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erastova [34]
4 years ago
7

Suppose a student completes an experiment with an average value of 2.9 mL and a calculated standard deviation of 0.71 mL. What i

s the minimum value within a 1 SD range of the average
Chemistry
1 answer:
Makovka662 [10]4 years ago
5 0

Answer:

The correct answer is 2.2 mL.

Explanation:

Given:

Average: 2.9 mL

SD: 0.71 mL

We can define a 1 SD range in which the value of volume (in mL) will be comprised:

Volume (mL) = Average ± SD = (2.9 ± 0.7) mL

Maximum value= Average + SD= 2.9 + 0.7 mL = 3.6 mL

Minimum value= Average - SD = 2.9 - 0.7 mL = 2.2 mL

Thus, the minimum value within a 1 SD range of the average is 2.2 mL

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The diagram below shows part of the rock cycle. (6 points)
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What is the volume of the rectangular prism below? A.38,832 mm3 B.480,000 mm3 C.499,200 mm3 D.501,120 mm3
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How many grams of sodium chloride are contained in 250.0 g of a 15% NaCl solution?
kramer

Given concentration of NaCl=15%

Means ,

In every 100g of Solution 15g of NaCl is present .

Now

  • Given mass=250g

So ,

\\ \Large\sf\longmapsto 250\times 15\%

\\ \Large\sf\longmapsto 250\times \dfrac{15}{100}

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<u>37.5g of NaCl present in 250g of solution.</u>

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3 years ago
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A solution is made by dissolving 0.0150 mol of HF in enough water to make 1.00 L of solution. At 26 °C, the osmotic pressure of
Alex777 [14]
Given:
M = 0.0150 mol/L HF solution
T = 26°C = 299.15 K
π = 0.449 atm

Required:
percent ionization

Solution:
First, we get the van't Hoff factor using this equation:
π = i MRT
0.449 atm = i (0.0150 mol/L) (0.08206 L atm / mol K) (299.15 K)
i = 1.219367

Next, calculate the concentration of the ions and the acid.
We let x = [H+] = [F-]
[HF] = 0.0150 - x

Adding all the concentration and equating to iM
x +x + 0.0150 - x = <span>1.219367 (0.0150)
x = 3.2905 x 10^-3

percent dissociation = (x/M) (100) = (3.2905 x 10-3/0.0150) (100) = 21.94%

Also,
percent dissociation = (i -1) (100) = (</span><span>1.219367 * 1) (100) = 21.94%</span>
7 0
3 years ago
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