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tankabanditka [31]
2 years ago
12

A student adds 0. 0030 mol of hcl to 100 ml of a 0. 10 m solution of a r2nh, a weak base. The ph of the solution is found to be

11. 10
Chemistry
1 answer:
egoroff_w [7]2 years ago
5 0

Since an acidic salt solution is produced when a strong acid neutralizes a weak base, the pH of the salt solution formed when HCl is added to R2NH will be less than 7.

<h3>What is a neutralization reaction?</h3>

A neutralization reaction is the react ion between an acid and a base to form salt and water only.

Neutralization reactions can either produce a neutral solution, an acidic solution or an alkaline solution at equivalence point.

When a strong acid is added to a weak base, the pH of the salt solution formed will be less than 7.

Therefore, the pH of the salt solution formed when HCl is added to R2NH will be less than 7.

Learn more about pH at: brainly.com/question/940314

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Explanation:

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3 years ago
What is the molar mass of h2
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3 years ago
The factor that is changed throughout an experiment is called the _______. A. apparatus B. constant C. variable D. hypothesis
ser-zykov [4K]
The variable is what changes during an experiment. Hopefully this helped! :)
4 0
3 years ago
Read 2 more answers
5)
Kryger [21]

The empirical formula : C₁₂H₄F₇

The molecular formula : C₂₄H₈F₁₄

<h3>Further explanation</h3>

mol C (MW=12 g/mol)

\tt \dfrac{18.24}{12}=1.52

mol H(MW=1 g/mol) :

\tt \dfrac{0.51}{1}=0.51

mol F(MW=19 g/mol)

\tt \dfrac{16.91}{19}=0.89

mol ratio of C : H : O =1.52 : 0.51 : 0.89=3 : 1 : 1.75=12 : 4 : 7

Empirical formula : C₁₂H₄F₇

(Empirical formula)n=molecular formula

( C₁₂H₄F₇)n=562 g/mol

(12.12+4.1+7.19)n=562

(281)n=562⇒ n =2

Molecular formula : C₂₄H₈F₁₄

6 0
3 years ago
If the titrant has a molarity of 0.1000 m and there are 45.00 ml of analyte present, what is the molarity of the analyte?
marysya [2.9K]
Given: 

Concentration of titrant = 0.1000 M
Volume of titrant = 45 mL

The molarity of analyte depends on the amount of the analyte present in the titrated solution. If the amount of analyte is 20 mL, then its concentration is:

45ml * 0.10 M = C analyte * 20 ml

C analyte = 0.225 M
4 0
3 years ago
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