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3 years ago

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CO + 2H2 CH3OH How many molecules of CO are needed to form 600g of CH3OH?

1 answer:

MA_775_DIABLO [31]3 years ago

8 0

Explanation:

molecules of Hydrogen gas to produce 1 molecule of ... 1 mol CO. 1 mol CO. 1 mol CH3OH. 2 mol H2. 1 mol CO. 1 mol CO. 1 mol CH3OH ... How many moles of hydrogen.

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Elements that are arranged in the same column on the periodic table belong to the same family of elements. the atoms of elements

Vlad [161]

The answer is: c. electrons

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Calculate the solubility of ( = ) in moles per liter. Ignore any acid–base properties. s = mol/L Calculate the solubility of ( =

BaLLatris [955]

This is an incomplete question, here is a complete question.

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties.

CaCO₃, Ksp = 8.7 × 10⁻⁹

Answer : The solubility of CaCO₃ is, $9.33\times 10^{-5}mol/L$

Explanation :

As we know that CaCO₃ dissociates to give $Ca^{2+}$ ion and $CO_3^{2-}$ ion.

The solubility equilibrium reaction will be:

$CaCO_3\rightleftharpoons Ca^{2+}+CO_3^{2-}$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ca^{2+}][CO_3^{2-}]$

Let solubility of CaCO₃ be, 's'

$K_{sp}=(s)\times (s)$

$K_{sp}=s^2$

$8.7\times 10^{-9}=s^2$

$s=9.33\times 10^{-5}mol/L$

Therefore, the solubility of CaCO₃ is, $9.33\times 10^{-5}mol/L$

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3 years ago

Is the atomic number the lesser or the greater of the numbers for each atom on the periodic table?

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Lesser. atomic number means proton number

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Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔

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Answer : The correct option is, (C) 1.1

Solution : Given,

Initial moles of $N_2O_4$ = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

The given equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initially c 0

At equilibrium $(c-c\alpha)$ $2c\alpha$

The expression of $K_c$ will be,

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

where,

$\alpha$ = degree of dissociation = 40 % = 0.4

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

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

$K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}$

$K_c=1.066\aprrox 1.1$

Therefore, the value of equilibrium constant for this reaction is, 1.1

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3 years ago

Boyles Law P1V1 = P2V2

arsen [322]

Answer:

A. The balloons will increase to twice their original volume.

Explanation:

Boyle's law states that the pressure exerted on a gas is inversely proportional to the volume occupied by the gas at constant temperature. That is:

P ∝ 1/V

P = k/V

PV = k (constant)

P = pressure, V = volume.

$P_1V_1=P_2V_2$

Let the initial pressure of the balloon be P, i.e. $P_1=P$ , initial volume be V, i.e. $V_1=V$ . The pressure is then halved, i.e. $P_2=\frac{P}{2}$

$P_1V_1=P_2V_2\\\\P*V=\frac{P}{2} *V_2\\\\V_2=\frac{2*P*V}{P}\\\\V_2=2V$

Therefore the balloon volume will increase to twice their original volume.

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3 years ago