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Chemistry 6 Online
OpenStudy (welder55):

Please Help! I will give medal to anyone who helps. Im on connections and I need help with a chemistry lab i already conducted the lab but i need help with the questions.

OpenStudy (welder55):

OpenStudy (mtalhahassan2):

@welder55 on which question do you need help

OpenStudy (welder55):

All of them i am really bad at chemistry

OpenStudy (welder55):

well 1-7 at the bottom the others should be filled in

OpenStudy (mtalhahassan2):

ok so lets start with second

OpenStudy (mtalhahassan2):

To answer this question, you need to know the molarity equation. Do you know the molarity equation?

OpenStudy (mtalhahassan2):

come one

OpenStudy (mtalhahassan2):

try to find in your notes

OpenStudy (mtalhahassan2):

sure take your time

OpenStudy (mtalhahassan2):

wait a sec

OpenStudy (mtalhahassan2):

well this is what i find on internet Molarity=(mol solute)/(L solvent)

OpenStudy (mtalhahassan2):

ok?

OpenStudy (welder55):

so first i would have to find the amount of moles in NaHCO3?

OpenStudy (mtalhahassan2):

Now, to find the moles of NaHCO3 used, divide the molarity of the solution by the number of Liters you used for the titration. (Don't forget to convert the mL to L) .

OpenStudy (mtalhahassan2):

yes @welder55

OpenStudy (mtalhahassan2):

making sence?

OpenStudy (mtalhahassan2):

For the second part, you need to know the equation: M1V1=M2V2

OpenStudy (welder55):

For the first part i got a crazy number and i dont think its right @MTALHAHASSAN2

OpenStudy (mtalhahassan2):

@welder55 can I help you little later because I am doing a test sorry

OpenStudy (mtalhahassan2):

sorry

OpenStudy (welder55):

@mathstudent55

OpenStudy (welder55):

@welshfella @skullpatrol @Conqueror

OpenStudy (welder55):

@Preetha

OpenStudy (welder55):

@myininaya

OpenStudy (welder55):

@Michele_Laino

OpenStudy (kkutie7):

1.Write the balanced equation for the reaction of acetic acid ( COOH3CH ) and sodium bicarbonate ( 3NaHCO ). What type of reaction(s) took place? \[CH_{3}COOH + NaHCO_{3} \rightarrow CH_{3}COONa + H_{2}O + CO_{2} \] I guess I'd call it an acid base reaction. I'm not sure that it fits well with anything else.

OpenStudy (kkutie7):

2. Calculate the number of moles of NaHCO3 that were required to neutralize the COOH3CH in the vinegar. using the balanced equation we find a molar ratio of 1:1 so one mole of NaHCO3 can neutralize one mole of CH3COOH in vinegar.

OpenStudy (kkutie7):

3.Calculate the molarity of the vinegar sample. (Don’t forget to convert mL to L.) \[molarity=\frac{moles}{L} \] I'm not sure how much vinegar was used. the other questions kind of build on this so this is as far as I'm going

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