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4 years ago

(WILL MARK BRAINLIEST) PLZ HELP ASAP

Chemistry

2 answers:

ArbitrLikvidat [17]4 years ago

3 0

Answer:

a) 2NaOH(aq) + CuSO4(aq) -------------> Cu(OH)2(s) + Na2SO4(aq)

b) Ca(OH)2(aq) + CO2(g) --------------> CaCO3 + H2O (this is already balanced)

c) Pb(NO3)2 + H2SO4 --------> PbSO4 + 2HNO3.

d) 2KNO3 ------> 2KNO2 + O2

e) H2SO4 + 2(NaOH) -----> Na2SO4 + 2(H2O)

f) Ca(NO3)2(aq) + (NH4)2CO3(aq) ----------------> CaCO3(s) + 2NH4NO3(aq)

ikadub [295]4 years ago

3 0

Answer:

a) 2NaOH(aq) + CuSO4(aq) ----> Cu(OH)2(s) + Na2SO4(aq)

b) Ca(OH)2(aq) + CO2(g) --> CaCO3 + H2O (the chemical equation is already balanced as is)

c) Pb(NO3)2 + H2SO4 --> PbSO4 + 2HNO3

d) 2KNO3 --> 2KNO2 + O2

e) H2SO4 + 2(NaOH) --> Na2SO4 + 2(H2O)

f) Ca(NO3)2(aq) + (NH4)2CO3(aq) --> CaCO3(s) + 2NH4NO3(aq)

Explanation:

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What volume (mL) of the partially neutralized stomach acid was neutralized by NaOH during the titration? (portion of 25.00 mL sa

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The question is incomplete, here is the complete question:

What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)

<u>Answer:</u> The volume of HCl neutralized is 1.25 mL

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

$n_1M_1V_1=n_2M_2V_2$

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

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=2M\\V_1=?mL\\n_2=1\\M_2=0.1M\\V_2=25mL$

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$1\times 2\times V_1=1\times 0.1\times 25\\\\V_1=\frac{1\times 0.1\times 25}{1\times 2}=1.25mL$

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