Write the number in scientific notation 1123

The particle size of solute particles does not affects the rate. Question 14 options: True False

stepan [7]

The answer is false

8 0

3 years ago

Helpppp nowww plsss!!!

gayaneshka [121]

Answer:

The Sun will appear to rise and set more slowly.

5 0

2 years ago

In the laboratory, a general chemistry student measured the pH of a 0.486 M aqueous solution of triethanolamine, C6H15O3N to be

STatiana [176]

Answer:

Kb = 6.22x10⁻⁷

Explanation:

Triethanolamine, C₆H₁₅O₃N, is in equilibrium with water:

C₆H₁₅O₃N(aq) + H₂O(l) ⇄ C₆H₁₅O₃NH⁺(aq) + OH⁻(aq)

Kb is defined from concentrations in equilibrium, thus:

Kb = [C₆H₁₅O₃NH⁺] [OH⁻] / [C₆H₁₅O₃N]

The equilibrium concentration of these compounds could be written as:

[C₆H₁₅O₃N] = 0.486M - X

[C₆H₁₅O₃NH⁺] = X

[OH⁻] = X

pH is -log [H⁺], thus, [H⁺] = 10^-pH = 1.820x10⁻¹¹M

Also, Kw = [OH⁻] ₓ [H⁺];

1x10⁻¹⁴ = [OH⁻] ₓ [H⁺]

1x10⁻¹⁴ = [OH⁻] ₓ [1.820x10⁻¹¹M]

5.495x10⁻⁴M = [OH⁻], that means X = 5.495x10⁻⁴M

Replacing in Kb formula:

Kb = [5.495x10⁻⁴M] [5.495x10⁻⁴M] / [0.486M-5.495x10⁻⁴M]

Kb = 6.22x10⁻⁷

5 0

2 years ago

The compound trimethylamine, (CH3)3N, is a weak base when dissolved in water. Write the Kb expression for the weak base equilibr

Rina8888 [55]

Answer: The expression of $K_b$ is written below.

Explanation:

We are given a chemical compound which is trimethylamine that acts as a weak base when dissolved in water.

It accepts a proton from the water to form trimethylammonium ion and hydroxide ion.

The chemical equation for the reaction of trimethylamine in water follows:

$(CH_3)_3N+H_2O\rightleftharpoons (CH_3)_3NH^++OH^-$

The expression of $K_b$ for above equation follows:

$K_b=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}$

Hence, the expression of $K_b$ is written above.

5 0

3 years ago

The density of pure copper is 8.96 at 20°C. If 5.00 g of pure copper pellets is added to a graduated cylinder containing 14.6 mL

Triss [41]

Answer:

15.2 mL

Explanation:

First we calculate the volume of the copper

$volume = \frac{mass}{density}= \frac{5.00 g}{8.96 g/mL} = 0.5580 mL\\$

when this volume of copper is added to the water, the water level will rise by the level of the volume of copper. So the final volume is:

14.6 mL + 0.5580 mL = 15.158 mL

Since the measuring cylinder is graduated to one decimal place , we can round this up to 15.2 mL

8 0

3 years ago

Write the number in scientific notation 1123​

Write the number in scientific notation 1123