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2 years ago

Does anyone have the data table answers for 8.03 - Solutions Lab Report?

Chemistry

1 answer:

choli [55]2 years ago

3 0

A laboratory report is an important part of the scientific process that communicate the important work that has been done.

<h3>How to depict the laboratory report?</h3>

Your information is incomplete. Therefore, an overview of a laboratory report will be given. The laboratory report is important for future studies a d experiments.

A laboratory report is broken down intosections such as title, abstract, introduction, methods, materials, results, discussion conclusion, and references.

Learn more about solutions lab report on:

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What is allotropy ?

notsponge [240]

An example of an allotrope is carbon:
Carbon can exist in graphite, diamond and amorphous

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2 years ago

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Assuming the compound is dissolved in water, what is the formula for phosphorous acid? h3p h3po4 hp h3po3

IrinaVladis [17]

I think the correct answer from the choices listed above is the last option. The formula for phosphorous acid is H3PO3. It also called orthophosphorous acid, one of several oxygen acids of phosphorus, used as reducing agent in chemical analysis. Hope this answers the question.

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2 years ago

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CaCl2 (picture) i don’t know how to do the problem

Vadim26 [7]

Answer:

Ca: 1

Cl: 2

Explanation:

Since there is no subscript but there is the element, its assumed that there is at least 1, otherwise it would'nt be there

8 0

3 years ago

The empirical formula of a chemical substance is C2H3. The molar mass of a molecule of the substance is 54.091 g/mol. What is th

Stella [2.4K]

Answer:- Molecular formula of the compound is $C_4H_6$ .

Solution:- An empirical formula and molar mass of the compound is given and we are asked to calculate it's molecular formula. First of all we calculate the empirical formula mass and then divide the molar mass by empirical formula mass to find out the number of empirical formula units.

Empirical formula is multiplied by the number of empirical formula units to get the molecular formula.

Empirical formula mass = 2(12.01) + 3(1.01)

= 24.02 + 3.03

= 27.05

number of empirical formula units = $\frac{54.091}{27.05}$

= 2

So, molecular formula of the compound = $2(C_2H_3)$

= $C_4H_6$

4 0

3 years ago

A 25.00 mL sample containing an unknown amount of Al3+ and Pb2+ required 17.14 mL of 0.04907 M EDTA to reach the end point. A 50

dolphi86 [110]

Answer:

pAl³⁺ = 1,699

pPb²⁺ = 1,866

Explanation:

In this problem, the first titration with EDTA gives the moles of Al³⁺ and Pb²⁺, these moles are:

Al³⁺ + Pb²⁺ in 25,00mL = 0,04907M×0,01714L = 8,411x10⁻⁴ moles of EDTA≡ moles of Al³⁺ + Pb²⁺.

The molar concentration is 8,411x10⁻⁴ moles/0,02500L = 0,0336M Al³⁺+Pb²⁺.

In the second part each Al³⁺ reacts with F⁻ to form AlF₃. Thus, you will have in solution just Pb²⁺.

The moles added of EDTA are:

0,02500L×0,04907M = 1,227x10⁻³ moles of EDTA

The moles of EDTA in excess that react with Mn²⁺ are:

0,02064M × 0,0265L = 5,470x10⁻⁴ moles of Mn²⁺≡ moles of EDTA

That means that moles of EDTA that reacted with Pb²⁺ are:

1,227x10⁻³ moles - 5,470x10⁻⁴ moles = 6,800x10⁻⁴ moles of EDTA ≡ moles of Pb²⁺.

The molar concentration of Pb²⁺ is:

6,800x10⁻⁴mol/0,0500L = 0,0136 M Pb²⁺

Thus, molar concentration of Al³⁺ is:

0,0336M Al³⁺+Pb²⁺ - 0,0136 M Pb²⁺ = 0,0200M Al³⁺

pM is -log[M], thus pAl³⁺ and pPb²⁺ are:

pAl³⁺ = 1,699

pPb²⁺ = 1,866

I hope it helps!

3 0

2 years ago