366.94100100​

Chemistry
1 answer:
Alla [95]3 years ago
5 0

Answer:

The annswer is 6.94

Explanation:

The answer is 6.94

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You are given a sulfuric acid solution of unknown concentration. You dispense 10.00 mL of the unknown solution into an Erlenmeye
dmitriy555 [2]

Answer:

"0.053457 M" of sulfuric acid.

Explanation:

The given values are:

V = 10 mL solution

V_{added} = 12.20 mL

V_{total} = 22.20 mL

then,

M 0.103 M of NaOH,

V_{rinsed} = experiment  will not be affected

V_{total \ base} = 10.38 mL

Now,

⇒  mol of NAOH = MV

                            = 0.103\times 10.38

                            =  1.06914  \ m

Whether Sulfuric acid, then

⇒  H_{2}SO_{4} + 2NaOH = Na_{2}SO_{4} + 2H_{2}O

⇒  mol \ of \ acid =\frac{1}{2}\times \ mol \   of  \ base

⇒  1.06914 \ m \ mol \ of \ base = \frac{1}{2}\times 1.06914 = 0.53457 \ m \ mol \ of \ acid

Before any dilution:

V_{sample} = 10  \ mL

⇒  M \ acid = \frac{m \ mol}{V}

                 =\frac{ 0.53457 }{10}

                 =0.053457 \ M (Sulfuric acid)

6 0
3 years ago
Why are group 1a (1) and group 2a (2) elements found in many compounds, but not group 8a (18) elements?
Rashid [163]
Because group 1a and 2a are most reactive due to their least number of electrons in their outmost shell compared to group 8a(18)
4 0
3 years ago
Read 2 more answers
Calculate the hydrogen-ion concentration [H+] for the aqueous solution in which [OH–] is 1 x 10–11 mol/L. Is this solution acidi
valina [46]
(H+)(OH-) = Kw
kw=  1 x10^-14
OH-=  1   x10 ^-11
(H+)=  KW / OH-

concentration   of H+  = (1x10^-14) /.(1  x 10 ^-11)   =  1  x10  ^-3

Ph=  -log (H+)

PH=-log (  1  x  10  ^-3)  =  3  therefore  the    solution  is  acidic  since  the  PH   less than  7

7 0
3 years ago
15A.4(a) What is the temperature of a two-level system of energy separation equivalent to 400 cm−1 when the population of the up
maksim [4K]

Answer:

T = 525K    

Explanation:

The temperature of the two-level system can be calculated using the equation of Boltzmann distribution:

\frac{N_{i}}{N} = e^{-\Delta E/kT}  (1)

<em>where Ni: is the number of particles in the state i, N: is the total number of particles, ΔE: is the energy separation between the two levels, k: is the Boltzmann constant, and T: is the temperature of the system </em>         

The energy between the two levels (ΔE) is:

\Delta E = hck    

<em>where h: is the Planck constant, c: is the speed of light and k: is the wavenumber</em>      

\Delta E = 6.63\cdot 10^{-34} J.s \cdot 3\cdot 10^{8}m/s \cdot 4 \cdot 10^{4}m^{-1} = 7.96 \cdot 10^{-21}J  

Solving the equation (1) for T:

T = \frac{-\Delta E}{k \cdot Ln(N_{i}/N)}  

<em>With Ni = N/3 and k = 1.38x10⁻²³ J/K, </em><em>the temperature of the two-level system is:</em><em> </em>

T = \frac{-7.96 \cdot 10^{-21}J}{1.38 \cdot 10^{-23} J/K \cdot Ln(N/3N)} = 525K                                  

I hope it helps you!

3 0
3 years ago
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Answer:

avery

Explanation:

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