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2 years ago

Which techniques would be best for separating a

Chemistry

1 answer:

pashok25 [27]2 years ago

6 0

Answer: centrifugation, boiling/heating , and long standing

Explanation:

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Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH

sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

$C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

$C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole$ ..[1]

$H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole$ ..[2]

$CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole$ ..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, Using Hess's law:

We get :

$C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole$ ..[1]

$2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol$ ..[2]

$CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole$ [3]

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3$

$\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)$

$\Delta H=-238.7kJ/mole$

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

4 0

2 years ago

Express 450 x 10-9 m (wavelength of blue light) in decimal notation (i.e., express

Llana [10]

Answer:

0.00000045 m is the wavelength of blue light in decimal

3 0

3 years ago

How closely a measurement reflects the actual value is precision. True or False??

Rina8888 [55]

Answer:

very very very verytrue

7 0

2 years ago

The length of a diagonal of a cube with edge length 3 in.

Lilit [14]

Answer:

Diagonal

5.196

-Hops

8 0

2 years ago

A patients blood is tested for growth hormone which has a molecular mass of 848.952g/mol. The test results show a normal concent

rusak2 [61]

Answer:

$\boxed{4.7\times 10^{-9}}$

Explanation:

Data:

MM = 848.952 g/mol

c = 4.0 x 10⁻⁶ g/L

V = 1 L

Calculations:

(a) Mass of growth hormone

$\text{Mass} = \text{1 L} \times \dfrac{4.0 \times 10^{-6} \text{ g}}{\text{1 L}} = 4.0 \times 10^{-6} \text{ g}$

(b) Moles of growth hormone

$\text{Moles} = 4.0 \times 10^{-6} \text{ g} \times \dfrac{\text{1 mol}}{\text{848.952 g}} = 4.7\times 10^{-9}\text{ mol}\\\\\text{There are $\boxed{\mathbf{4.7\times 10^{-9}}\textbf{ mol}}$ of growth hormone in 1 L of blood.}$

8 0

3 years ago