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scoray [572]
2 years ago
14

Which of the following reactions would result in the formation of a gaseous product?

Chemistry
2 answers:
emmainna [20.7K]2 years ago
8 0

The  reaction that would result  in formation  of gaseous product  is


KHCO3  +  HBr  ( third  answer)


<em><u>explanation</u></em>

KHCO3  react with  HBr  to form  KBr ,  CO2  and H2O  according  to  the  below equation.


KHCO3 + HBr →  KBr  + CO2 + H2O

CO2 ( carbon (IV)  oxide)  is the gas that  is  formed   during   reaction.


other reaction does not produce a   a gas for example  NaOH +   HCl forms  NaCl  and H2O


Yanka [14]2 years ago
3 0
The third reaction will form a gaseous product, CO₂
KHCO₃ + HBr → KBr + H₂O + CO₂
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Which of the following statements is not true regarding acids? (3 points)
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Answer:

Lewis acid is a substance that donates a lone-pair of electrons.

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What is said in the statement corresponds to a Lewis base, not an acid. For example, NH3 is a Lewis base, since it is capable of donating its pair of electrons. Trimethylborane (Me3B) is a Lewis acid, since it is capable of accepting a solitary pair.

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3 years ago
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3 years ago
Which of these is NOT true about electromagnets?
Alex17521 [72]

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Permanent = false

Explanation:

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4 0
1 year ago
The reaction C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ/mol.262 kJ/mol. At 600.0 K,600.0 K, the rate c
crimeas [40]

Answer : The rate constant at 785.0 K is, 1.45\times 10^{-2}s^{-1}

Explanation :

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 600.0K = 6.1\times 10^{-8}s^{-1}

K_2 = rate constant at 785.0K = ?

Ea = activation energy for the reaction = 262 kJ/mole = 262000 J/mole

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 600.0K

T_2 = final temperature = 785.0K

Now put all the given values in this formula, we get:

\log (\frac{K_2}{6.1\times 10^{-8}s^{-1}})=\frac{262000J/mole}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{785.0K}]

K_2=1.45\times 10^{-2}s^{-1}

Therefore, the rate constant at 785.0 K is, 1.45\times 10^{-2}s^{-1}

7 0
3 years ago
What is the maximum concentration of Ag⁺ that can be added to a 0.00750 M solution of Na₂CO₃ before a precipitate will form? (Ks
Firlakuza [10]

Answer:

\large \boxed{1.64\times 10^{-5}\text{ mol/L }}

Explanation:

Ag₂CO₃(s) ⇌2Ag⁺(aq) + CO₃²⁻(aq); Ksp = 8.10 × 10⁻¹²

                           2x      0.007 50 + x

K_{sp} =\text{[Ag$^{+}$]$^{2}$[CO$_{3}^{2-}$]} = (2x)^{2}\times 0.00750 = 8.10 \times 10^{-12}\\0.0300x^{2} = 8.10 \times 10^{-12}\\x^{2} = 2.70 \times 10^{-10}\\x = \sqrt{2.70 \times 10^{-10}} = \mathbf{1.64\times 10^{5}} \textbf{ mol/L}\\\text{The maximum concentration of Ag$^{+}$ is $\large \boxed{\mathbf{1.64\times 10^{-5}}\textbf{ mol/L }}$}

 

3 0
3 years ago
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