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Korolek [52]
3 years ago
9

The pH scale is called the logarithmic scale what does this mean?

Chemistry
1 answer:
Ratling [72]3 years ago
6 0
A logarithmic scale is a nonlinear scale used when there is a large range of quantities
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The chemical equation shows iron(III) phosphate reacting with sodium sulfate. 2FePO4 + 3Na2SO4 Fe2(SO4)3 + 2Na3PO4 What is the t
slava [35]

<u>Answer:</u> The theoretical yield of iron(III) sulfate is 26.6 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of iron(III) phosphate = 20.00 g

Molar mass of iron(III) phosphate = 150.82 g/mol

Putting values in equation 1, we get:

\text{Moles of iron(III) phosphate}=\frac{20g}{150.82g/mol}=0.133mol

The given chemical equation follows:

2FePO_4+3Na_2SO_4\rightarrow Fe_2(SO_4)_3+2Na_3PO_4

As, sodium sulfate is present in excess. So, it is considered as an excess reagent.

Thus, iron(III) phosphate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of iron(III) phosphate produces 1 mole of iron(III) sulfate

So, 0.133 moles of iron(III) phosphate will produce = \frac{1}{2}\times 0.133=0.0665moles of iron(III) sulfate

Now, calculating the mass of iron(III) sulfate from equation 1, we get:

Molar mass of iron(III) sulfate = 399.9 g/mol

Moles of iron(III) sulfate = 0.0665 moles

Putting values in equation 1, we get:

0.0665mol=\frac{\text{Mass of iron(III) sulfate}}{399.9g/mol}\\\\\text{Mass of iron(III) sulfate}=(0.0665mol\times 399.9g/mol)=26.6g

Hence, the theoretical yield of iron(III) sulfate is 26.6 grams

8 0
3 years ago
Jill graphs the speed of a model train that is travelling at a speed of 15 kilometers per hour. Which graph shows the speed of t
Tasya [4]

Answer:c

Explanation:

8 0
3 years ago
10.0 g of calcium carbonate was heated. The mass of calcium oxide left was 5.6 g. Calculate the mass of
Kipish [7]

Answer:

4.4g

Explanation:

Mass of CaCO3 = 10g

Mass of CaO = 5.6g

Mass of CO2 =?

Mass of CaCO3 = Mass of CaO + Mass of CO2

Mass of CO2 = Mass of CaCO3 — Mass of CaO

Mass of CO2 = 10 — 5.6

Mass of CO2 = 4.4g

4 0
3 years ago
Read 2 more answers
Please help, I'll give 5 stars and brainliest!
Bezzdna [24]
Huh. Is this supposed to be biology?
8 0
2 years ago
Read 2 more answers
What is the empirical formula of a compound containing 5.03 grams carbon, 0.42 grams hydrogen, and 44.5 grams chlorine?
djverab [1.8K]

The empirical formula of the compound is CHCl₃.

<h3>Calculation:</h3>

Given,

Mass of carbon = 5.03 g

Mass of hydrogen = 0.42 g

Mass of chlorine = 44.5 g

Molecular weight of carbon = 12 g

Molecular weight of hydrogen = 1 g

Molecular weight of chlorine = 35.4 g

First, calculate the moles of each element,

                      Moles = given mass/ molecular weight

Moles of carbon = 5.03/12 = 0.42

Moles of hydrogen = 0.42/1 = 0.42

Moles of chlorine = 44.5/ 35.4 = 1.26

Divide the moles of each element by the smallest number of moles,

0.42 mol of C/ 0.42 = 1 C

0.42 mol of H/ 0.42 = 1 H

1.26 mol of Cl/0.42 = 3 Cl

The ratio of elements is 1:1:3

Therefore the empirical formula of the compound will be CHCl₃.

Learn more about empirical formula here:

brainly.com/question/20708102

#SPJ4

5 0
2 years ago
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