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3 years ago

Which example is a weak base? ethyl alcohol, ammonia, potassium hydroxide or vinegar?

Chemistry

1 answer:

BlackZzzverrR [31]3 years ago

7 0

Answer:

potassium

Explanation:

b/c electronegativitu

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At standard temperature and pressure. 0.500 mole of xenon gas occupies

ANEK [815]

Answer:

0.500 mole of Xe (g) occupies 11.2 L at STP.

General Formulas and Concepts:

Gas Laws

STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

Stoichiometry

Mole ratio
Dimensional Analysis

Explanation:

Step 1: Define

Identify.

0.500 mole Xe (g)

Step 2: Convert

[DA] Set up: $\displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg)$
[DA] Evaluate: $\displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg) = 11.2 \ \text{L Xe}$

Topic: AP Chemistry

Unit: Stoichiometry

3 0

2 years ago

If a 1.00 mL sample of the reaction mixture for the equilibrium constant experiment required 32.40 mL of 0.258 M NaOH to titrate

andrey2020 [161]

Answer:

The concentration of acetic acid is 8.36 M

Explanation:

Step 1: Data given

Volume of acetic acid = 1.00 mL = 0.001 L

Volume of NaOH = 32.40 mL = 0.03240 L

Molarity of NaOH = 0.258 M

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate the concentration of the acetic acid

b*Ca*Va = a*Cb*Vb

⇒with b = the coefficient of NaOH = 1

⇒with Ca = the concentration of CH3COOH = TO BE DETERMINED

⇒with Va = the volume of CH3COOH = 1.00 mL = 0.001L

⇒with a = the coefficient of CH3COOH = 1

⇒with Cb = the concentration of NaOH = 0.258 M

⇒with Vb = the volume of NaOH = 32.40 mL = 0.03240 L

Ca * 0.001 L = 0.258 * 0.03240

Ca = 8.36 M

The concentration of acetic acid is 8.36 M

6 0

3 years ago

Isoflavones, quercetin and anthocyanins are all what type of phytochemicals

kykrilka [37]

Isoflavones, quercetin, and anthocyanin are all Flavonoids.

8 0

3 years ago

The reaction H2SO4 + 2 NaOH -> Na2SO4 + 2H2O represents an acid-base tritration. The equivalence point occurs when 24.75 mL o

artcher [175]

            moles NaOH = c · V = 0.2432 mmol/mL · 24.75 mL = 6.0192 mmol
            moles H2SO4 = 6.0192 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 3.0096 mmol
Hence
            [H2SO4]= n/V = 3.0096 mmol / 38.94 mL = 0.07729 M
The answer to this question is [H2SO4] = 0.07729 M

6 0

3 years ago

The metric system of units is known as

andre [41]

Answer:

International System of Units (SI)

Explanation:

I hope it helps :)

6 0

3 years ago