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3 years ago

A 1 mL pipette delivers a measured volume of .98 mL (calculate the percentage error)

Chemistry

2 answers:

Natalija [7]3 years ago

4 0

Answer:

Explanation:

.98 is 98% of one and therefore they are missing 2%

r-ruslan [8.4K]3 years ago

4 0

Answer:

Explanation:

The answer is 2%.

100-98 is 2.

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What is the mass of 1.58 moles of CH4

HACTEHA [7]

<h3>Answer:</h3>

25.4 g CH₄

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

1.58 mol CH₄

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of C - 12.01 g/mol

[PT] Molar Mass of H - 1.01 g/mol

Molar Mass of CH₄ - 12.01 + 4(1.01) = 16.05 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 1.58 \ mol \ CH_4(\frac{16.05 \ g \ CH_4}{1 \ mol \ CH_4})$
Multiply/Divide: $\displaystyle 25.359 \ g \ CH_4$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

25.359 g CH₄ ≈ 25.4 g CH₄

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I think B. As an idea is just a way that could be possible

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One of the reactions that occurs in a blast furnace, in which iron ore is converted to cast iron, is Fe2O3 + 3CO → 2Fe + 3CO2 Su

Tpy6a [65]

Answer : The percent purity of $Fe_2O_3$ in the original sample is 87.94 %

Explanation :

The given balanced chemical reaction is:

$Fe_2O3+3CO\rightarrow 2Fe+3CO_2$

First we have to calculate the mass of Fe.

$\text{Moles of }Fe=\frac{\text{Mass of }Fe}{\text{Molar mass of }Fe}$

Molar mass of Fe = 55.8 g/mole

$\text{Moles of }Fe=\frac{1.79\times 10^3kg}{55.8g/mole}=\frac{1.79\times 10^3\times 1000g}{55.8g/mole}=3.15\times 10^4mole$

Now we have to calculate the moles of $Fe_2O_3$

From the balanced chemical reaction we conclude that,

As, 2 moles of Fe produced from 1 mole of $Fe_2O_3$

So, $3.15\times 10^4mole$ of Fe produced from $\frac{3.15\times 10^4}{2}=15750$ mole of $Fe_2O_3$

Now we have to calculate the mass of $Fe_2O_3$

$\text{ Mass of }Fe_2O_3=\text{ Moles of }Fe_2O_3\times \text{ Molar mass of }Fe_2O_3$

Molar mass of $Fe_2O_3$ = 159.69 g/mole

$\text{ Mass of }Fe_2O_3=(15750moles)\times (159.69g/mole)=2.515\times 10^6g=2.515\times 10^3kg$

Now we have to calculate the percent purity of $Fe_2O_3$ in the original sample.

Mass of original sample = $2.86\times 10^3kg$

$\text{Percent purity}=\frac{\text{Mass of }Fe_2O_3}{\text{Mass of sample}}\times 100$

$\text{Percent purity}=\frac{2.515\times 10^3kg}{2.86\times 10^3kg}\times 100=87.94\%$

Therefore, the percent purity of $Fe_2O_3$ in the original sample is 87.94 %

3 0

3 years ago