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3 years ago

8

Bases increase the concentration of hydronium ions by donating hydroxide ions to water molecules.

1 answer:

Pani-rosa [81]3 years ago

7 0

Answer: False

Explanation: the answer is false

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At this temperature, 0.300 mol H 2 0.300 mol H2 and 0.300 mol I 2 0.300 mol I2 were placed in a 1.00 L container to react. What

Alekssandra [29.7K]

Answer:

The concentration of HI present at equilibrium is 0.471 M.

Explanation:

5 0

3 years ago

What is true about both sodium and gold?

vladimir2022 [97]

They all are correct , so with that being said anyone of them can be right

5 0

4 years ago

Read 2 more answers

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

The following reaction shows calcium chloride reacting with silver nitrate.

adell [148]

Grams of Ca(NO₃)₂ produced : 0.985 g

<h3>Further explanation</h3>

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

Reaction

CaCl₂ + 2AgNO₃ → 2AgCl + Ca(NO₃)₂

MW AgNO₃ : 107.9+14+3.16=169.9

mol AgNO₃ :

$\tt \dfrac{2}{169.9}=0.012$

mol ratio Ca(NO₃)₂ : AgNO₃ = 1 : 2, so mol Ca(NO₃)₂ :

$\tt \dfrac{1}{2}\times 0.012=0.006$

MW Ca(NO₃)₂ : 40.1+2.14+6.16=164.1 g/mol

mass Ca(NO₃)₂ :

$\tt 0.006\times 164.1=0.985$

7 0

3 years ago

It takes 495.0 kJ of energy to remove 1 mole of electron from an atom on the surface of sodium metal. How much energy does it ta

Zigmanuir [339]

Answer:

$\lambda=241.9\ nm$

Explanation:

The work function of the sodium= 495.0 kJ/mol

It means that

1 mole of electrons can be removed by applying of 495.0 kJ of energy.

Also,

1 mole = $6.023\times 10^{23}\ electrons$

So,

$6.023\times 10^{23}$ electrons can be removed by applying of 495.0 kJ of energy.

1 electron can be removed by applying of $\frac {495.0}{6.023\times 10^{23}}\ kJ$ of energy.

Energy required = $82.18\times 10^{-23}\ kJ$

Also,

1 kJ = 1000 J

So,

Energy required = $82.18\times 10^{-20}\ J$

Also, $E=\frac {h\times c}{\lambda}$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

So,

$79.78\times 10^{-20}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda}$

$\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{82.18\times 10^{-20}}$

$\lambda=\frac{10^{-26}\times \:19.878}{10^{-20}\times \:82.18}$

$\lambda=\frac{19.878}{10^6\times \:82.18}$

$\lambda=2.4188\times 10^{-7}\ m$

Also,

1 m = 10⁻⁹ nm

So,

$\lambda=241.9\ nm$

6 0

3 years ago