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spayn [35]
1 year ago
13

1. The following results were obtained in analysis of iron ore for iron: 34.79%, 34.86%, 34.80%, 34.53% and 34.76% Calculate the

Standard Deviation for the analysis ​
Chemistry
1 answer:
lana66690 [7]1 year ago
3 0
<h2>Standart Deviation Steps:</h2>

1. Work out the mean (the simple average of the numbers).

2. Then for each number: subtract the mean and square the result.

3. Then work out the mean of those squared differences.

4. Take the square root of that.

<h2>Result:</h2>

Number of Elements in Array: 5

Series Average: 34,748

Median: 34,795

Smallest Number: 34,53

Biggest Number: 34.86

Standard Deviation: 0.12716131487209

Variance (Standard Deviation): 0.01617

Standard Deviation of the Population: 0.1137365376649

Variance (Standard Deviation of Population): 0.012936

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Calculate the percent yield of a reaction where a student collected 16.8 grams of product and was expecting to collect 49.7 gram
masya89 [10]

Answer:

\boxed{\text{33.8 \%}}

Explanation:

\text{\% yield} = \dfrac{\text{actual yield}}{\text {theoretical yield}} \times\text{100 \%}

Data:

        Actual yield = 16.8 g

Theoretical yield = 49.7 g

Calculation:

\text{\% yield} = \dfrac{\text{16.8 g}}{\text {49.7 g}} \times\text{100 \%} = \boxed{\textbf{33.8 \%}}

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2 years ago
For a concave mirror, incident light beams through C will reflect:
Katena32 [7]

Answer:

c

Explanation:

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Burka [1]
C. Tetraphosphorus trisulfide
8 0
2 years ago
Use the following information to determine the empirical formula if a compound is found to have 18.7% Li, 16.3% C, and 65.5% O A
Elina [12.6K]

Answer:

The empirical formula is Li2CO3

Explanation:

Step 1: Data given

Suppose the mass of the compound = 100 grams

The compound contains :

18.7% Li = 18.7 grams of Li

16.3 % C = 16.3 grams of C

65.0% O = 65.0 grams of O

Total = 100%

Molar mass of Li = 6.94 g/mol

Molar mass of C = 12.01 g/mol

Molar mass of O = 16 g/mol

Step 2: Calculate moles

Moles = Mass / molar mass

Moles Li = 18.7 grams / 6.94 g/mol = 2.65 moles

Moles C = 16.3 grams / 12.01 = 1.36 moles

Moles O = 65.0 grams / 16.0 g/mol = 4.06 moles

Step 3: Calculate the mol ratio

We divide by the smallest number of moles

Li: 2.65/1.36 ≈ 2

C: 1.36/1.36 = 1

O: 4.06/1.36 ≈ 3

The empirical formula is Li2CO3

3 0
3 years ago
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