If it takes 65.3 min for the concentration of a reactant to drop to 20.0% of its initial value in a first-order reaction, what is the rate constant for the reaction in the units min-1?

Answers

Answer 1

Answer:

$65.3 / 20$

equals 3.265

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How many molecules are in 9.8 grams of calcium

Answers

Hmm
One mole of calcium is usually equal to 40 grams so divide 9.8 by 40 which would be 4.0 molecules

True or false Diprotic acids lose both hydrogens at the same time

Answers

Best Answr : Falsee im sureee .

Describe a polyhydroxyl alcohol.

Answers

To know that a certain compound is classified as an alcohol is through it molecular equation. Alcohols have at least 1 -OH or hydroxyl attachment. When we say polyhydroxyl that means there is 2 or more hydroxyl group in the structure of an alcohol. Ethylene glycol is an example.

A polyhydroxyl alcohol is an alcohol that has two OH molecules attached. A regular alcohol has only one.

Which colligative property is extremely important to the functioning of living organisms?a. osmotic pressure
b. boiling-point elevation
c. vapor-pressure lowering
d. freezing-point depression

Answers

I think the correct answer is A. Osmotic pressure is the colligative property that is extremely important to the functioning of living organisms. The osmotic pressure is the pressure required to prevent osmosis from happening.

Answer:

its D

Explanation:

i just got it correct on usatestprep

The value of ΔH° for the reaction below is -126 kJ. The amount of heat that is released by the reaction of 25.0 g of Na2O2 with water is __________ kJ. 2Na2O2(s) + 2H2O(l) → 4NaOH(s) + O2(g) The value of ΔH° for the reaction below is -126 kJ. The amount of heat that is released by the reaction of 25.0 g of Na2O2 with water is __________ kJ. 2Na2O2(s) + 2H2O(l) → 4NaOH(s) + O2(g) 40.4 80.8 -126 67.5 20.2

Answers

The amount of heat that is released by the chemical reaction of 25.0 g of $Na_(2)O_2$ with water is -20.223 Joules.

Given the following data:

Mass of $Na_(2)O_2$ = 25.0 grams

Enthalpy of combustion = -126 kJ/mol

To find the amount of heat that is released by the chemical reaction of 25.0 g of $Na_(2)O_2$ with water:

First of all, we would determine the number of moles of $Na_(2)O_2$ in this chemical reaction:

$2Na_2O_2_((s)) + 2H_2O_((l))$ ------> $4NaOH_((s)) + O_2_((g))$

$Number\;of\;moles \;(Na_(2)O_2)= (Mass\; of\;Na_(2)O_2)/(Molar\;mass\;of\;Na_(2)O_2)$

Substituting the values into the formula, we have;

$Number\;of\;moles \;(Na_(2)O_2)= (25.0)/(77.98)$

Number of moles ( $Na_(2)O_2$ ) = 0.321 moles.

Now, we can find the quantity of heat released when $Na_(2)O_2$ reacts with water:

2 mole of $Na_(2)O_2$ = -126 kJ/mol

0.321 mole of $Na_(2)O_2$ = X kJ/mol

Cross-multiplying, we have:

$2X = 0.321$ × $(-126)$

$2X = -40.446\n\nX = (-40.446)/(2)$

X = -20.223 Joules.

Answer : The amount of heat released by the reaction is, 20.2 kJ

Explanation :

First we have to calculate the number of moles of $Na_2O_2$ .

$\text{Moles of }Na_2O_2=\frac{\text{Mass of }Na_2O_2}{\text{Molar mass of }Na_2O_2}$

Molar mass of $Na_2O_2$ = 77.98 g/mole

$\text{Moles of }Na_2O_2=(25.0g)/(77.98g/mole)=0.320mole$

Now we have to calculate the heat released during the reaction.

The balanced chemical reaction is:

$2Na_2O_2(s)+2H_2O(l)\rightarrow 4NaOH(s)+O_2(g)$

From the reaction we conclude that,

As, 2 moles of $Na_2O_2$ releases heat = 126 kJ

So, 0.320 moles of $Na_2O_2$ releases heat = $(0.320)/(2)* 126=20.2kJ$

Therefore, the amount of heat released by the reaction is, 20.2 kJ

Based on the Arrhenius theory, when potassiumhydroxide dissolves in water, the only negative
ion in the aqueous solution is
(1) O2 (aq) (3) H (aq)
(2) OH2 (aq) (4) OH (aq)

Answers

Answer : The correct option is, (4) $OH^-(aq)$ .

Explanation :

According to the Arrhenius theory, an acid is a substance that ionizes in water to gives $H^+$ ions (hydronium ions or hydrogen ions) and a base is a substance which ionizes in water to gives $OH^-$ ions (hydroxide ions).

As we know that when the potassium hydroxide (KOH) dissolves in water or ionizes in water to gives $OH^-$ ions (hydroxide ions) and $K^+$ ions (potassium ion). That means $OH^-$ ion is a negative ion in aqueous solution and $K^+$ ions is a positive ion in aqueous solution.

Hence, the correct option is, (4) $OH^-(aq)$ .

(4) $OH^(-)$ (aq)

That's the answer.

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