Trial ITrial IIFinal reading HCI26. 40 ml27.10 mlInitial reading HCI1.2 ml1.8 mlVolume of HCl used??Final reading NaOH33.7 ml32.8 mlInitial reading NaOH0.8 mL0.5 mlVolume of NaOH used??For trial I:Normality of NaOH : 0.25 NVol. of HCl used = final volume - initial volumeSolve for the average Normality of HCI : ?NHCI x VHCI = NNAOH x VNAOHVol. of NaOH used = final volume - initial volumeNNAOH x VNAOHFor trial II:NHCI =Vol. of HCl used = final volume - initial volumeVHCIActivaVol. of NaOH used = final volume - initial volumeGo to S Trial I.NHCITrial II.NHCI =Average NHCI =

Answered: Image /qna-images/answer/ddb13543-f489-4886-93c9-ea22fb9561f7.jpg

Trial I Trial II Final reading HCI 26. 40 ml 27.10 ml Initial reading HCI 1.2 ml 1.8 ml Volume of HCl used ? ? Final reading NaOH 33.7 ml 32.8 ml Initial reading NaOH 0.8 ml 0.5 ml Volume of NaOH used ? ?

Trial I Trial II Final reading HCI 26. 40 ml 27.10 ml Initial reading HCI 1.2 ml 1.8 ml Volume of HCl used ? ? Final reading NaOH 33.7 ml 32.8 ml Initial reading NaOH 0.8 ml 0.5 ml Volume of NaOH used ? ?

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Concept explainers

UV Visible Spectroscopy

Spectroscopy refers to the absorption or emission of light or other radiations by matter. When a substance or chemical absorbs light or radiation, they reach an excited state followed by de-excitation or emission of the light or radiation forming distinct spectra. UV-visible spectroscopy or ultraviolet-visible refers to the absorption and emission in the ultraviolet and visible regions of the electromagnetic spectrum.

Question

Trial I
Trial II
Final reading HCI
26. 40 ml
27.10 ml
Initial reading HCI
1.2 ml
1.8 ml
Volume of HCl used
?
?
Final reading NaOH
33.7 ml
32.8 ml
Initial reading NaOH
0.8 mL
0.5 ml
Volume of NaOH used
?
?
For trial I:
Normality of NaOH : 0.25 N
Vol. of HCl used = final volume - initial volume
Solve for the average Normality of HCI : ?
NHCI x VHCI = NNAOH x VNAOH
Vol. of NaOH used = final volume - initial volume
NNAOH x VNAOH
For trial II:
NHCI =
Vol. of HCl used = final volume - initial volume
VHCI
Activa
Vol. of NaOH used = final volume - initial volume
Go to S

Trial I.
NHCI
Trial II.
NHCI =
Average NHCI =

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