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3 years ago

Consider the following reaction where K. = 9.52 10 2 at 350 K.

Chemistry

1 answer:

Harman [31]3 years ago

3 0

Answer:

The correct answer is A :))

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A fusion reaction takes place between carbon and another element neutrons are released and a different element is formed the dif

Andreas93 [3]

A fusion reaction takes place between carbon and another element. Neutrons are released, and a different element is formed. The different element is Lighter than helium.

7 0

3 years ago

What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (–97.6˚C) and 1 atm

Rudiy27

Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K

Explanation :

Formula used :

$\Delta S=\frac{\Delta H_{freezing}}{T_f}$

where,

$\Delta S$ = change in entropy

$\Delta H_{fus}$ = change in enthalpy of fusion = 3.17 kJ/mol

As we know that:

$\Delta H_{fus}=-\Delta H_{freezing}=-3.17kJ/mol=-3170J/mol$

$T_f$ = freezing point temperature = $-97.6^oC=273+(-97.6)=175.4K$

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

$\Delta S=\frac{\Delta H_{freezing}}{T_m}$

$\Delta S=\frac{-3170J/mol}{175.4K}$

$\Delta S=-18.07J/mol.K$

Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K

4 0

4 years ago

What is Lewis acid and Lewis base? give examples

AleksAgata [21]

Explanation:

example is copper iron...........

6 0

3 years ago

How many moles of NH3 can be produced by the reaction of 2.00 g of N2 with 3.00 g H2?

yulyashka [42]

0.235 if I’m not mistaken!

8 0

3 years ago

10 points!! Please give me work with it and I will mark as brainlist.

Georgia [21]

Answer: The molecular formula will be $C_2H_4O_2$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 40.0 g

Mass of O = 53.3 g

Mass of H = 6.66 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{40.0g}{12g/mole}=3.33moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{53.3g}{16g/mole}=3.33moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.66g}{1g/mole}=6.66moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.33}{3.33}=1$

For O = $\frac{3.33}{3.33}=1$

For H = $\frac{6.66}{3.33}=2$

The ratio of C : O : H = 1: 1: 2

Hence the empirical formula is $COH_2$

The empirical weight of $COH_2$ = 1(12)+1(16)+2(1)= 30g.

The molecular weight = 60 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{60}{30}=2$

The molecular formula will be= $2\times CH_2O=C_2H_4O_2$

8 0

3 years ago