Answer:O B. Weak acid breaks down minerals by reacting with them.
Explanation:
Chemical weathering is the process whereby rain water which sometimes can be acidic ( As rain falls down it reacts with CO2 in the atmosphere and form acid rain), This reacts with the minerals contained in rocks, dissolves and degrades them further to form entire new minerals.
An example of a rock that is greatly affected by overexposure of acid rain is limestone containing calcite which easily degrades by acid rain.
Other processes whereby Chemical weathering can occur are through reaction with water and oxygen.
Hi there! :)
Both are compounds.
contains 1 carbon (C) atom and 2 oxygen (O) atoms.
It contains two different elements, so it is a COMPOUND.
is an IONIC compound with elements Sodium (Na) and Oxygen (O), which are different elements. Therefore, it is also a COMPOUND.
For this reaction: ΔG⁰>0.
Balanced chemical reaction A(g) ⇌ (g)
ΔG° indicates that all reactants and products are in their standard states.
ΔG° = R·T·lnK.
ΔG° is Gibbs free energy
T is the temperature on the Kelvin scale
R is the ideal gas constant
The equilibrium constant (K) is the ratio of the partial pressures or the concentrations of products to reactants.
Gibbs free energy (G) determines if reaction will proceed spontaneously, nonspontaneously or in equilibrium processes.
If K < 1, than ΔG° > 0.
Reactants (in this example A) are favored over products (in this example B) at equilibrium.
More about equilibrium: brainly.com/question/25651917:
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Answer:
ΔS° = 180.5 J/mol.K
Explanation:
Let's consider the following reaction.
4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)
The standard molar entropy of the reaction (ΔS°) can be calculated using the following expression.
ΔS° = ∑np × S°p - ∑nr × S°r
where,
ni are the moles of reactants and products
S°i are the standard molar entropies of reactants and products
ΔS° = 4 mol × S°(NO(g)) + 6 × S°(H₂O(g)) - 4 mol × S°(NH₃(g)) - 5 mol × S°(O₂(g))
ΔS° = 4 mol × 210.8 J/K.mol + 6 × 188.8 j/K.mol - 4 mol × 192.5 J/K.mol - 5 mol × 205.1 J/K.mol
ΔS° = 180.5 J/K
This is the change in the entropy per mole of reaction.
<span>C2H5
First, you need to figure out the relative ratios of moles of carbon and hydrogen. You do this by first looking up the atomic weight of carbon, hydrogen, and oxygen. Then you use those atomic weights to calculate the molar masses of H2O and CO2.
Carbon = 12.0107
Hydrogen = 1.00794
Oxygen = 15.999
Molar mass of H2O = 2 * 1.00794 + 15.999 = 18.01488
Molar mass of CO2 = 12.0107 + 2 * 15.999 = 44.0087
Now using the calculated molar masses, determine how many moles of each product was generated. You do this by dividing the given mass by the molar mass.
moles H2O = 11.5 g / 18.01488 g/mole = 0.638361 moles
moles CO2 = 22.4 g / 44.0087 g/mole = 0.50899 moles
The number of moles of carbon is the same as the number of moles of CO2 since there's just 1 carbon atom per CO2 molecule.
Since there's 2 hydrogen atoms per molecule of H2O, you need to multiply the number of moles of H2O by 2 to get the number of moles of hydrogen.
moles C = 0.50899
moles H = 0.638361 * 2 = 1.276722
We can double check our math by multiplying the calculated number of moles of carbon and hydrogen by their respective atomic weights and see if we get the original mass of the hydrocarbon.
total mass = 0.50899 * 12.0107 + 1.276722 * 1.00794 = 7.400185
7.400185 is more than close enough to 7.40 given rounding errors, so the double check worked.
Now to find the empirical formula we need to find a ratio of small integers that comes close to the ratio of moles of carbon and hydrogen.
0.50899 / 1.276722 = 0.398669
0.398669 is extremely close to 4/10, so let's reduce that ratio by dividing both top and bottom by 2 giving 2/5.
Since the number of moles of carbon was on top, that ratio implies that the empirical formula for this unknown hydrocarbon is
C2H5</span>
