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Wewaii [24]
3 years ago
15

1.

Chemistry
1 answer:
xenn [34]3 years ago
4 0
The answer is OH.

Hope this helps!
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Calculate the formula unit mass of Sodium hydroxide.
Ksivusya [100]

Answer:

MNa=22.989 g/mol.

MO=15.999 g/mol.

MH=1.0079 g/mol

4 0
3 years ago
Molecular chlorine and molecular fluorine combine to form a gaseous product. Under the same conditions of temperature and pressu
uranmaximum [27]

Answer:

  • Cl F₃

Explanation:

<u>1) Reactants:</u>

The reactants are:

  • <em>Molecular chlorine</em>: this is a gas diatomic molecule, i.e. Cl₂ (g)

  • <em>Molecular fluorine</em>: this is also a gas diatomic molecule: F₂ (g)

<u>2) Stoichiometric coefficients:</u>

  • <em>One volume of Cl₂ react with three volumes of F₂</em> means that the reaction is represented with coefficients 1 for Cl₂ and 3 for F₂. So, the reactant side of the chemical equation is:

        Cl₂ (g) + 3F₂ (g) →

<u>3) Product:</u>

  • It is said that the reaction yields <em>two volumes of a gaseous product;</em> then, a mass balance indicates that the two volumes must contain 2 parts of Cl and 6 parts of F. So, one volume must contain 1 part of Cl and 3 parts of F. That is easy to see in the complete chemical equation:

       Cl₂ (g) + 3F₂ (g) → 2Cl F₃ (g)

        As you see, that last equation si balanced: 2 atoms of Cl and 6 atoms of F on each side, and you conclude that the formula of the product is ClF₃.

8 0
3 years ago
which of the following methods is most appropriate to use to determine the number of different-colored components in a sample of
Andru [333]

The answer is paper chromatography using different solvents with a range of polarities as the mobile phase.

Paper chromatography- Low-molecular-mass molecules can be separated using paper chromatography based on how evenly they are distributed in the stationary and mobile phases. Paper chromatography is regarded as a potent analytical technique because of its low cost and the availability of numerous procedures for the separation of chemicals.

A small amount of a sample solution is poured onto a strip of chromatography paper in a paper chromatography experiment. After that, a solvent is used to suspend the chromatography paper. The sample solution's constituent components split out into bands of distinct hue as the solvent goes up the paper.

The speed of the chromatography process is influenced by the solvent's polarity. Therefore, we may conclude that all of the other components in the mixture move more quickly during the chromatography experiment if the solvent's polarity is increased.

Thus, answer is paper chromatography using different solvents with a range of polarities as the mobile phase.

To learn more about paper chromatography refer- brainly.com/question/1394204

#SPJ4

3 0
1 year ago
How many miles of calcium oxide will be produced when 1.6 miles of iron (III) oxide react with calcium phosphate
Hitman42 [59]
What is this for. Is it history
5 0
4 years ago
How many grams of sodium acetate ( molar mass = 83.06 g/mol ) must be added to 1.00 Liter of a 0.200 M acetic acid solution to m
Pie

<u>Answer:</u> The mass of sodium acetate that must be added is 30.23 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of acetic acid solution = 0.200 M

Volume of solution = 1 L

Putting values in above equation, we get:

0.200M=\frac{\text{Moles of acetic acid}}{1L}\\\\\text{Moles of acetic acid}=(0.200mol/L\times 1L)=0.200mol

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[\text{salt}]}{[\text{acid}]})  

pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})

We are given:

pK_a = negative logarithm of acid dissociation constant of acetic acid = 4.74

[CH_3COONa]=?mol  

[CH_3COOH]=0.200mol

pH = 5.00

Putting values in above equation, we get:

5=4.74+\log(\frac{[CH_3COONa]}{0.200})

[CH_3COONa]=0.364mol

To calculate the mass of sodium acetate for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of sodium acetate = 83.06 g/mol

Moles of sodium acetate = 0.364 moles

Putting values in above equation, we get:

0.364mol=\frac{\text{Mass of sodium acetate}}{83.06g/mol}\\\\\text{Mass of sodium acetate}=(0.364mol\times 83.06g/mol)=30.23g

Hence, the mass of sodium acetate that must be added is 30.23 grams

7 0
3 years ago
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