The partial pressure is the amount of linguistic compound there is which makes the lagitude of the element 64.663
Answer:
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Answer:
(E) changing temperature
Explanation:
Consider the following reversible balanced reaction:
aA+bB⇋cC+dD
If we know the molar concentrations of each of the reaction species, we can find the value of Kc using the relationship:
Kc = ([C]^c * [D]^d) / ([A]^a * [B]^b)
where:
[C] and [D] are the concentrations of the products in the equilibrium; [A] and [B] reagent concentrations in equilibrium; already; b; c and d are the stoichiometric coefficients of the balanced equation. Concentrations are commonly expressed in molarity, which has units of moles / 1
There are some important things to remember when calculating Kc:
- <em>Kc is a constant for a specific reaction at a specific temperature</em>. If you change the reaction temperature, then Kc also changes
- Pure solids and liquids, including solvents, are not considered for equilibrium expression.
- The reaction must be balanced with the written coefficients as the minimum possible integer value in order to obtain the correct value of Kc
We can determine the empirical formula by first converting each of the grams to moles. remember to do this, first, we need the molar mass of the molecules which can be calculated by adding the mass of the atoms from the periodic table.
molar mass of CO2= 44.0 g/mol
molar mass of H2O= 18.02 g/mol
now, lets determine the grams of each atom
Carbon: 23.98 g x (12.011 g / 44.01 g) = 6.54 g C
Hydrogen: 4.91 g x (2.0158 g / 18.02 g) = 0.55 g H
Oxygen: 10.0 - (6.54 + 0.55) = 2.91 g O
Now let's convert each mass to moles.
C: 6.54 g / 12.01 g / mol = 0.54 mol
H: 0.55 g / 1.01 g/mol = 0.54 mol
O: 2.91 g / 16.00 g/mol = 0.18 mol
now that we have the moles of each atom, we need to divide them by the smallest value to find the ration. If you do not get the whole number, you need to multiply until to get a whole number.
C: 0.54 mol / 0.18 mol = 3
H: 0.54 mol / 0.18 mol = 3
O: 0.18 mol / 0.18 mol = 1
empirical formula--> C₃H₃O
There is no gaseous product when NaNO3 reacts with 6M HCl.
We must note that gas evolution is one of the signs that a chemical reaction has taken place. Hence, we must consider the options and look out for an option in which reaction with HCl does not lead to a gaseous product. This is the option that does not lead to evolution of a gas.
Let us consider the reaction; NaNO3(s) + HCl(aq) ---> NaCl(s) + HNO3(aq), there is no gaseous product hence NaNO3 does not result in gas evolution upon addition of 6M HCl.
Missing parts;
Addition of 6 M HCl to which substance will NOT result in gas evolution?
(A) Al
(B) Zn
(C) K2CO3
(D) NaNO3
Learn more about evolution of gas: brainly.com/question/2186340