If you take a 15.0-mL portion of the stock solution and dilute it to a total volume of 0.600 L , what will be the concentration

Iteams in house made of thermosetting plastics?

Thermoplastics and thermosetting polymers Examples include: polyethylene (PS) and polyvinyl choline (PVC). Common thermoplastics range from 20,000 to 50,000 amu, while thermosets are assumed to have infinite molecular weight.

3 0

3 years ago

Lab report on flock flock y flock z insects and fruits

n200080 [17]

Answer:

In this lab, you will learn the anatomy of an insect, how to identify an insect to order, how to collect and curate insects,

Explanation:

7 0

2 years ago

Help me please <br> Answer & exp.

nordsb [41]

Explanation:

Some Rules Regarding Oxidation Numbers:

- Hydrogen has oxidation number of + 1 except in hydrides where it is -1

- Oxygen has oxidation number of -2 except in peroxides where it is -1

- Some elements have fixed oxidation numbers. E.g Halogen group elements has oxidation number of -1

- Oxidation number of a compound is the sum total of the individual elements and a neutral compound has oxidation number of 0.

A. HI

Hydrogen has oxidation of + 1

Oxidation number of I:

1 + x = 0

x = -1

B. PBr3

Br has oxidation number of - 1

Oxidation number of Pb:

x + 3 (-1) = 0

x = + 3

C. KH

Hydrogen has oxidation of + 1

Oxidation number of K:

1 + x = 0

x = -1

D. H3PO4

Hydrogen has oxidation number of + 1

Oxygen has oxidation number of -2

Oxidation number of P:

3(1) + x + 4(-2) = 0

3 + x - 8 =0

x = 5

8 0

3 years ago

What is the molarity of a 1.44 (m/v)% solution of an HCOOOH (Formic acid)?

AlexFokin [52]

This problem is providing the mass-volume percent of a formic acid solution so its molarity is required and found to be 0.313 M after the following calculations.

<h3>Molarity</h3>

In chemistry, units of concentration provide a measurable understanding of the relationship between the relative amounts of both solute and solvent. In the case of molarity, one must relate moles of solute and liters of solution as follows:

$M=\frac{mol\ solute}{Volume\ solution \ in \ L}$

In such a way, when given this mass-volume percent of 1.44% for the formic acid in the solution, one can assume there is 100 mL of solution and 1.44 g of solute (formic acid), which means one must convert the volume to liters and the mass to moles with:

$mol\ solute=\frac{1.44g}{46.03g/mol} =0.0313mol\\\\Volume\ in \liters: 100mL*\frac{1L}{1000mL}=0.100L$