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3 years ago

5

2. Soil is made of A. air B. rocks C. decomposed leaves D. all of the above

2 answers:

Irina18 [472]3 years ago

3 0

Answer:

B

Explanation:

<h2>Soil is a mixture of </h2><h2 /><h2>tiny particles of rock,</h2>

<h2>dead plants and animals </h2>

<h2>,air and water.</h2>

BARSIC [14]3 years ago

3 0

I would say B rocks because soil usually has tiny pices of rock

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An aluminum can weighing 10 g absorbs 106.8 J of heat and warms by 12 degrees C. What is the specific heat of the aluminum can?

Sladkaya [172]

Answer:

$c_{Al} = 0.89\,\frac{J}{kg\cdot ^{\circ}C}$

Explanation:

The heating process is modelled after the First Law of Thermodynamics:

$Q = m \cdot c_{Al}\cdot \Delta T$

The specific heat of the aluminium can is:

$c_{Al} = \frac{Q}{m\cdot \Delta T}$

$c_{Al} = \frac{106.8\,J}{(10\,g)\cdot (12^{\circ}C)}$

$c_{Al} = 0.89\,\frac{J}{kg\cdot ^{\circ}C}$

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3 years ago

The decomposition of copper(II) nitrate on heating is endothermic reaction. 2Cu(NO3)2(s) → 2C10(s) + 4NO2(g) + O2(g) Calculate t

Basile [38]

Answer:

The enthalpy change for the given reaction is 424 kJ.

Explanation:

$2Cu(NO_3)_2(s)\rightarrow 2CuO(s) + 4NO_2(g) + O_2(g),\Delta H_{rxn}=?$

We have :

Enthalpy changes of formation of following s:

$\Delta H_{f,Cu(NO_3)_2}=-302.9 kJ/mol$

$\Delta H_{f,CuO}=-157.3 kJ/mol$

$\Delta H_{f,NO_2}= 33.2 kJ/mol$

$\Delta H_{f,O_2}= 0 kJ/mol$ (standard state)

$\Delta H_{rxn}=\sum [\Delta H_f(product)]-\sum [\Delta H_f(reactant)]$

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

$\Delta H_{rxn}$ =

$=(2 mol\times \Delta H_{f,CuO}+4\times \Delta H_{f,NO_2}+1 mol\times \Delta H_{f,O_2})-(2mol\times \Delta H_{f,Cu(NO_3)_2})$

$\Delta H_{rxn}$ =

$(2mol\times (-157.3 kJ/mol)+4\times 33.2 kJ/mol=1 mol\times 0 kJ/mol)-(2 mol\times (-302.9 kJ/mol)$

$\Delta H_{rxn}=424 kJ$

The enthalpy change for the given reaction is 424 kJ.

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4 years ago

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Molodets [167]

Answer:

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Explanation:

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3 years ago

I need help with this question .

salantis [7]

Answer:

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Explanation:

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Calculate the no. of each atoms presents in 10 grams of calcium carbonate.

raketka [301]

Molar Mass of Calcium carbonate:-

$\\ \sf\longmapsto CaCO_3$

$\\ \sf\longmapsto 40u+12u+3(16u)$

$\\ \sf\longmapsto 52u+48u$

$\\ \sf\longmapsto 100u$

$\\ \sf\longmapsto 100g/mol$

Given Mass=10g

$\boxed{\sf No\:of\;moles=\dfrac{Given\:Mass}{Molar\:Mass}}$

$\\ \sf\longmapsto No\:of\;moles=\dfrac{10}{100}$

$\\ \sf\longmapsto No\:of\:moles=0.1mol$

Now

$\boxed{\sf No\:of\:molecules=No\;of\:moles\times Avagadro\: No}$

$\\ \sf\longmapsto 0.1\times 6.023\times 10^{23}$

$\\ \sf\longmapsto 6.023\times 10^{22}molecules$

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3 years ago