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Answers (54)
What about sqrt (1000) x sqrt(1000) = 1000
Those numbers contain zeros also
I don't agree.
C'mon Bot. 200 contains a zero
50 contains a zero. 20 contains a zero
Hey Bot, with regard to question 1, if color does not affect the temperature then why are black objects warmer than white objects.
This is a better answer than the former one.
Bot, would you like to reconsider your answer?
Let's look at the two answers Bot gave 41/90 = 0.4555555555.........on my calculator and 9/20 = 0.45 on my calculator. What's wrong with the first answer? What's wrong with the second answer?
Let me point out that trioxonitrate V acid is not the name of choice for nitric acid, HNO3.
I don't think so>
You're guessing.
I don't think so
bOT==Did you look up the inverse sin of sin 70/1.45 correctly
Bot--You're right that powdered sugar contains starch and that will react with the iodine to give the iodine-starch deep blue color. The cornstarch will give that color also. Actually, then there are two answers; i.e., corn starch and powdered sugar.
Bot, are you sure that baking soda is the correct response?
Bot, that is the absolute value of subtracting -22 and -112
Another typo in the problem is 0.02 mol. We ASSUME that is the molar concentration but the problem uses the wrong terminology. The problem should have been 0.02 M and not mol.
I would like to point out that this is a flawed question. 1. The correct Ka value for HCOOH is 1.78 x 10^-4 and not 10^-14; i.e. it is a much strong acid that indicated in the problem and 2, Note that the answer for your solution is correct BUT that a pH
Bot, when I worked the problem I came up with 15.9. Is that correct?
That was a struggle
You've made three mistakes. 16 + 3 = 11 + 8------19 - 19 is true 16 = 16 16 - 4 = 7 + 6 ------12 = 13 is not true 16 = 19 - 3 ------- 16 = 16 is true Now which one is not true?
Bot, you can't add and subtract
guess again
finally guess correctly
try again
try that again bot
5:15 + 40 = 5:55 p.m.
Here is the correct solution. moles NaOH = M x L = 0.180 x 0.070 = 0.0126 2NaOH + H2SO4 ==> Na2SO4 + 2H2O Convert moles NaOH to mols H2SO4 this way. 0.0126 mols NaOH x (1 mol H2SO4/2 moles NaOH) = 0.0063 Then M H2SO4 = mols/L = 0.0063 mols/0.030 L = 0.21 M
I don't believe it Bot.
Still not right. Here is the correct balanced equation. 2C5H10O2(l) + 13O2(g) → 10CO2(g) + 10H2O(g)
The O atoms still don't balance. I see 16 O atoms on the left and 15 on the right.
Bot, the O atoms aren't balanced. I see 18 O atoms on the left and only 15 on the right.
Bot, this is incorrect. None of the work with H2SO4 is needed. moles NaOH = M x L = 0.180 x 0.070 = 0.0126 grams NaOH = moles NaOH x molar mass NaOH = 0.0126 mol x 40 g/mol = 0.0504 g NaOH
From your statement: Moles of KOH used = 0.00749 mol (calculated using M x L). so M = 0.249 and L = 0.030 = 0.00747 and not 0.00749
Bot, that is incorrect. M KOH is correct @ 0.249 mols/dm^3. H2SO4 + 2KOH --> K2SO4 + 2H2O mols KOH used = M x L = 0.249 x 0.030 L = 0.00749 mols H2SO4 = 0.00749 moles KOH x (1 mol H2SO4/2 moles NaOH) = 0.003735 M H2SO4 = moles H2SO4/dm^3 H2SO4 or dm^3
Bot, you're wrong again. moles Li3N formed from 0.432 mols Li follows: 0.432 mol Li × (1/3 mol Li3N/1 mol Li) = 0.144 mol Li3N This next step is your error which I've corrected. moles Li3N formed from 0.107 moles N2 follows: 0.107 mol N2 × (2 mole Li3N /
Hey Bot. No, moles Li3N formed by 3 g N2 is GREATER than moles formed by 3 g Li; therefore, Li is the limiting reagent. I think you should rethink and recalculate.
If I look up the heat of combustion of methane I find 890 kJ/mol which doesn't fit any of the answers. I think the question is flawed.
Hey Bot. Your answer of 86.60 meters is correct as is the 28.87 meters. The total distance is 86.60 m + 28.87 m = 115.47 m which is answer D
Bot, I think you rational is incorrect but that Pt is +4.
I don't believe it. I think Pt must be +4
a).[Pt(en)2Cl2]2+ Hey Bot. If Pt is +2 and 2*-1 = -2 for Cl2 that is 0 but it should be +2 for total charge on the complex. What do you think?
That's right. (rate He/rate N2) = sqrt (molar mass N2/molar mass He) (rate He/rate N2 = sqrt (28/4) rate He/rate N2 = sqrt 7 = 2.64. Therefore, rate He = 2.64*rate N2.
Given that the He atoms move faster than N2 atoms then how much faster does a He atom move than a N2 atom.
Bot---Doesn't a He atom move faster than a N2 atom?
The Bot is wrong again. The molar mass of Al2(SO4)3.2H2O is correctly listed as 378. How many O atoms are in the molecule. There are 3*4 in the (SO4)3 + 2 in 2H2O = 14 The mass of the 14 O atoms = 14*16 = 224 mass % O in Al2(SO4)3.2H2O = (224/378)*100 =
Bot, I disagree. When I took my first class in analytical chemistry over 75 years ago I was told to record my data in a bound notebook AND to use permanent ink. Under no circumstance was I to use a graphite pencil because graphite was not permanent.
Hey Bot--You have a problem here. You didn't factor in making only 200 mL of solution.
Bot==Still not right. The molar mass of Na2CO3.10H2O is Na: 2(22.99 g/mol) = 45.98 g/mol C: 1(12.01 g/mol) = 12.01 g/mol O: 3(16.00 g/mol) = 48.00 g/mol H2O: 10(18.01) = 180.1
