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marin [14]
3 years ago
12

Which one of the following is not a valid expression for the rate of the reaction below? 4NH3 + 7O2 → 4NO2 + 6H2O Which one of t

he following is not a valid expression for the rate of the reaction below? 4NH3 + 7O2 4NO2 + 6H2O 16 Δ[H2O]Δt 14 Δ[NO2]Δt - 14 Δ[NH3]Δt - 17 Δ[O2]Δt All of the above are valid expressions of the reaction rate.
Chemistry
1 answer:
Veseljchak [2.6K]3 years ago
3 0

Answer : All of the above are valid expressions of the reaction rate.

Explanation :

The given rate of reaction is,

4NH_3+7O_2\rightarrow 4NO_2+6H_2O

The expression for rate of reaction for the reactant :

\text{Rate of disappearance of }NH_3=-\frac{1}{4}\times \frac{d[NH_3]}{dt}

\text{Rate of disappearance of }O_2=-\frac{1}{7}\times \frac{d[O_2]}{dt}

The expression for rate of reaction for the product :

\text{Rate of formation of }NO_2=+\frac{1}{4}\times \frac{d[NO_2]}{dt}

\text{Rate of formation of }H_2O=+\frac{1}{6}\times \frac{d[H_2O]}{dt}

From this we conclude that, all the options are correct.

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Question<br> What is the molarity of a 400 mL solution containing 0.60 moles of NaCl?
Mumz [18]

Answer:

0.24 M

Explanation:

Molarity = Moles solute / Liters solution

Step 1: Identify variables

400 mL = Liters solution

0.60 moles = Moles solute

Step 2: Identify conversions

1 L = 1000 mL

Step 3: Convert mL to L

400mL(1 L/1000mL) = 0.4 L

Step 4: Find molarity

M = (0.4 L)(0.60 mol) = 0.24 M

6 0
3 years ago
The Sun always shines on half of the Moon. During a new moon, the Moon looks dark.
Sveta_85 [38]
You’re answer is C Explaination if the pit side was facing earth it wouldn’t be a new moon
8 0
3 years ago
What is the partial pressure of radon if the total pressure is 780 torr and the
Brilliant_brown [7]

Answer:

0.03atm

Explanation:

Given parameters:

Total pressure  = 780torr

Partial pressure of water vapor  = 1.0atm

Unknown:

Partial pressure of radon  = ?

Solution:

A sound knowledge of Dalton's law of partial pressure will help solve this problem.

The law states that "the total pressure of a mixture of gases is equal to the sum of the partial pressures of the constituent gases".

Mathematically;

           P_{t}   =   P_{1}  + P_{2}   +   P_{n}

Since the total pressure is 780torr, convert this to atm;

                       760torr = 1 atm

                       780torr  = \frac{780}{760} atm  = 1.03atm

     For this problem;

Total pressure  = Partial pressure of radon + Partial pressure of water vapor

            1.03 = Partial pressure of radon + 1.0

  Partial pressure of radon  = 1.03 - 1.00  = 0.03atm

6 0
3 years ago
How much 10.0 M HNO must be added to 1.00 L of a buffer that is 0.0100 M acetic acid and 0.100 M sodium acetate to reduce the pH
Naya [18.7K]

Explanation:

It is given that molarity of acetic acid = 0.0100 M

Therefore, moles of acetic acid = molarity of acetic acid × volume of buffer

            Moles of acetic acid = 0.0100 M × 1.00 L

                                              = 0.0100 mol

Similarly, moles of acetate = molarity of sodium acetat × volume of buffer

                                           = 0.100 mol

When HNO_{3} is added, it will convert acetate to acetic acid.

Hence, new moles acetic acid = (initial moles acetic acid) + (moles HNO_{3})

                                                = 0.0100 mol + x

New moles of sodium acetate = (initial moles acetate) - (moles HNO_{3})

                                        = 0.100 mol - x

According to Henderson - Hasselbalch equation,

           pH = pK_{a} + log\frac{[conjugate base]}{[weak acid]}

             pH = pKa + log\frac{(new moles of sodium acetate)}{(new moles of acetic acid)}

           4.95 = 4.75 + log\frac{(0.100 mol - x)}{(0.0100 mol + x)}

       log\frac{(0.100 mol - x)}{(0.0100 mol + x)}  = 4.95 - 4.75

                                            = 0.20

\frac{(0.100 mol - x)}{(0.0100 mol + x)}  = antilog (0.20)

                                           = 1.6

Hence,    x = 0.032555 mol

Therefore, moles of HNO_{3} = 0.032555 mol

volume of HNO_{3} = \frac{moles HNO_{3}}{molarity of HNO_{3}}

                                = \frac{0.032555 mol}{10.0 M}

                                 = 0.0032555 L

or,                             = 3.25           (as 1 L = 1000 mL)

Thus, we can conclude that volume of HNO_{3} added is 3.26 mL.

5 0
3 years ago
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morpeh [17]

Answer:

ver explicacion

Explanation:

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El zinc se oxida y el hidrógeno se reduce. HCl es el agente oxidante mientras que Zn es el agente reductor.

2Na (s) + H2SO4 (ac) ----> Na2SO4 (ac) + H2 (g)

El Na se oxida y el hidrógeno se reduce. El ácido sulfúrico es el agente oxidante, mientras que el sodio metálico es el agente reductor de tge.

Ca (OH) 2 (ac) + H2CO3 (ac) ------> CaCO3 (s) + 2H2O (l)

Esta no es una reacción redox ya que no hay cambios en el número de oxidación de izquierda a derecha.

4 0
3 years ago
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