What is one thing chemical properties and physical proprieties of matter have in common

A 0.75M solution of CH3OH is prepared in 0.500 kg of water. How many moles of CH3OH are needed?

4vir4ik [10]

Answer:

We need 0.375 mol of CH3OH to prepare the solution

Explanation:

For the problem they give us the following data:

Solution concentration 0,75 M

Mass of Solvent is 0,5Kg

knowing that the density of water is 1g / mL, we find the volume of water:

$d = \frac{g}{mL} \\\\ V= \frac{g}{d} = \frac{500g}{1 \frac{g}{mL} } = 500mL = 0,5 L$

Now, find moles of $CH_{3} OH$ are needed using the molarity equation:

$M = \frac{ moles }{ V (L)} \\\\\\molesCH_{3}OH = M . V(L) = 0,75 M . 0,5 L\\\\molesCH_{3}OH = 0,375 mol$

therefore the solution is prepared using 0.5 L of H2O and 0.375 moles of CH3OH, resulting in a concentration of 0,75M

5 0

3 years ago

Change the bond between the two carbon atoms in each molecule to a double or triple bond as needed to complete the structure. If

Georgia [21]

Answer:

Your question is complex, because I think you wrote it wrong.

Although in front of this what I can help you is that the carbons are associated between a single, double or triple union.

This depends on whether they are attached to more or less carbons or hydrogens, the carbons have the possibility of joining 4 radicals, both other carbons and hydrogens.

Simple junctions talks about compound organisms called ALKANS.

The double unions, in organic these compounds are called as ALQUENOS.

And as for the tertiary unions, the organic chemistry names them as ALQUINOS.

These compounds that we write, a simple union, the less energy, the less this union, that is why the triple bond is the one that contains the most energy when breaking or destroying it in a reaction.

Explanation:

In a chemical compound the change of these unions if we modified them we would generate changes even in the classifications naming them as well as different compounds and not only that until they change their properties

6 0

3 years ago

In one of the routine analyses, Dr Entropy found that the concentration of phosphate had increased from 0.0090 mg/L to 0.220 mg/

wariber [46]

Answer:

The final volume should be 22 mL

Explanation:

For this problem, we will use the dilution equation:

C1*V1 = C2*V2

<u>Step 1</u>: Data given

with C1 = the initial concentration C1 = 0.220 mg/L

with V1 = the initial volume = 10 mL = 10 * 10^-3 L

with C2 = the final concentration = 0.100 mg/L

with V2 = the final volume = TO BE DETERMINED

<u>Step 2</u>: Calculating the final volume

C1*V1 = C2*V2

0.220 mg/L * 10*10^-3 L = 0.100 mg/L * V2

V2 = (0.220 mg/L * 10*10^-3 L) / 0.100 mg/L

V2 =0.022 L = 22 mL

The final volume should be 22 mL

3 0

4 years ago

What is the charge of phosphate in AlPO4?

Dennis_Churaev [7]

The charge of phosphate : 3-

<h3>Further explanation</h3>

Oxidation is an increase in oxidation number, while reduction is a reduction in oxidation number.

The formula for determining Oxidation Numbers in general:

1. Single element atomic oxidation number = 0.

Group IA (Li, Na, K, Rb, Cs, and Fr): +1

Group IIA (Be, Mg, Ca, Sr and Ba): +2

Group IIIA(Al, B) : +3

H in compound = +1, except metal hydride compounds (Hydrogen which binds to groups IA or IIA) oxidation number H = -1, for example, LiH, MgH₂, etc.

2. Oxidation number O in compound = -2, except OF2 = + 2 and in peroxide (Na₂O₂, BaO₂) = -1 and superoxide, for example KO₂ = -1/2.

3 The oxidation number in an uncharged compound = 0,

Total oxidation number in ion = ion charge, Example NO₃⁻ = -1

Compound AlPO₄ :

The oxidation number in an uncharged compound = 0

Since the charge of Al as group 3A (or group 13) is +3, while the AlPO₄ compound is a polyatomic compound (consisting of polyatomic ions PO₄ and Al), the PO₄ charge:

charge Al + charge PO₄ = 0

+ 3 + charge PO₄ = 0

<em>charge PO₄ = -3</em>

6 0

3 years ago

To find S, the number of shared

Anton [14]

Answer:

It’s 6 :)

Explanation:

3 0

3 years ago