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3 years ago

7

What is chemical weathering of rock

2 answers:

balu736 [363]3 years ago

7 0

Chemical weathering is the weakening and subsequent disintegration of rock by chemical reactions.

Drupady [299]3 years ago

3 0

It’s the weakening and subsequent disintegration of rock by chemical reactions

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Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,

WITCHER [35]

Answer: Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

$2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]$

We are given:

$\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ$

Enthalpy of combustion (per mole) of $C_4H_{10} (g)$ is -2657.5 kJ

6 0

3 years ago

The cylinder of aluminum in class was approximately 13mm in circumference and 42mm in length. What would be it's approximate mas

Roman55 [17]

Answer:

Mass, m = 1.51 grams

Explanation:

It is given that,

The circumference of Aluminium cylinder, C = 13 mm = 1.3 cm

Length of the cylinder, h = 4.2 cm

We know that the density of the Aluminium is 2.7 g/cm³

Circumference, C = 2πr

$r=\dfrac{C}{2\pi}\\\\r=\dfrac{1.3}{2\pi}\\\\r=0.206\ cm$

Density is equal to mass per unit volume.

$d=\dfrac{m}{V}$

m is mass of the cylinder

V is the volume of the cylinder

$V=\pi r^2h\\\\V=\dfrac{22}{7}\times0.206^2\times 4.2\\\\V=0.5601\ cm^3$

So,

$m=d\times V\\\\m=2.7\times 0.5601\\\\m=1.51\ g$

So, the mass of the cylinder is 1.51 grams.

5 0

4 years ago

How many half-lives are required for the concentration of reactant to decrease to 1.56% of its original value?4247.56.56

neonofarm [45]

Answer:

6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.

Explanation:

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given:

Concentration is decreased to 1.56 % which means that 0.0156 of $[A_0]$ is decomposed. So,

$\frac {[A_t]}{[A_0]}$ = 0.0156

Thus,

$\frac {[A_t]}{[A_0]}=e^{-k\times t}$

$0.0156=e^{-k\times t}$

kt = 4.1604

The expression for the half life is:-

Half life = 15.0 hours

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$\frac{4.1604}{t}=\frac {ln\ 2}{t_{1/2}}$

$t = 6\times t_{1/2}$

<u>6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.</u>

6 0

3 years ago

What is the electron structure of sodium

yarga [219]

Electron structure of sodium:
₁₁Na: 1s²2s²2p⁶3s¹

8 0

3 years ago

Read 2 more answers

**CHEMISTRY!! PLEASE HELP ASAP!!**

Minchanka [31]

Calcium = 1
Sulfur = 1
Oxygen = 4

6 0

3 years ago