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stepladder [879]
2 years ago
14

A certain first-order reaction has a half-life of 25.2 s at 20°C. What is the value of the rate constant k at 60°C if the activa

tion energy, Ea=80 kJ mol-1?
Chemistry
1 answer:
olasank [31]2 years ago
3 0

Answer:

hi

Explanation:

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xenn [34]

Answer:

Its C hope it helped

Explanation:

3 0
2 years ago
Match the following aqueous solutions with the appropriate letter from the column on the right.1. 0.19 m AgNO3 2. 0.17 m CrSO4 3
vichka [17]

Answer:

0.13 m of Mn(NO_3)_2 → Highest boiling point

0.19 m of AgNO_3 → Second  Highest boiling point

0.17 m of CrSO_4 → Third highest boiling point

0.31 m Sucrose (nonelectrolyte)  → Lowest boiling point

Explanation:

Elevation in boiling is given by :

\Delta T_b=i\times k_b\times m

Where :

i = van't Hoff factor

k_b= Molal Elevation constant of solvent

m = molaity of the solution

1) 0.19 m of AgNO_3

AgNO_3\rightarrow Ag^++NO_3^{-}

i = 2 (electrolyte)

Molality of the solution = 0.19

Elevation is boiling point of solution:

\Delta T_b=2\times k_b\times 0.19 m

\Delta T_b=0.38 m\times k_b

2) 0.17 m of CrSO_4

CrSO_4\rightarrow Cr^{2+}+SO_4^{2-}

i = 2 (electrolyte)

Molality of the solution = 0.17

Elevation is boiling point solution :

\Delta T_b=2\times k_b\times 0.17 m

\Delta T_b=0.34 m\times k_b

3) 0.13 m of Mn(NO_3)_2

Mn(NO_3)_2\rightarrow Mn^{2+}+2NO_3^{-}

i = 3 (electrolyte)

Molality of the solution = 0.13

Elevation is boiling point solution :

\Delta T_b=3\times k_b\times 0.13 m

\Delta T_b=0.39 m\times k_b

4) 0.31 m Sucrose (nonelectrolyte)

i = 1 ( non electrolyte)

Molality of the solution = 0.31 m

Elevation is boiling point solution :

\Delta T_b=1\times k_b\times 0.31 m

\Delta T_b=0.31 m\times k_b

Higher the value of elevation in temperature higher will be the boiling point of the solution .

The decreasing order of solution from highest boiling point to lowest boiling point is :

0.39 m\times k_b>0.38 m\times k_b>0.34 m\times k_b>0.31 m\times k_b

0.13 m of Mn(NO_3)_2 → Highest boiling point

0.19 m of AgNO_3 → Second  Highest boiling point

0.17 m of CrSO_4 → Third highest boiling point

0.31 m Sucrose (nonelectrolyte)  → Lowest boiling point

6 0
3 years ago
if it takes 54 mL of 0.1 NaOH to neutralize 125 mL of an HCL solution, what is the concentration of HCL?
OleMash [197]

Answer:

0.0432M

Explanation:

We begin by writing a balanced equation for the reaction. This is illustrated below:

NaOH + HCl —> NaCl + H2O

From the equation above,

The number of mole of the acid (nA) = 1

The number of mole of the base (nB) = 1

Data obtained from the question include:

Vb (volume of the base) = 54mL

Cb (concentration of the base) = 0.1M

Va (volume of the acid) = 125mL

Ca ( concentration of the acid) =?

Using CaVa/CbVb = nA/nB, the concentration of the acid can easily be obtained as shown below:

CaVa/CbVb = nA/nB

Ca x 125 / 0.1 x 54 = 1

Cross multiply to express in linear form:

Ca x 125 = 0.1 x 54

Divide both side by 125

Ca = (0.1 x 54) / 125

Ca = 0.0432M

Therefore, the concentration of the acid is 0.0432M

3 0
3 years ago
Send Message
Ganezh [65]

Answer:

1 Atm

Explanation:

Dalton's law

The total pressure is 3 Atm so all you have to do is subtract the other partial pressures from 3

4 0
3 years ago
Read 2 more answers
The law of conservation of mass states that mass is neither created nor destroyed during a chemical reaction. This can be gleane
svet-max [94.6K]

Answer:

Explanation:

The law of conservation of mass states that mass is neither created nor destroyed during a chemical reaction. This can be gleaned from the third postulate in Dalton's series. Magnesium oxide decomposes into magnesium and oxygen. If 4.03 g of magnesium oxide decomposes to form 2.43 g of magnesium, what mass of oxygen gas is also released in the reaction

The word say magnesium oxide decomposes to magnesium and oxygen

the chemical symbols say

MgO-----------> Mg + O2  (since natural oxygen is diatomic)

the balanced  equation says

2MgO-------------->2Mg + O2

4.03 gm----------> 2.43 + ?0 gms

tour high school Algebra I class says

? = 4.03 -2.43 =1.60

your chemical analytcal lab says %mO in MgO = 16/40.3 = 39.7%

your calculator says

39.7/100 X 4.03 = 1.60

all of these prove the law of conservation of mass

3 0
2 years ago
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