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Rasek [7]
3 years ago
5

Pls help me ASAP thank you!!

Chemistry
1 answer:
Stolb23 [73]3 years ago
7 0
Sorry i cant even read that
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A 25.00 mL sample of of a sulfurous acid solution is completely neutralized using 17.12 mL of 0.13 M sodium hydroxide. What is t
Kryger [21]

Answer:

0.0445 M is the initial concentration of sulfurous acid.

Explanation:

H_2SO_3+2NaOH\rightarrow Na_2SO_3+2H_2O

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_3

are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=?\\V_1=25.00mL\\n_2=1\\M_2=0.13 M\\V_2=17.12 mL

Putting values in above equation, we get:

2\times M_1\times 25.00 mL=1\times 0.13 M\times 17.12 mL

x=\frac{1\times 0.13 M\times 17.12 mL}{2\times 25.00}=0.0445 M

0.0445 M is the initial concentration of sulfurous acid.

8 0
3 years ago
What can be produced by splitting unraniam 235
Ainat [17]

It splits into two equal nuclei of approx. equal mass.

7 0
3 years ago
A family pool holds 10,000.0 gallons of water. How many cubic meters is this?
Kobotan [32]

Answer:

Hey there!

10000 gallons would be 37.85 cubic meters.

Let me know if this helps :)

5 0
3 years ago
Read 2 more answers
Can someone solve this plz<br> βeta decay<br> carbon atomic mass 14 <br> atomic # 6
nordsb [41]

Answer:

14/6C your welcome

Explanation:

8 0
3 years ago
Determine the ph of a 0.188 m nh3 solution at 25°c. the kb of nh3 is 1.76 × 10-5.
Ber [7]
When NH3 is dissolved in water, it dissociates  partially producing NH4+ ions and OH- ions. It has an equation:
NH3 + H2O → NH4+ + OH- 

<span>We use the Kb expression to determine the [OH-] concentration,
</span>
<span>Kb = [NH4+] [OH-] /* [NH3] </span>

We can write NH4+ as OH- since they are of equal ratio. 
<span>(1.76*10^-5) = [OH-]² / 0.188 
</span><span>[OH-]² = 3.3088*10^-6 </span>
<span>[OH-] = 1.819*10^-3 </span>

We calculate for H+ concentration as follows:

<span>[H+] [OH-] = 10^-14 </span>
<span>[H+] = 10^-14 / [OH-] </span>
<span>[H+] = 10^-14 / (1.819*10^-3) </span>
<span>[H+] = 5.50*10^-12 </span>

<span>pH = -log [H+] </span>
<span>pH = -log (5.5*10^-12) </span>
<span>pH = 11.26</span>
4 0
3 years ago
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