Den
This page lists questions and answers that were posted by visitors named Den.
Questions
The following questions were asked by visitors named Den.
Write each power as a product of the same factor. 5^4 a. 5 × 5 × 5 × 5 b. 4 × 4 × 4 × 4 × 4 c. 5 × 5 × 5 × 5 × 5 d. 5 × 5 × 5 My answer c.
15 years ago
Define a variable and write an expression for the phrase. the quotient of 6 times a number and 16 a.(16x) / 6 b.(6x) / 16 c.96x d.x / 96
15 years ago
Evaluate each expression. 9 ¸ 3 + 3 ´ 4 a. 24 b. 15 c. 17 d. 6 MY ANSWER 15??????
15 years ago
Evaluate each expression. 7 + (3 + 12 / 2)^2 a. 16 b. 88 c. 46 d. 256 I think 88 ????????
15 years ago
3 - 15 × 5 a. 90 b. 0 c. –72 d. 78
15 years ago
Simply reporting measures of central tendency or measures of variability will not tell the whole story. Using the following information, what else does a psychologist need to know or think about when interpreting this information? A school psychologist de...
13 years ago
Arrange the following atoms in order of increasing size from smallest to largest. Phosphorus Fluorine Chlorine Sulfur
13 years ago
Weight of the mustard package Sample: 3.02 (g) Weight of the mustard package Solution: 33.3 (g) Trial #1 Trial #2 Trial #3 Weight of Mustard Package Solution Delivered (g) : 1.09 .948 .909 Weight of NaOH Solution Delivered (g) : .354 .304 .269 Concentrati...
13 years ago
n the titration of 25.00 ml of a water sample, it took 20.590 ml of 4.995x10-3 M EDTA solution to reach the endpoint. Complete the questions below: Calculate the number of moles of EDTA required to titrate the water sample. (enter your answer with 3 signi...
13 years ago
10 A color candy was chosen randomly out of a bag. Below are the results: Color Probability Blue 0.30 Red 0.10 Green 0.15 Yellow 0.20 Orange ??? a. What is the probability of choosing a yellow candy? b. What is the probability that the candy is blue, red,...
10 years ago
If a strong acid or base is dissolved in water, will all of its parts have the concentration? For example in Ca(OH)2 = Ca + 2OH if the concentration of Ca(OH)2 is 8.11x10^-3 M, would the concentrations of Ca and 2OH be the same? or in the case of OH, the...
8 years ago
"For 4HCl + O2 <-> 2Cl2 + 2H2O : If in a container of 1.0 L we start with 2 moles of O2 and 4 moles of HCl, the molar fraction of O2 is: a) x/5 - x/2 b) 1-x / 6+x c) 5x / 5-x d) 1-x / 5-x e) 2x / 6-x " I tried solving this, but I wasn't sure if all 2 mole...
8 years ago
Answers
The following answers were posted by visitors named Den.
Ok!Thanks
15 years ago
I DON'T KNOW. b.???
15 years ago
15 ITS RIGHT?
15 years ago
I don't understand.????
15 years ago
Sorry! Which shows the expression below simplified? (3.6 × 1015) + (1.8 × 1012) a. 3.60018 × 10^15 b. 3.902 × 10^14 c. 3.618 × 10^15 d. 3.6018 × 10^15
15 years ago
-72 ?????
15 years ago
60 mph, i think. i first found the radius (since it's the same for both) then calculated the speed for curve b
13 years ago
Forget it i got the answer
13 years ago
mustard solution, how would i calculate the mass of naOH used?
13 years ago
After getting this, in order to convert the three masses into moles, what do I divide each mass by
13 years ago
299.49 J
8 years ago
They want O2 after equilibrium is reached. The problem doesn't give me the equilibrium constant. And I suppose x would be the unknown value you put when you're writing the ICE table? for example there are 2 moles of O2 at the beginning, x would be the cha...
8 years ago
I went back to look at my procedure and I noticed I made an error in the second one, I actually got (1-x / 5-x) , which is one of the choices. Could that be the answer? Or do we have to take into account all moles of O2 despite the proportion given in the...
8 years ago
So all 2 moles of O2 are used for the molar fraction formula even though only one of its moles can react to the 4 moles of HCl because '-x' is being added to it?
8 years ago
I see, thank you so much for the help! I didn't expect anyone to answer and less reply continuously haha. I'm very grateful! Thank you once again and have a great day!
8 years ago