Kelsy

This page lists questions and answers that were posted by visitors named Kelsy.

Questions

The following questions were asked by visitors named Kelsy.

Hy Dr Bob222 this will be the last time I will bother you about this type of question. Can you please tell what I am doing wrong? This is the best answer I could come up with: Question: Strychnine, C21H22N2O2(aq) is a weak base but a powerful poison. Calc...

In which sentence are the conjunctions used correctly? A. The top award was given to neither Steve or Jim. B. Either Fred nor John knew about the accident. C. Mary enjoyed neither roller skating or ice skating. D. Neither beauty nor popularity was conside...

I need to calculate the amount of NaH2PO4 (in grams) and 1.00M NaOH solution (in ml) needed to make 250 ml of 0.20M H2PO4- at pH 8.00. My work is lengthy and involved but the answer I got was 7.00 ml of the NaOH and 0.0070 moles of the NaH2PO4 needed. Can...

I would like to check my answer to the following question: The permanganate ion oxidizes bromide in acidic conditions to form bromine, manganese(II) ion and water. Give the balanced equation of the reaction. My answer: 3e- + 8H+ (aq) + MnO4- (aq) + 2Br- (...

The Martian moon Phobos travels once around the planet Mars every 7 hours and 39 minutes. What is the angular velocity of Phobos in radians/second?

Answers

The following answers were posted by visitors named Kelsy.

I get 4.0 x 10^-3 but im incorrect

No Sir, thank you :)

This would be the cell membrane or in a plant cell, the cell wall.

y= 587,000(0.88)^x so.. y= 587,000(0.88)^10 y=587000 x 0.278500976 y=163480.0729

x=-7 and x=-4 x+7=0 and x+4=0 (x+7)(x+4) = x^2 + 4x + 7x +28 the equation would be x^2 + 11x + 28

It seems to me that it is a hyperbole (exaggeration )

It sounds to me like the calming reaction would be due to her parasympathetic nervous system, as she was originally in fight or flight mode and had a sympathetic nervous system reaction.

Hmmm, ok. True. Ok I will try again. Thank you.

Can you suggest a different pka or what I might have done wrong? I got a pka of 7.2 by using pKa = -log(Ka) and Ka for H2PO4- is 6.3e10^-8. I know that a pKa of 7 will yield a buffer in the range of 6.2 to 8.2. I calculated I needed 6.1 g of the NaH2PO4 a...

Could you show me how you got it? I'm not sure what I did wrong.

Thanks on the arrow. I don't know. In acid solutions I need the H+ ions. Charges have to balance. I balanced using the oxidation states of each species. I did half reactions. Perhaps the water is messing me up.

Should I use H3O+ rather than H+?

No, that won't work.

I'm lost. I really don't see what I am missing in this.

I think I have it... 10e- + 16H+ + 2MnO4- + 10 Br- ---> 5 Br2 + 8H2O + 10 e- Now remove the 10e- from both sides: 16H+ + 2MnO4- + 10 Br- ---> 5 Br2 + 8H2O

Ooops I left some out: 16H+ + 2MnO4- + 10 Br- ---> 5 Br2 + 2Mn 2+ + 8H2O :) Is this it???

We overlapped. Yes I got it. Thank you.

Yes. Thank you. I totally forgot to multiply the half reactions after balancing them. Lame. I appreciate the help.

Tanyas heart beats 68 times per minute how many times does it beat in one hour