1J /(g*C), 1*C, 1 J, m=g

Roman55 [17]

Answer:

Weathering, Erosion

Explanation:

Plants and animals can be agents of mechanical weathering. The seed of a tree may sprout in soil that has collected in a cracked rock. As the roots grow, they widen the cracks, eventually breaking the rock into pieces. Over time, trees can break apart even large rocks.

Tree root systems have a handful of large roots that branch out into a network of smaller roots that often extend out far beyond their branches do. These root systems prevent erosion by holding the soil in place and improving drainage which helps water get absorbed into the soil instead of just running over the top.

Hope this helps

All the love, Ya boi Fraser :)

7 0

2 years ago

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If 100cm3 of O2 diffused in 4s and 50cm3 of gas Y diffused in 3s. calculate the relative molecular mass of gas X (O=16).

sertanlavr [38]

Answer:

T2/T=√M1√M2

T2=3 T1=4 M1=O2=32

M2=?

4/2=√M2/√32

=1.3

Explanation:

8 0

3 years ago

In examining several amino acids, you want to identify the feature or features that differentiate one amino acid from another. W

sergiy2304 [10]

Your answer is C
amino acids are created by a bond of two carbon atoms and each carbon is part of a different group, one negative and one positive. when the chemical reactions occurs, it creates the several diff amino acids

4 0

3 years ago

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Sulfuryl dichloride is formed when sulfur dioxide reacts with chlorine.

zubka84 [21]

<u>Answer:</u> The value of $\Delta G^o$ of the reaction is 28.38 kJ/mol

<u>Explanation:</u>

For the given chemical reaction:

$SO_2(g)+Cl_2(g)\rightarrow SO_2Cl_2(g)$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

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

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(SO_2Cl_2(g))})]-[(1\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(Cl_2(g))})]$

We are given:

$\Delta H^o_f_{(SO_2Cl_2(g))}=-364kJ/mol\\\Delta H^o_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H^o_f_{(Cl_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-364))]-[(1\times (-296.8))+(1\times 0)]=-67.2kJ/mol=-67200J/mol$

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_f_{(product)}]-\sum [n\times \Delta S^o_f_{(reactant)}]$

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

$\Delta S^o_{rxn}=[(1\times \Delta S^o_{(SO_2Cl_2(g))})]-[(1\times \Delta S^o_{(SO_2(g))})+(1\times \Delta S^o_{(Cl_2(g))})]$

We are given:

$\Delta S^o_{(SO_2Cl_2(g))}=311.9J/Kmol\\\Delta S^o_{(SO_2(g))}=248.2J/Kmol\\\Delta S^o_{(Cl_2(g))}=223.0J/Kmol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(1\times 311.9)]-[(1\times 248.2)+(1\times 223.0)]=-159.3J/Kmol$

To calculate the standard Gibbs's free energy of the reaction, we use the equation:

$\Delta G^o_{rxn}=\Delta H^o_{rxn}-T\Delta S^o_{rxn}$

where,

$\Delta H^o_{rxn}$ = standard enthalpy change of the reaction =-67200 J/mol

$\Delta S^o_{rxn}$ = standard entropy change of the reaction =-159.3 J/Kmol

Temperature of the reaction = 600 K

Putting values in above equation, we get:

$\Delta G^o_{rxn}=-67200-(600\times (-159.3))\\\\\Delta G^o_{rxn}=28380J/mol=28.38kJ/mol$

Hence, the value of $\Delta G^o$ of the reaction is 28.38 kJ/mol

7 0

3 years ago

What volume is occupied by 26.6 g of argon gas at a pressure of 1.29 atm and a temperature of 355 K ?

agasfer [191]

Answer:

V = 15 L

Explanation:

PV = nRT

P = 1.29 atm, T= 355 K mass argon 40g/mol

n = mass/ molecular mass; n = 26.6 g/40 g/mol; n = 0.665 mol

V = (0,665 mol x 0.082 x 355 K) / 1.29 atm = 15 L

4 0

3 years ago