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2 years ago

Which of the following is a product when Li2O and H2O react together?

Chemistry

2 answers:

Katarina [22]2 years ago

7 0

Answer:

C - LiOH

Explanation:

To balance this, first, you need to put all things in a row.

substances --> products

Li2O + H20 --> products

You know Li is an alkaline metal and with water will form a hydroxide so, LiOH will be the product.

Li2O + H20 --> LiOH

Now, you have to balance this to have the right amount in the equation:

Li2O + H20 --> 2 LiOH

And that's it.

sleet_krkn [62]2 years ago

3 0

Just LiOH, i also think Li2O + H2O = LiOH

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lys-0071 [83]

Answer:

Multiply 1.25 by 0.04 and divide the result obtained by 1,000

Explanation:

Since there is .04 oz in 1 gram, we take the 1.25 gram and multiply by .04.

Next, to convert per liter to per milliliter, we divide by 1,000.

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3 years ago

What volume of a 2.25 M sodium chloride solution will contain 4.58 moles of sodium chloride

Rzqust [24]

Answer:

Option E. 2.04 L

Explanation:

Data obtained from the question include:

Molarity of NaCl = 2.25 M

Mole of NaCl = 4.58 moles

Volume =..?

Molarity is simply defined as the mole of solute per unit litre of the solution. It is represented mathematically as:

Molarity = mole /Volume

With the above formula, we can obtain the volume of the solution as follow:

Molarity = mole /Volume

2.25 = 4.58/volume

Cross multiply

2.25 x volume = 4.58

Divide both side by 2.25

Volume = 4.58/2.25

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2 years ago

An electron in a hydrogen atom moves from level 3 to level 1. In a second hydrogen atom, an electron drops from level 2 to level

topjm [15]

I think the answer that you are looking for is B but im not a 100% sure

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3 years ago

Read 2 more answers

Why is Hawaii made of volcanoes? <br>

Vadim26 [7]

Answer:

Explanation:

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2 years ago

2) A common "rule of thumb" -- for many reactions around room temperature is that the

babunello [35]

The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

$T_1=300 \ K$

$T_2=310 \ K$

The rate constant = $K_1 \text{ and } K_2$ respectively.

The activation energy and the Arhenius factor is same.

So by the arhenius equation,

$K_1 = Ae^{-\frac{E_a}{RT_1}}$ and $K_2 = Ae^{-\frac{E_a}{RT_2}}$

$\Rightarrow \frac{K_1}{K_2}= \frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}}$

$\Rightarrow \frac{K_1}{K_2}= e^{-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

Given, $E_a = 0.269$ J/mol

R = 8.314 J/mol/K

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \left(\frac{1}{300} -\frac{1}{310} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \times \frac{10}{300 \times 310}$

$\Rightarrow \ln \frac{K_2}{K_1}= 3.479 \times 10^{-6}$

$\Rightarrow \frac{K_2}{K_1}= e^{3.479 \times 10^{-6}}$

$\Rightarrow \frac{K_2}{K_1}= 1$

∴ $K_2=K_1$

So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.

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2 years ago